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# Go Math Grade 8 Answer Key Chapter 6 Functions

Go Math Grade 8 Answer Key Chapter 6 Functions PDF is provided here for download. The best-loved math experts are provided an easy way of solving maths questions along with the explanations. Every problem is simplified with a clear explanation with many tricks. Great problems with a high level of difficulty also solved in an easy way. Refer to Go Math Grade 8 Answer Key Chapter 6 Functions now to begin your practice.

## Go Math Grade 8 Chapter 6 Functions Answer Key

Prefer learning using Go Math Grade 8 Chapter 6 Functions Solution Key. To get the good results in an exam you must prepare with the Go Math Grade 8 Chapter 6 Answer Key. Use HMH Go Math Grade 8 Answer Key for the best practice of maths. Step by step explanation is provided for every question below. Build your confidence in solving math problems by practicing with Go Math Grade 8 Answer Key Chapter 6 Functions.

Lesson 1: Identifying and Representing Functions

Lesson 2: Describing Functions

Lesson 3: Comparing Functions

Lesson 4: Analyzing Graphs

Lesson 5: Model Quiz

Lesson 6: Mixed Review

### Guided Practice – Identifying and Representing Functions – Page No. 158

Complete each table. In the row with x as the input, write a rule as an algebraic expression for the output. Then complete the last row of the table using the rule.

Question 1.

Type below:
_______________

Answer:

Explanation:
Unit Cost of ticket = 40/2 = 20
Total cost = 20x where x is the number of tickets.
x = 20x
10 = 20(100) = 200

Question 2.

Type below:
_______________

Answer:

Explanation:
Number of pages per minute = 1/2 = 0.5
Total cost = 0.5x where x is the number of minutes.
x = 0.5x
30 = 0.5(30) = 15

Question 3.

Type below:
_______________

Answer:

Explanation:
Units cost of Muffins = 2.25/1 = 2.25
Total cost = 2.25x where x is the number of muffins
x = 2.25x
12 = 2.25(12) = 27

Determine whether each relationship is a function.

Question 4.

_______________

Answer:
Function

Explanation:
Each input is assigned to exactly one output.

Question 5.

_______________

Answer:
Not a function

Explanation:
The input value is 4 is paired with two outputs 25 and 35

Question 6.
The graph shows the relationship between the weights of 5 packages and the shipping charge for each package. Is the relationship represented by the graph a function? Explain.

_______________

Answer:
Function

Explanation:
Each input is assigned to exactly one output.

Essential Question Check-In

Question 7.
What are four different ways of representing functions? How can you tell if a relationship is a function?
Type below:
_______________

Answer:
The function can be represented by an equation, table, graph, and Venn diagram.
If a relationship is a function, each input is paired with exactly one output.

### Independent Practice – Identifying and Representing Functions – Page No. 159

Determine whether each relationship represented by the ordered pairs is a function. Explain.

Question 8.
(2, 2), (3, 1), (5, 7), (8, 0), (9, 1)
_______________

Answer:
Function

Explanation:
Each input value is paired with exactly one output value.

Question 9.
(0, 4), (5, 1), (2, 8), (6, 3), (5, 9)
_______________

Answer:
Not a function

Explanation:
The input value is 5 is paired with two outputs 1 and 9

Question 10.
Draw Conclusions
Joaquin receives $0.40 per pound for 1 to 99 pounds of aluminum cans he recycles. He receives$0.50 per pound if he recycles more than 100 pounds. Is the amount of money Joaquin receives a function of the weight of the cans he recycles? Explain your reasoning.
_______________

Answer:
Yes

Explanation:
The amount of money increases with the weight of the cans. No weight will result in the same amount of money earned.

Question 11.
A biologist tracked the growth of a strain of bacteria, as shown in the graph.

a. Explain why the relationship represented by the graph is a function.
Type below:
_______________

Answer:
The relationship is a function as each input has been assigned exactly one output. There is only one number of bacteria for each number of hours.

Question 11.
b. What If?
Suppose there was the same number of bacteria for two consecutive hours. Would the graph still represent a function? Explain.
Type below:
_______________

Answer:
Yes. If the number of bacteria for two consecutive hours is the same, one input will still be paired with one output, hence the relationship is still a function.

Question 12.
Multiple Representations
Give an example of a function in everyday life, and represent it as a graph, a table, and a set of ordered pairs. Describe how you know it is a function.

Type below:
_______________

Answer:
The cost of a bouquet of flowers and the number of flowers in the bouquet is a function. The unit cost of flowers = $0.85 and x the number of flowers. Hence, C= 0.85x (2, 1.7), (4, 3.4), (6, 5.1), (8, 6.8), (10, 8.5) Each value of the input is paired with exactly one output. ### Identifying and Representing Functions – Page No. 160 The graph shows the relationship between the weights of six wedges of cheese and the price of each wedge. Question 13. Is the relationship represented by the graph a function? Justify your reasoning. Use the words “input” and “output” in your explanation, and connect them to the context represented by the graph. _______________ Answer: Yes, the relationship represented by the graph is a function. Each input (weight) in the graph is paired with exactly one output(price). Question 14. Analyze Relationships Suppose the weights and prices of additional wedges of cheese were plotted on the graph. Might that change your answer to question 13? Explain your reasoning. Type below: _______________ Answer: No. As the weight of the cheese will increase, the cost of wedges of cheese will increase as well. Hence, for each input (weight), there would be exactly one output (price). H.O.T. Focus on Higher Order Thinking Question 15. Justify Reasoning A mapping diagram represents a relationship that contains three different input values and four different output values. Is the relationship a function? Explain your reasoning. _______________ Answer: No. Since there are three inputs and four outputs, one of the inputs will have more than one output, hence the relationship cannot be a function. Question 16. Communicate Mathematical Ideas An onion farmer is hiring workers to help harvest the onions. He knows that the number of days it will take to harvest the onions is a function of the number of workers he hires. Explain the use of the word “function” in this context. Type below: _______________ Answer: Number of days = f(number of workers) Explanation: We know that the more the number of workers will be involved in the harvesting of onion, the lesser days it will take to complete. Thus the number of workers becomes the independent variable and the number of days becomes the dependent variable. Here the word function is used to describe that the number of days is dependent on the number of workers. Number of days = f(number of workers) ### Guided Practice – Describing Functions – Page No. 164 Plot the ordered pairs from the table. Then graph the function represented by the ordered pairs and tell whether the function is linear or nonlinear. Question 1. y = 5 − 2x _______________ Answer: Graph of a linear function is a straight line Linear relationship Explanation: Given y = 5 – 2x y = 5 – 2(-1) = 5 + 2 = 7 y = 5 – 2(1) = 5 – 2 = 3 y = 5 – 2(3) = 5 – 6 = -1 y = 5 – 2(5) = 5 – 10 = -5 Question 2. y = 2 − x2 _______________ Answer: y = 2 − x2 Graph the ordered pairs. Then draw a line through the points to represent the solution. Graph of a linear function is not a straight line Non-linear relationship Explanation: y = 2 − x2 y = 2 – 4 = -2 y = 2 – 1 = 1 y = 2 – 0 = 2 y = 2 – 1 = 1 y = 2 – 4 = -2 Explain whether each equation is a linear equation. Question 3. y = x2 – 1 _______________ Answer: The equation is not in the form of a linear equation, hence is not a linear equation. Explanation: Compare the equation with the general linear equation y = mx + b. The equation is not in the form of a linear equation, hence is not a linear equation. Question 4. y = 1 – x _______________ Answer: The equation is in the form of a linear equation, hence is a linear equation. Explanation: Compare the equation with the general linear equation y = mx + b. The equation is in the form of a linear equation, hence is a linear equation. Essential Question Check-In Question 5. Explain how you can use a table of values, an equation, and a graph to determine whether a function represents a proportional relationship. Type below: _______________ Answer: From a table, determine the ratio y/x. If it is constant the relationship is proportional. From a graph, note if the graph passes through the origin. The graph of a proportional relationship must pass through the origin (0, 0). From an equation, compare with general linear form of equation, y = mx + b. If b = 0, the relationship is proportional. ### Independent Practice – Describing Functions – Page No. 165 Question 6. State whether the relationship between x and y in y = 4x – 5 is proportional or nonproportional. Then graph the function. _______________ Answer: Explanation: First, we compare the equation with the general linear equation y = mx + b. y = 4x – 5 is in the form of y = mx + b, with m = 4 and b = -5. Therefore, the equation is a linear equation. Since b is not equal to 0, the relationship is non-proportional. Then, we choose several values for the input x. We substitute these values of x in the equation to find the output y. y = 4x – 5 If x = 0; y = 4(0) – 5 = -5; (0, -5) If x = 2; y = 4(2) – 5 = 3; (2, 3) If x = 4; y = 4(4) – 5 = 11; (4, 11) If x = 6; y = 4(6) – 5 = 19; (6, 19) We graph the ordered pairs and we draw a line through the points to represent the solutions of the function. Question 7. The Fortaleza telescope in Brazil is a radio telescope. Its shape can be approximated with the equation y = 0.013x2. Is the relationship between x and y linear? Is it proportional? Explain. ____________ ____________ Answer: Compare the equation with the general linear equation y = mx + b. The equation is not in the form of a linear equation, hence it is not a linear equation. Since x is squared, it is not proportional. Question 8. Kiley spent$20 on rides and snacks at the state fair. If x is the amount she spent on rides, and y is the amount she spent on snacks, the total amount she spent can be represented by the equation x + y = 20. Is the relationship between x and y linear? Is it proportional? Explain.
____________
____________

Answer:
x + y = 20
Rewriting the equation
y = 20 – x
Compare the equation with the general linear equation y = mx + b.
It is linear
Since b is not equal to 0, the relationship is not proportional.

Question 9.
Represent Real-World Problems
The drill team is buying new uniforms. The table shows y, the total cost in dollars, and x, the number of uniforms purchased.

a. Use the data to draw a graph. Is the relationship between x and y linear? Explain.
____________

Answer:

The graph of a linear relationship is a straight line.
x and y are linear.

Question 9.
b. Use your graph to predict the cost of purchasing 12 uniforms.
$________ Answer:$720

Explanation:

The cost of 12 uniforms is $720 Question 10. Marta, a whale calf in an aquarium, is fed a special milk formula. Her handler uses a graph to track the number of gallons of formula y the calf drinks in x hours. Is the relationship between x and y linear? Is it proportional? Explain. ____________ ____________ Answer: The relationship is linear The relationship is proportional Explanation: As the data lies on a straight line, the relationship is linear As the graph passes through the origin, the relationship is proportional ### Describing Functions – Page No. 166 Question 11. Critique Reasoning A student claims that the equation y = 7 is not a linear equation because it does not have the form y=mx + b. Do you agree or disagree? Why? ____________ Answer: Disagree; The equation can be written in the form y = mx + b Where m is 0. The graph of the solutions is a horizontal line. Question 12. Make a Prediction Let x represent the number of hours you read a book and y represent the total number of pages you have read. You have already read 70 pages and can read 30 pages per hour. Write an equation relating x hours and y pages you read. Then predict the total number of pages you will have read after another 3 hours. _______ pages Answer: 160 pages Explanation: Let x represent the number of hours you read a book and y represents the total number of pages you have read. You have already read 70 pages and can read 30 pages per hour. m = 30; b = 70 pages y = 30x + 70 x = 3 hrs y = 30(3) + 70 = 160 H.O.T. Focus on Higher Order Thinking Question 13. Draw Conclusions Rebecca draws a graph of a real-world relationship that turns out to be a set of unconnected points. Can the relationship be linear? Can it be proportional? Explain your reasoning. Type below: ______________ Answer: The relationship is linear if all the points lie on the same line. If the relationship is linear and passes through the origin, it is proportional. Question 14. Communicate Mathematical Ideas Write a real-world problem involving a proportional relationship. Explain how you know the relationship is proportional. Type below: ______________ Answer: The amount of money earned at a car wash is a proportional relationship. When there are 0 cars washed,$0 are earned. The amount of money earned increases by the unit cost of a car wash.

Question 15.
Justify Reasoning
Show that the equation y + 3 = 3(2x + 1) is linear and that it represents a proportional relationship between x and y.
Type below:
______________

Answer:
y + 3 = 3(2x + 1)
y +3 = 6x + 3
y = 6x
As b = 0, it is a Proportional Relationship.

### Guided Practice – Comparing Functions – Page No. 170

Doctors have two methods of calculating maximum heart rate. With the first method, maximum heart rate, y, in beats per minute is y = 220 − x, where x is the person’s age. Maximum heart rate with the second method is shown in the table.

Question 1.
Which method gives the greater maximum heart rate for a 70-year-old?
____________ method

Answer:
Second

Explanation:
y = 220 – x
y = 220 – 70 = 150
Find the slope using two points from the grapgh by
Slope m = (y2 -y1)/(x2 – x1) where (x1, y1) = (20, 194) and (x2, y2) = (30, 187)
Slope m = (y2 -y1)/(x2 – x1) = (187 – 194)/(30 – 20) = -7/10 = -0.7
197 = -0.7(20) + b
y-intercept b = 208
Substituting the value of the slope m and y-intercept in the slope-intercept form. y = mx + b where, m = -0.7 and b = 208.
y = -0.7x +208
x = 70yrs
y = -0.7(70) + 208 = 159
150 < 159
Second method gives the greater maximum heart rate for a 70 year ols.

Question 2.
Are heart rate and age proportional or nonproportional for each method?
____________

Answer:
For method 1, the relationship is non-propotional.
For method 2, the relationship is non-propotional.

Explanation:
Compare the equation with the general linear equation y = mx + b.
It is linear
Since b is not equal to 0, the relationship is not proportional.

Aisha runs a tutoring business. With Plan 1, students may choose to pay $15 per hour. With Plan 2, they may follow the plan shown on the graph. Question 3. Describe the plan shown on the graph. Type below: ______________ Answer: Choose two points on the graph to find the slope. Find the slope m = (y2 -y1)/(x2 – x1) m = (60 – 40)/(4 – 0) = 20/4 = 5 Read the y-intercept from the graph: b = 40 Use your slope and y-intercept values to write an equation in slope-intercept form. y = 5x + 40 Plan 2 has an intial cost of$40 and a rate of $5 per hour. Question 4. Sketch a graph showing the$15 per hour option.
Type below:
______________

Answer:

Question 5.
What does the intersection of the two graphs mean?
Type below:
______________

Answer:
The intersection of the two graphs represents the number of hours for which both plans will cost the same,

Question 6.
Which plan is cheaper for 10 hours of tutoring?
______________

Answer:
Plan 1
y = 15x
x = 10 hrs
y = 15(10) = $150 Plan 2 y = 5x + 40 y = 5(10) + 40 =$90
$150 >$90
Plane 2 is cheaper

Question 7.
Are cost and time proportional or nonproportional for each plan?
Type below:
______________

Answer:
Comparing with the general linear form of equation y = mx + b. Since b = 0, the relationship is proportional
The cost and time are proportional for Plan 1
Comparing with the general linear form of equation y = mx + b. Since b is not equal to 0, the relationship is proportional
The cost and time are not proportional for Plan 2

Essential Question Check-In

Question 8.
When using tables, graphs, and equations to compare functions, why do you find the equations for tables and graphs?
Type below:
______________

Answer:
The tables and graphs represent a part of the solution of the function. By writing the equation, any value can be a substitute to evaluate the function and compared it with the equations.

### Independent Practice – Comparing Functions – Page No. 171

The table and graph show the miles driven and gas used for two scooters.

Question 9.
Which scooter uses fewer gallons of gas when 1350 miles are driven?
______________

Answer:
Scooter B uses fewer gallons of gas when 1350 miles are driven

Explanation:
The equation for Scooter A Slope m = m = (y2 -y1)/(x2 – x1) where (x1, y1) = (150, 2) and (x2, y2) = (300, 4)
Slope m = (y2 -y1)/(x2 – x1) = (4 – 2)/(300 – 150) = 2/150 = 1/75
2 = 1/75(150) + b
y-intercept b = 0
Substituting the value of the slope m and y-intercept in the slope-intercept form. y = mx + b where, m = 1/75 and b = 0.
y = 1/75x
x = 1350 miles
y = 1/75(1,350)
y = 18gal
The equation for Scooter B Slope m = m = (y2 -y1)/(x2 – x1) where (x1, y1) = (0, 0) and (x2, y2) = (90, 1)
Slope m = (y2 -y1)/(x2 – x1) = (1 – 0)/(90 – 0) = 1/90
2 = 1/90(90) + b
y-intercept b = 0
Substituting the value of the slope m and y-intercept in the slope-intercept form. y = mx + b where, m = 1/90 and b = 0.
y = 1/90x
x = 1350 miles
y = 1/90(1,350)
y = 15gal
Compare the gallons of gas to drive 1,350 miles
18 > 15
Scooter B uses fewer gallons of gas when 1,350 miles are driven.

Question 10.
Are gas used and miles proportional or nonproportional for each scooter?
______________

Answer:
The gas used and miles are proportional to both scooters.

Explanation:
Compare with general linear form of an equation, y = mx + b. If b = 0, the relationship is proportional.
The gas used and miles are proportional to both scooters.

A cell phone company offers two texting plans to its customers. The monthly cost, y dollars, of one plan is y = 0.10x + 5, where x is the number of texts. The cost of the other plan is shown in the table.

Question 11.
Which plan is cheaper for under 200 texts?
______________

Answer:
Plane 1 is cheaper

Explanation:
Plan 1
y = 0.10x + 5
Subsitute x = 199
y = 0.10(199) + 5 = $24.90 Find the slope using two points from the graph by m = (y2 -y1)/(x2 – x1) where (x1, y1) = (100, 20), (x2, y2) =(200, 25) Substitute the value of m and (x1, y1) = (100, 20), (x2, y2) =(200, 25) Slope m = (y2 -y1)/(x2 – x1) = (25 – 20)/(200 – 100) = 5/100 = 0.05 20 = 0.05(100) + b y-intercept b = 15 Substituting the value of slope (m) and (x, y) in the slope intercept form to find y intercept (b): y = 0.05x + 15 x = 199 y = 0.05(199) + 15 =$24.95
Compare the cost for two plans for text < 200
$24.90 <$24.95
Plane 1 is cheaper

Question 12.
The graph of the first plan does not pass through the origin. What does this indicate?
Type below:
______________

Answer:
Plan 1
y = 0.10x + 5
The graph that does not pass through the origin indicates that there is a base price of $5 for the plan. Question 13. Brianna wants to buy a digital camera for a photography class. One store offers the camera for$50 down and a payment plan of $20 per month. The payment plan for a second store is described by y = 15x + 80, where y is the total cost in dollars and x is the number of months. Which camera is cheaper when the camera is paid off in 12 months? Explain. ______________ Answer: For first store, the slope interecept form y = mx + b where m = 20 dollars per month and b = 50 dollar. y = 20x + 50 x = 12 months y = 20(12) + 50 =$290
Second store
y = 15x + 80
x = 12 months
y = 15(12) + 80 = $260 Compare the cost of camera if it paid off in 12 months$290 > $260 Camera is cheaper at second store ### Comparing Functions – Page No. 172 Question 14. The French club and soccer team are washing cars to earn money. The amount earned, y dollars, for washing x cars is a linear function. Which group makes the most money per car? Explain. ______________ Answer: Explanation: Find the slope using two points from the grapgh by Slope m = (y2 -y1)/(x2 – x1) where (x1, y1) = (2, 10) and (x2, y2) = (4, 20) Slope m = (y2 -y1)/(x2 – x1) = (20 – 10)/(4 – 2) = 10/2 = 5 French Club makes$5 per car.
Find the slope using two points from the grapgh by
Slope m = (y2 -y1)/(x2 – x1) where (x1, y1) = (0, 0) and (x2, y2) = (2, 16)
Slope m = (y2 -y1)/(x2 – x1) = (16 – 0)/(2 – 0) = 16/2 = 8
Soccer Club makes $8 per car. Compare the money earned for washing one car$5 < $8 Soccer club makes the most money per car H.O.T. Focus on Higher Order Thinking Question 15. Draw Conclusions Gym A charges$60 a month plus $5 per visit. The monthly cost at Gym B is represented by y = 5x + 40, where x is the number of visits per month. What conclusion can you draw about the monthly costs of the gyms? __________ is more expensive Answer: Gym A is more expensive Explanation: Since the rate per visit is the same, the monthly cost of Gyn A is always more than Gym B. Question 16. Justify Reasoning Why will the value of y for the function y = 5x + 1 always be greater than that for the function y = 4x + 2 when x > 1? Type below: ______________ Answer: y1 = 5x + 1 and y2 = 4x + 2 Subtracting y2 from y1 y1 – y2 = 5x + 1 – (4x + 2) y1 – y2 = x -1 For x>= 1 we get x – 1 >= 0 So y1 – y2 >= 0 or y1 >= y2 Question 17. Analyze Relationships The equations of two functions are y = −21x + 9 and y = −24x + 8. Which function is changing more quickly? Explain. ______________ Answer: y = -21x + 9 y = -24x + 8 y = -24x + 8 is changing more quickly as the absolute value of -24 is greater than the absolute value of -21. ### Guided Practice – Analyzing Graphs – Page No. 176 In a lab environment, colonies of bacteria follow a predictable pattern of growth. The graph shows this growth over time. Question 1. What is happening to the population during Phase 2? ______________ Answer: For Phase 2, the graph is increasing quickly. This shows a period of rapid growth. Question 2. What is happening to the population during Phase 4? ______________ Answer: In Phase 4, the graph is decreasing, hence the number of bacterias is decreasing. The graphs give the speeds of three people who are riding snowmobiles. Tell which graph corresponds to each situation. Question 3. Chip begins his ride slowly but then stops to talk with some friends. After a few minutes, he continues his ride, gradually increasing his speed. ______________ Answer: Graph 2 Explanation: The slope of the graph is increasing, then it becomes constant and starts increasing again. Graph 2 Question 4. Linda steadily increases her speed through most of her ride. Then she slows down as she nears some trees. ______________ Answer: Graph 3 Explanation: The slope of the graph is increasing and then decreasing. Graph 3 Question 5. Paulo stood at the top of a diving board. He walked to the end of the board, and then dove forward into the water. He plunged down below the surface, then swam straight forward while underwater. Finally, he swam forward and upward to the surface of the water. Draw a graph to represent Paulo’s elevation at different distances from the edge of the pool. Type below: ______________ Answer: ### Independent Practice – Analyzing Graphs – Page No. 177 Tell which graph corresponds to each situation below. Question 6. Arnold started from home and walked to a friend’s house. He stayed with his friend for a while and then walked to another friend’s house farther from home. ______________ Answer: Graph 3 Explanation: The graph increases (as Arnold walks from home to friend’s house), then becomes constant (when he stayed with his friend) and then increases again (when he walk to another friend’s house farther away). Graph 3 Question 7. Francisco started from home and walked to the store. After shopping, he walked back home. ______________ Answer: Graph 1 Explanation: The graph increases (as Francisco walked from home to store), becomes constant (when he shops), and then decreases (as he walked back home) Graph 1 Question 8. Celia walks to the library at a steady pace without stopping. ______________ Answer: Graph 2 Explanation: The graph increases at a constant rate (as Celia walks to library without any stops) Graph 2 Regina rented a motor scooter. The graph shows how far away she is from the rental site after each half hour of riding. Question 9. Represent Real-World Problems Use the graph to describe Regina’s trip. You can start the description like this: “Regina left the rental shop and rode for an hour…” Type below: ______________ Answer: Regina left the rental shop and rode for an hour. She rested for half an hour and then started back. After half an hour, she changed her mind and rode for another half an hour. She rest for half an hour. Then she started back and ranched the rental site in 2 hours. Question 10. Analyze Relationships Determine during which half hour Regina covered the greatest distance. Type below: ______________ Answer: Regina covered the greatest distance between 0.5 to 1hr of the journey. She covered 12 miles. ### Analyzing Graphs – Page No. 178 The data in the table shows the speed of a ride at an amusement park at different times one afternoon. Question 11. Sketch a graph that shows the speed of the ride over time. Type below: ______________ Answer: Question 12. Between which times is the ride’s speed increasing the fastest? Type below: ______________ Answer: The speed is increasing the fastest during the 3: 21 and 3: 22 Question 13. Between which times is the ride’s speed decreasing the fastest? Type below: ______________ Answer: The speed is decreasing the fastest during the 3: 23 and 3: 24 H.O.T. Focus on Higher Order Thinking Question 14. Justify Reasoning What is happening to the fox population before time t? Explain your reasoning. Type below: ______________ Answer: The population decreases and then increases before time t Question 15. What If? Suppose at time t, a conservation organization moves a large group of foxes to the island. Sketch a graph to show how this action might affect the population on the island after time t. Type below: ______________ Answer: Explanation: Population is decreasing at first, then it is increasing rapidly. Question 16. Make a Prediction At some point after time t, a forest fire destroys part of the woodland area on the island. Describe how your graph from problem 15 might change. Type below: ______________ Answer: The population would dramatically decrease if there was a fire due to lack of food supply and good land. ### 6.1 Identifying and Representing Functions – Model Quiz – Page No. 179 Determine whether each relationship is a function. Question 1. __________ Answer: Not a function Explanation: A relationship is a function when each input is paired with exactly one output. The input 5 has more than one output. Not a function Question 2. __________ Answer: Function Explanation: A relationship is a function when each input is paired with exactly one output. Each input is paired with only one output. Function Question 3. (2, 5), (7, 2), (−3, 4), (2, 9), (1, 1) __________ Answer: Not a function Explanation: A relationship is a function when each input is paired with exactly one output. Input 2 has more than one output. Not a function 6.2 Describing Functions Determine whether each situation is linear or nonlinear, and proportional or nonproportional. Question 4. Joanna is paid$14 per hour.
__________
__________

Answer:
Linear
Proportional

Explanation:
Writing the situation as an equation, where x is the number of hours.
y = 14x
Compare with general linear equation y = mx + b
Linear
Since b = 0, the relationship is proportional.
Proportional

Question 5.
Alberto started out bench pressing 50 pounds. He then added 5 pounds every week.
__________
__________

Answer:
Linear
Non-proportional

Explanation:
Writing the situation as an equation, where x is the number of hours.
y = 5x + 50
Compare with general linear equation y = mx + b
Linear
Since b is not equal to 0, the relationship is non-proportional.
Non-proportional

6.3 Comparing Functions

Question 6.
Which function is changing more quickly? Explain.

__________

Answer:
Function 2 is changing more quickly.

Explanation:
Find the rate of change for function 1
Rate of Change = (20 – 0)/(0 – 5) = -4
Find the rate of change for function 1
Rate of Change = (6.5 – 11)/(3 – 2) = -4.5
Althogh -4.5 < -4, the absolute value of -4.5 s greater than -4.
Function 2 is changing more quickly.

6.4 Analyzing Graphs

Question 7.
Describe a graph that shows Sam running at a constant rate.
Type below:
______________

Answer:
The graph would be a straight line

Explanation:
Since Sam is running at a constant rate, distance covered per unit of time remains the same and the relationship is linear and proportional.
The graph would be a straight line

Essential Question

Question 8.
How can you use functions to solve real-world problems?
Type below:
______________

Answer:
If in the equation the power of x is 1 then it is linear otherwise nonlinear.
In a graph, if the points form a line it is linear if they form a curve it is a nonlinear function.

### Selected Response – Mixed Review – Page No. 180

Question 1.
Which table shows a proportional function?

Options:
a. A
b. B
c. C
d. D

Answer:
c. C

Explanation:
It contains the ordered pair of the origin (0, 0)
Option C represents a proportional relationship.

Question 2.
What is the slope and y-intercept of the function shown in the table?

Options:
a. m = -2; b = -4
b. m = -2; b = 4
c. m = 2; b = 4
d. m = 4; b = 2

Answer:
c. m = 2; b = 4

Explanation:
Find the slope using two points from the grapgh by
Slope m = (y2 -y1)/(x2 – x1) where (x1, y1) = (1, 6) and (x2, y2) = (4, 12)
Slope m = (y2 -y1)/(x2 – x1) = (12 – 6)/(4 – 1) = 6/3 = 2
Substituting the value of the slope m and (x, y) to find the slope-intercept form.
12 = 4(2) + b
y-intercept b = 4

Question 3.
The table below shows some input and output values of a function.

What is the missing output value?
Options:
a. 20
b. 21
c. 22
d. 23

Answer:
b. 21

Explanation:
Find the rate of change = (17.5 – 14)/(5 – 4) = 3.5
Since the missing output is corresponding to x = 6 and 3.5 to 17.5 (for x = 5)
Output = 17.5 + 3.5 = 21

Question 4.
Tom walked to school at a steady pace, met his sister, and they walked home at a steady pace. Describe this graph.
Options:
a. V-shaped
b. upside down V-shaped
c. Straight line sloping up
d. Straight line sloping down

Answer:
b. upside-down V-shaped

Explanation:
The graph would increase at a constant rate and would decrease at a constant rate.
The graph would be the upside-down V-shaped

Mini-Task

Question 5.
Linear functions can be used to find the price of a building based on its floor area.Below are two of these functions.
y = 40x + 15,000

a. Find and compare the slopes.
Type below:
____________

Answer:
Compare the slopes
The slope for the first function is less than the slope of the second function.
y = 40x + 15000
Compare with slope intercept form y = mx + b where m is the slope m = 40
Second function find the slope using given points by Slope m = (y2 -y1)/(x2 – x1) where (x1, y1) = (7, 3) and (x2, y2) = (6, 4)
Slope m = (y2 -y1)/(x2 – x1) = (56000 – 32000)/(700 – 400) = 24000/300 = 80
m = 80

Question 5.
b. Find and compare the y-intercepts.
Type below:
____________

Answer:
y = 40x + 15,000
Compare with slope-intercept form y = mx + b where m is the slope b = 15000
The second function find the slope using given points by Slope m and (x, y) in the slope-intercept form to fins y-intercept b
y = mx + b where (x, y) = (700, 56000) and m = 80
56000 = 80(700) + b
b = 0
Compare y intercepts
The y-intercept of the first function is greater than the y-intercept of the second function

Question 5.
c. Describe each function as proportional or nonproportional.
Type below:
____________

Answer:
Comparable to slope interecept form y = mx + b
First function: y = 40x + 15000
Second function: y = 80x
Since b is not equal to 0
First function is non-proportional
Since b = 0
The second function is proportional.

### Conclusion:

Download Go Math Grade 8 Answer Key Chapter 6 Functions PDF for free. The famous and great math experts are worked to provide the best way of solving math problems. Immediately, start your practice and learn maths easily.


text:
### Artificial IntelligenceAIMA Exercises

Consider a robot with two simple manipulators, as shown in figure figRobot2. Manipulator A is a square block of side 2 which can slide back and on a rod that runs along the x-axis from x=$-$10 to x=10. Manipulator B is a square block of side 2 which can slide back and on a rod that runs along the y-axis from y=-10 to y=10. The rods lie outside the plane of manipulation, so the rods do not interfere with the movement of the blocks. A configuration is then a pair ${\langle}x,y{\rangle}$ where $x$ is the x-coordinate of the center of manipulator A and where $y$ is the y-coordinate of the center of manipulator B. Draw the configuration space for this robot, indicating the permitted and excluded zones.

Consider a robot with two simple manipulators, as shown in figure figRobot2. Manipulator A is a square block of side 2 which can slide back and on a rod that runs along the x-axis from x=$-$10 to x=10. Manipulator B is a square block of side 2 which can slide back and on a rod that runs along the y-axis from y=-10 to y=10. The rods lie outside the plane of manipulation, so the rods do not interfere with the movement of the blocks. A configuration is then a pair ${\langle}x,y{\rangle}$ where $x$ is the x-coordinate of the center of manipulator A and where $y$ is the y-coordinate of the center of manipulator B. Draw the configuration space for this robot, indicating the permitted and excluded zones.

Submit Solution

format: plain


text:
# Importing from Excel: SAS Newbie

#### raywood

##### New Member
I have an Excel table. In SAS 9.2, I can use File > Import Data to bring that table into the SAS Explorer pane: contents of Work now includes that file. If I double-click on it, I get a VIEWTABLE pane showing the data.

I am under the impression that it exists as a temporary file that I could save to some permanent location. But as far as using it in SAS, I'm at a loss.

My PC mindset tells me that I need appropriate syntax for making references to that file -- that, perhaps, a simple command would run a batch of univariate procedures on it, or whatever. But I don't know if that belief is accurate, and I don't know what the syntax is. I have been searching for hours and I am stuck.

#### hlsmith

##### Less is more. Stay pure. Stay poor.
When you imported the data you gave a name for the temporary file. It should be the exact same name on the temporary dataset you opened. So use that name:

Code:
[B][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080]Proc [/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/B][B][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080]Freq [/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/B][FONT=Courier New][SIZE=2][COLOR=#0000ff][FONT=Courier New][SIZE=2][COLOR=#0000ff][FONT=Courier New][SIZE=2][COLOR=#0000ff]data[/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][FONT=Courier New][SIZE=2][FONT=Courier New][SIZE=2]=tempname_I_referenced;[/SIZE][/FONT]
[/SIZE][/FONT][FONT=Courier New][SIZE=2][COLOR=#0000ff][FONT=Courier New][SIZE=2][COLOR=#0000ff][FONT=Courier New][SIZE=2][COLOR=#0000ff]table[/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT]
[SIZE=2][FONT=Courier New][SIZE=2][FONT=Courier New]variable_names_from_table_you_can_open[/FONT][/SIZE]
[SIZE=2][FONT=Courier New];[/FONT][/SIZE]
[/FONT][/SIZE][B][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080]run[/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/B][FONT=Courier New][SIZE=2][FONT=Courier New][SIZE=2]; [/SIZE][/FONT][/SIZE][/FONT]

[B][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080]proc [/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/B][B][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080]univariate [/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/B][FONT=Courier New][SIZE=2][COLOR=#0000ff][FONT=Courier New][SIZE=2][COLOR=#0000ff][FONT=Courier New][SIZE=2][COLOR=#0000ff]data[/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][FONT=Courier New][SIZE=2][FONT=Courier New][SIZE=2]=temp_name_again [/SIZE][/FONT][/SIZE][/FONT][FONT=Courier New][SIZE=2][COLOR=#0000ff][FONT=Courier New][SIZE=2][COLOR=#0000ff][FONT=Courier New][SIZE=2][COLOR=#0000ff]normal[/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][FONT=Courier New][SIZE=2][FONT=Courier New][SIZE=2];[/SIZE][/FONT]
[/SIZE][/FONT][FONT=Courier New][SIZE=2][COLOR=#0000ff][FONT=Courier New][SIZE=2][COLOR=#0000ff][FONT=Courier New][SIZE=2][COLOR=#0000ff]var[/COLOR][/SIZE][/FONT]
[/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][FONT=Courier New][SIZE=2][FONT=Courier New][SIZE=2]variable_names_from_that_dataset_again;[/SIZE][/FONT]
[/SIZE][/FONT][B][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080][FONT=Courier New][SIZE=2][COLOR=#000080]run[/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/B][FONT=Courier New][SIZE=2][FONT=Courier New][SIZE=2];[/SIZE][/FONT]
[/SIZE][/FONT]

You cannot run data steps when the temp file is open within SAS, so make sure it is closed - that can cause problems, but otherwise you typically just have data=the_files_name for the temp file. Let us know if you still are having problems.

#### noetsi

##### Loves R
This is a case where enterprise guide is really useful, because it is so simple. Once you import the file you can save it to a permament location (once you created that with a libname) simply by using PROC SQL

PROC SQL;
CREATE TABLE SASUSER.Qdata1 AS
SELECT t1.'Work Climate - 1. DVR Headq_0001'n AS x1,
t1.'Work Climate - 10. My super_0001'n AS x10,
t1.'Work Climate - 11. Managers, sup'n AS x11,
t1.'Work Climate - 12. I am abl_0001'n AS x12,
t1.'Work Climate - 13. I know w_0001'n AS x13
FROM WORK.USETHIS t1;
QUIT;

Do a libname for SASUSER first (once you do this you will never have to do the libname again, its a one time event).

Although this is done in SQL you can then use it with any SAS procedure.

#### raywood

##### New Member
OK. I ran this:

proc freq data = Q1;
table
var IQ Arithmetic;
run;

proc corr;
var IQ Arithmetic;
run;

and it gave me some success, some failure: some useful numbers in the Output window but also this:

71
72 proc freq data = Q1;
ERROR: File WORK.Q1.DATA does not exist.
73 table
74 var IQ Arithmetic;
ERROR: No data set open to look up variables.
ERROR: No data set open to look up variables.
ERROR: No data set open to look up variables.
75 run;

NOTE: The SAS System stopped processing this step because of errors.
NOTE: PROCEDURE FREQ used (Total process time):
real time 0.01 seconds
cpu time 0.01 seconds

76

77 proc corr;
78 var IQ Arithmetic;
79 run;

But Q1 was the file I imported, and it's where the data were. So that was puzzling.

#### hlsmith

##### Less is more. Stay pure. Stay poor.
Perhaps:

Is Q1 blue when you type it? Q1 can also mean quantile 1, and SAS may not know it is your dataset name. Even if this is not your problem, I would rename the dataset.

Also, you won't need the 'var' in the proc freq, just 'table'. It is also a good idea to also have a 'data=' in the proc corr as well. SAS uses the previously referenced set, but when you start writing alot of code or moving it around, you may run into issues in the future if you don't have the "data=".

Are you ready, do the following to find your dataset name:

Click View on the top banner,

Then "Exporer",

Then "Work",

Then right-click on the file,

Look at 'properties',

Location: will tell you the file location and name. The name is what is after 'work.'

#### raywood

##### New Member
Thanks! I am in the process of a writeup, incorporating this much-appreciated information.


text:
# functional calculus for Hermitian matrices

Let $I\subset\mathbb{R}$ be a real interval, $f$ a real-valued function on $I$, and let $M$ be an $n\times n$ real symmetric (and thus Hermitian) matrix whose eigenvalues are contained in $I$.

By the spectral theorem, we can diagonalize $M$ by an orthogonal matrix $O$, so we can write $M=ODO^{-1}$ where $D$ is the diagonal matrix consisting of the eigenvalues $\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\}$. We then define

 $\displaystyle f(A)=Of(D)O^{-1},$

where $f(D)$ denotes the diagonal matrix whose diagonal entries are given by $f(\lambda_{i})$.

It is easy to verify that $f(A)$ is well-defined, i.e. a permutation of the eigenvalues corresponds to the same definition of $f(A)$.

Title functional calculus for Hermitian matrices FunctionalCalculusForHermitianMatrices 2013-03-22 14:40:12 2013-03-22 14:40:12 mathcam (2727) mathcam (2727) 4 mathcam (2727) Definition msc 47C05 FunctionalCalculus


text:
# Compile Error (tableofcontenents, maketitle) [closed]

I'm trying to compile this file:

\documentclass[12pt, a4paper]{book}
\usepackage{doc}
\usepackage[italian]{babel}
\begin{document}
\author{Me}
\title{Test}
\date{\today}
\maketitle
\tableofcontenents
\frontmatter
\section{Preface}
\section{Nomenclature}
\mainmatter
\section{Introduction}
\section{Result and discussion}
\subsection{Section}
\subsection{Another section}
\section{Conclusion}
\end{document}

But i get these two error:

! Undefined control sequence.
\maketitle ...ge \global \@topnum \z@ \@maketitle
\fi \thispagestyle {titlep...
l.29 \maketitle

?

! Undefined control sequence.
l.30 \tableofcontenents

?

What could I do? Thank you in advice!

## closed as off-topic by user31729, Stefan Pinnow, Kurt, user13907, ZarkoJul 19 '16 at 22:39

• This question does not fall within the scope of TeX, LaTeX or related typesetting systems as defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• Welcome to TeX.SX! You can't use the doc package here And the command is called \tableofcontents, not \tableofcontenents – user31729 Jul 17 '16 at 20:19
• What do you think \usepackage{doc} is for? If somebody recommended it to you, it's a good occasion for breaking a friendship. ;-) – egreg Jul 17 '16 at 21:57
• Thank you, now at least compile! The problem now is that the resulting .dvi has an empty Index and the sections are numbered as subsections. – sworwitz Jul 18 '16 at 9:04
• For the index I found out I have to compile two or three times but the section problems still remain – sworwitz Jul 18 '16 at 9:08
• I'm voting to close this question as off-topic because the issue could be solved in comments (suggesting to remove the doc package from being included, correcting of typos and compiling twice) – user31729 Jul 19 '16 at 21:27


text:
# How do you find all the real and complex roots of x^4 + 3x^2 + 2 = 0?

Feb 6, 2016

$x = \pm i , \pm i \sqrt{2}$

#### Explanation:

Note that ${x}^{2} + 3 x + 2$ can be factored into $\left(x + 1\right) \left(x + 2\right)$, so

${x}^{4} + 3 {x}^{2} + 2 = \left({x}^{2} + 1\right) \left({x}^{2} + 2\right)$

So, we have the equation

$\left({x}^{2} + 1\right) \left({x}^{2} + 2\right) = 0$

Set both of these equal to $0$:

${x}^{2} + 1 = 0$

${x}^{2} = - 1$

$x = \pm i$

and

${x}^{2} + 2 = 0$

${x}^{2} = - 2 i$

$x = \pm i \sqrt{2}$

The equation has $4$ complex roots and no real roots (meaning it never crosses the $x$-axis:

graph{x^4+3x^2+2 [-13.83, 14.65, -2.31, 11.93]}


text:
Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript.

# Atomic-scale fragmentation and collapse of antiferromagnetic order in a doped Mott insulator

## Abstract

Disentangling the relationship between the insulating state with a charge gap and the magnetic order in an antiferromagnetic Mott insulator remains difficult due to inherent phase separation as the Mott state is perturbed1,2,3,4,5,6,7. Measuring magnetic and electronic properties at atomic length scales would provide crucial insight, but this is yet to be experimentally achieved. Here, we use spin-polarized scanning tunnelling microscopy (SP-STM) to visualize the periodic spin-resolved modulations originating from the antiferromagnetic order in a relativistic Mott insulator Sr2IrO4 (refs. 8,9), and how they change as a function of doping. We find that near the insulator-to-metal transition (IMT), the long-range antiferromagnetic order melts into a fragmented state with short-range correlations. Crucially, we discover that the short-range antiferromagnetic order is locally uncorrelated with the observed spectral gap magnitude. This suggests that static short-range antiferromagnetic correlations are unlikely to be the cause of the inhomogeneous closing of the spectral gap and the emergence of pseudogap regions near the IMT. Our work establishes SP-STM as a powerful tool for revealing atomic-scale magnetic information in complex oxides.

This is a preview of subscription content

## Access options

\$32.00

All prices are NET prices.

## Data availability

The data represented in Figs. 1f,g, 2g,h and 4e,f are available as Supplementary information files. All other data that support the plots within this paper and the other findings of this study are available from the corresponding author on reasonable request.

## Code availability

The computer code used for data analysis is available on request from the corresponding author.

## References

1. Emery, V., Kivelson, S. & Lin, H. Phase separation in the t-J model. Phys. Rev. Lett. 64, 475–478 (1990).

2. Yee, C.-H. & Balents, L. Phase separation in doped Mott insulators. Phys. Rev. X 5, 021007 (2015).

3. Qazilbash, M. M. et al. Mott transition in VO2 revealed by infrared spectroscopy and nano-imaging. Science 318, 1750–1753 (2007).

4. Cai, P. et al. Visualizing the evolution from the Mott insulator to a charge-ordered insulator in lightly doped cuprates. Nat. Phys. 12, 1047–1051 (2016).

5. Battisti, I. et al. Universality of pseudogap and emergent order in lightly doped Mott insulators. Nat. Phys. 13, 21–25 (2017).

6. Chen, X. et al. Influence of electron doping on the ground state of (Sr1–xLax)2IrO4. Phys. Rev. B 92, 075125 (2015).

7. Yan, Y. J. et al. Electron-doped Sr2IrO4: an analogue of hole-doped cuprate superconductors demonstrated by scanning tunneling microscopy. Phys. Rev. X 5, 041018 (2015).

8. Kim, B. et al. Novel J eff = 1/2 Mott state induced by relativistic spin-orbit coupling in Sr2IrO4. Phys. Rev. Lett. 101, 076402 (2008).

9. Cao, G. & Schlottmann, P. The challenge of spin–orbit-tuned ground states in iridates: a key issues review. Rep. Prog. Phys. 81, 042502 (2018).

10. Mott, N. F. The basis of the electron theory of metals, with special reference to the transition metals. Proc. Phys. Soc. A 62, 416–422 (1949).

11. Imada, M., Fujimori, A. & Tokura, Y. Metal-insulator transitions. Rev. Mod. Phys. 70, 1039–1263 (1998).

12. Lee, P. A., Nagaosa, N. & Wen, X.-G. Doping a Mott insulator: physics of high-temperature superconductivity. Rev. Mod. Phys. 78, 17–85 (2006).

13. Rau, J. G., Lee, E. K.-H. & Kee, H.-Y. Spin-orbit physics giving rise to novel phases in correlated systems: iridates and related materials. Annu. Rev. Condens. Matter Phys. 7, 195–221 (2016).

14. Wiesendanger, R. Spin mapping at the nanoscale and atomic scale. Rev. Mod. Phys. 81, 1495–1550 (2009).

15. Jeon, S. et al. Distinguishing a Majorana zero mode using spin-resolved measurements. Science 358, 772–776 (2017).

16. Natterer, F. D. D. et al. Reading and writing single-atom magnets. Nature 543, 226–228 (2017).

17. Hirjibehedin, C. F. F., Lutz, C. P. P. & Heinrich, A. J. J. Spin coupling in engineered atomic structures. Science 312, 1021–1024 (2006).

18. Enayat, M. et al. Real-space imaging of the atomic-scale magnetic structure of Fe1+yTe. Science 345, 653–656 (2014).

19. Manna, S. et al. Interfacial superconductivity in a bi-collinear antiferromagnetically ordered FeTe monolayer on a topological insulator. Nat. Commun. 8, 14074 (2017).

20. de la Torre, A. et al. Collapse of the Mott gap and emergence of a nodal liquid in lightly doped Sr2IrO4. Phys. Rev. Lett. 115, 176402 (2015).

21. Cao, Y. et al. Hallmarks of the Mott-metal crossover in the hole-doped pseudospin-1/2 Mott insulator Sr2IrO4. Nat. Commun. 7, 11367 (2016).

22. Kim, Y. K., Sung, N. H., Denlinger, J. D. & Kim, B. J. Observation of a d-wave gap in electron-doped Sr2IrO4. Nat. Phys. 12, 37–41 (2016).

23. Kim, Y. K. et al. Fermi arcs in a doped pseudospin-1/2 Heisenberg antiferromagnet. Science 345, 187–190 (2014).

24. Chen, X. et al. Unidirectional spin density wave state in metallic (Sr1−xLax)2IrO4. Nat. Commun. 9, 103 (2018).

25. Wang, F. & Senthil, T. Twisted hubbard model for Sr2IrO4: magnetism and possible high temperature superconductivity. Phys. Rev. Lett. 106, 136402 (2011).

26. Ge, M. et al. Lattice-driven magnetoresistivity and metal-insulator transition in single-layered iridates. Phys. Rev. B 84, 100402 (2011).

27. Zhao, H. et al. Charge-stripe crystal phase in an insulating cuprate. Nat. Mater. 18, 103–107 (2019).

28. Guevara, J. M. et al. Spin-polaron ladder spectrum of the spin–orbit-induced Mott insulator Sr2IrO4 probed by scanning tunneling spectroscopy. Phys. Rev. B 99, 121114 (2019).

29. Kyung, B. et al. Pseudogap induced by short-range spin correlations in a doped Mott insulator. Phys. Rev. B 73, 165114 (2006).

30. Ye, F. et al. Magnetic and crystal structures of Sr2IrO4: a neutron diffraction study. Phys. Rev. B 87, 140406 (2013).

31. Kim, J. et al. Magnetic excitation spectra of Sr2IrO4 probed by resonant inelastic X-ray scattering: establishing links to cuprate superconductors. Phys. Rev. Lett. 108, 177003 (2012).

32. Zhou, S., Jiang, K., Chen, H. & Wang, Z. Correlation effects and hidden spin–orbit entangled electronic order in parent and electron-doped iridates Sr2IrO4. Phys. Rev. X 7, 041018 (2017).

33. Solovyev, I. V., Mazurenko, V. V. & Katanin, A. A. Validity and limitations of the superexchange model for the magnetic properties of Sr2IrO4 and Ba2IrO4 mediated by the strong spin-orbit coupling. Phys. Rev. B 92, 235109 (2015).

34. Zhao, L. et al. Evidence of an odd-parity hidden order in a spin–orbit coupled correlated iridate. Nat. Phys. 12, 32–36 (2016).

35. Lawler, M. J. et al. Intra-unit-cell electronic nematicity of the high-T c copper-oxide pseudogap states. Nature 466, 347–351 (2010).

36. Huang, D., Liu, S., Zeljkovic, I., Mitchell, J. F. & Hoffman, J. E. Etching of Cr tips for scanning tunneling microscopy of cleavable oxides. Rev. Sci. Instrum. 88, 023705 (2017).

## Acknowledgements

We thank J. Hoffman, P. Lee and V. Madhavan for valuable discussions, and G. Gu for supplying the FeTe single crystals for characterizing spin-polarized STM tips. The spin-polarized STM measurements were supported by the US Department of Energy Early Career Award DE-SC0020130. I.Z. also acknowledges the support from the Army Research Office grant number W911NF-17-1-0399 (H.Z.) and the National Science Foundation grant number NSF-DMR-1654041 (A.U.) for developing the spin-polarized STM capability. Z.W. acknowledges the support from the US Department of Energy, Basic Energy Sciences grant no. DE-FG02-99ER45747. S.M. and J.M. would like to acknowledge the Office of Naval Research grant N00014-16-1-2657, the National Science Foundation grant DMR-1700137 and a grant from the John Templeton Foundation. S.D.W. acknowledges the support from the National Science Foundation award no. DMR-1905801 (S.D.W.), and additional funding support from the Army Research Office award W911NF-16-1-0361 (Z.P.).

## Author information

Authors

### Contributions

STM experiments were carried out by H.Z. H.Z., S.M. and A.U. were responsible for fabrication and initial characterization of spin-polarized STM tips. Iridate single crystals were grown by Z.P. and X.C., supervised by S.D.W. H.Z. analysed the STM data with guidance from I.Z. Z.W. provided theoretical input on the interpretation of the STM data. I.Z., S.D.W., Z.W., H.Z. and J.M. wrote the manuscript with input from all the authors. I.Z. supervised the project.

### Corresponding author

Correspondence to Ilija Zeljkovic.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

Peer review information Nature Physics thanks Milan Allan and the other, anonymous, reviewer(s) for their contribution to the peer review of this work.

Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Supplementary information

### Supplementary Information

Supplementary Figs. 1–13 and refs. 1–16.

## Source data

### Source Data Fig. 1

Data plotted in Fig. 1f,g.

### Source Data Fig. 2

Data plotted in Fig. 2g,h.

### Source Data Fig. 4

Data plotted in Fig. 4e,f.

## Rights and permissions

Reprints and Permissions

Zhao, H., Manna, S., Porter, Z. et al. Atomic-scale fragmentation and collapse of antiferromagnetic order in a doped Mott insulator. Nat. Phys. 15, 1267–1272 (2019). https://doi.org/10.1038/s41567-019-0671-9

• Accepted:

• Published:

• Issue Date:

• DOI: https://doi.org/10.1038/s41567-019-0671-9

• ### Rotation symmetry breaking in the normal state of a kagome superconductor KV3Sb5

• Hong Li
• He Zhao
• Ilija Zeljkovic

Nature Physics (2022)

• ### Robust Insulating State in Nb-Doped Sr$$_2$$IrO$$_4$$

• Hui Huang
• Ping Ji
• Binghui Ge

Journal of Superconductivity and Novel Magnetism (2022)


text:
# “Fractally self-similar” numbers

This is another question about visualization of Ford circles, the previous one being Confusion with practically implementing rational approximations. Here is an output of zooming into Ford circles at $\frac1{\sqrt2}$

Empirically I found that the pattern repeats periodically and arranged the number of frames so that one gets impression of a continuous infinite zoom (however if one looks very attentively there is a slight jump where the animation restarts).

What I want to know is whether the pattern indeed repeats rigorously for quadratic irrationalities, and whether for other irrational numbers the pattern occasionally becomes "almost" the same (clearly if the number is rational then the picture eventually starts degenerating into the horizontal line).

Since I do not actually know how exactly to formulate it, let me ask the question in form that practically occurred to me: for a real $x$ denote by $P_r(c)$ the picture of Ford circles in the rectangle $(x-c,x+c)\times(0,c)$ with resolution $r$. That is, features of size less than $rc$ cannot be observed; in particular, only the Ford circles of radius $>rc$ are visible, and moreover a circle cannot be distinguished from another one if they are both contained in an annulus of width $<rc$.

For which $x$ do there for any resolution $r$ exist two different $c$ and $c'$ such that the pictures $P_r(c)$ and $P_r(c')$ cannot be distinguished? (Well, they must be "sufficiently different" - say, there is still another $c<c''<c'$ such that $P_r(c'')$ can be distinguished from both.)

And the question about this question - what is (if any) a rigorous mathematical statement behind it?

(Later - decided to add the zoom for the golden ratio, here the jump is almost impossible to notice

Circles closest to the center alternate between left and right, ratios of their sizes must be something like consecutive Fibonacci numbers...

Before you can formulate your question precisely, you need a better notion of distance between two pictures of a set $S \subset \mathbb{C}$ (where in this case, $S$ is the union of Ford circles). If $\iota_1, \iota_2$ are pictures (bijections from square subsets $A_1, A_2$ of $\mathbb{C}$ to the unit square), then we can define distance as the Hausdorff distance between $\iota_1(A_1 \cap S)$ and $\iota_2(A_2 \cap S)$.

(With your original definition, we can trivially find two indistinguishable pictures by the pigeonhole principle, since there are only finitely many distinguishable pictures.)

The entire tessellation of Ford circles can be reconstructed by specifying any two tangent circles together with the horizontal line. Hence any Moebius transformation which maps a pair of tangent Ford circles to another such pair must extend to an automorphism of the entire tessellation. Since the Ford circles do not possess scale-invariance, this answers your first question (about whether the self-similarity is exact) negatively.

There is, on the other hand, a rigorous way in which we can say that your zoom sequence about a quadratic irrational is 'asymptotically' scale-invariant.

Specifically, the Ford circles (viewed as living in the upper half-plane) are invariant under the modular group $PSL(2, \mathbb{Z})$ consisting of all fractional linear transformations:

$$z \mapsto \dfrac{az + b}{cz + d}$$

where $a,b,c,d \in \mathbb{Z}$ and $ad - bc = 1$. Moreover, these are the only such conformal automorphisms of the Ford circles.

Now, for a quadratic irrational $\phi$ with Galois conjugate $\psi$, we can find a $2 \times 2$ matrix $A \in SL(2, \mathbb{Z})$ such that its eigenvectors are $(\phi, 1)$ and $(\psi, 1)$. This corresponds in the obvious way to a conformal automorphism $f$ of the Ford circles which fixes the two points $\phi$ and $\psi$.

Since $f$ is conformal, fixes $\mathbb{R}$ and fixes the point $\phi$, then to linear order it is given by a scaling $g$ about $\phi$. That is to say, if a point $x \in \mathbb{C}$ satisfies $|x - \phi| < \varepsilon$, then the displacement $|f(x) - g(x)| = o(\varepsilon)$.

You asked whether all other irrationals have the property that your zoom sequence contains two pictures that appear indistinguishable. If we instead pose the condition 'there exists an infinite convergent subsequence of pictures', then a sufficient condition is that if the sequence of continued fraction convergents is given by:

$\dfrac{a_1}{b_1}, \dfrac{a_2}{b_2}, \dfrac{a_3}{b_3}, \cdots$

then the sequence of ratios between successive denominators:

$\dfrac{b_2}{b_1}, \dfrac{b_3}{b_2}, \dfrac{b_4}{b_3}, \cdots$

contains a subsequence which converges, as then we have a sequence of automorphisms $f_i$ with the property that:

• $f_i(\dfrac{a_{k_i}}{b_{k_i}}) = \dfrac{a_{k_{i+1}}}{b_{k_{i+1}}}$
• $f_i(\dfrac{a_{k_i+1}}{b_{k_i+1}}) = \dfrac{a_{k_{i+1}+1}}{b_{k_{i+1}+1}}$

and for all sufficiently small balls around $\phi$, the automorphisms converge to a scaling about $\phi$ and the result follows as before.

On the other hand, if the ratios of successive denominators grow without bound, then you can actually find a sequence of pictures which converges to a zoomed-out picture of Ford circles. In your animation of $\frac{1}{\pi}$, you can see this because the next convergent after $\frac{355}{113}$ has a huge denominator.

Hence for every irrational real, you can find a convergent sequence of pictures.

• Still trying to digest your answer. The key point seems to be much the same as with my previous question - that two circles determine the rest. Except more precisely they determine the portion of the tesselation confined between them, I don't quite see how to reconstruct anything outside. – მამუკა ჯიბლაძე Nov 30 '15 at 18:07
• You can reconstruct all the circles. Specifically, given three externally tangent discs in $\mathbb{C} \cup \{ \infty \}$ (two of which are the interiors of those two circles; the other is the lower half-plane), you can uniquely construct an Apollonian disc packing. The Ford circles are precisely the discs in the Apollonian gasket which are tangent to the real line. – Adam P. Goucher Nov 30 '15 at 18:13
• Yes I see now - there is no real difference between "outside" and "inside" given transformations that interchange them. One more question though, about the penultimate paragraph - I don't quite see why in case of growing denominator ratios the pictures will converge to a zoomed-out picture: in case of quadratic irrationality there was that unique conjugate and iterates of the inverse of $f$ accumulated to it. But if now $\phi$ is not quadratic anymore, not even algebraic, why will turning the process backwards yield convergence to anything? – მამუკა ჯიბლაძე Nov 30 '15 at 18:57

Define the parent of a Ford circle to be the smaller of its two larger neighbors; then this parent relation defines the Stern–Brocot tree on the points of tangency of the circles. The path in this tree to a given real number is closely related to its continued fraction expansion. A path in the tree can be represented combinatorially by a binary sequence that determines whether to go left or right at each step, and this sequence is periodic exactly for the real numbers that have periodic continued fractions. That is, for the quadratic irrationals. It seems very likely to me that this periodicity in left-right steps is what is leading to your perceived periodicity in the relative sizes and positions of the circles, and that (once you determine an appropriate measure of shape similarity) this can be proven more rigorously: the shapes become approximately periodic when the left-right sequence in the tree is (after finitely many steps) exactly periodic.

• This is what confuses me, and it also appeared in the another answer: the continued fraction expansion is just periodic; both it and the succession of circles corresponds to a path in the Stern-Brocot tree and its turns are periodic too; yet, the picture itself turns out to be only "asymptotically periodic". Why? Is there a modification of the metric in which it is also strictly periodic? – მამუკა ჯიბლაძე Dec 1 '15 at 5:44
• Because the first two Ford circles you start with have no reason to have the correct proportions to be part of an exactly periodic sequence of circle shapes. And you can't suddenly jump into a periodic sequence of circle shapes from a sequence that does not have the same shapes, because the system is reversible. But the more times you repeat, the closer it converges to that exact periodicity. – David Eppstein Dec 1 '15 at 5:57
• So actually no pair of touching circles produces exactly homothety-periodic sequence? Somehow I feel there must be a natural metric enhancing strict periodicity of continued fractions / Stern-Brocot paths to strict periodicity in the metric homothety sense – მამუკა ჯიბლაძე Dec 1 '15 at 6:03
• Hm on the afterthought the first part is obvious - ratios of rational radii cannot become constant if they must gradually approach an irrational number. But what about the second? – მამუკა ჯიბლაძე Dec 1 '15 at 6:09