# Thread:

1. ## Is the number a^(1/2) + b^(1/2) always irrational with the following conditions?

Is the following number (with the added conditions taken into account) always irrational? And how would one go about proving this? Heres the number and the conditions:

Is the number $\sqrt{a} + \sqrt{b}$ always irrational? Given these conditions:

$a \neq b$

$a = 1,2,3.... \;\;\; and \;\;\;b=1,2,3,....$

$a \neq x^m$ for some numbers $x=1,2,3....$ and $m=1,2,3...$

$b \neq y^n$ for some numbers $y=1,2,3....$ and $n=1,2,3..$

There, I think I've covered all the proper conditions. But if there is one I missed, I'm sure you catch the drift of the exact question I am asking, and can probably guess whatever the missing condition is, not that there is a missing condition, just considering the possibility. Anyway, theres the question. Also, I understand the possibility that there is a simple example that shows the number $\sqrt{a} + \sqrt{b}$ to be rational in some obvious case, and if there is such a case, then this question is solved simply. And if your answer is yes, I"d be interested in seeing a proof, or atleast the "gist" of the proof. Thanks in advance

2. Originally Posted by mfetch22
Is the following number (with the added conditions taken into account) always irrational? And how would one go about proving this? Heres the number and the conditions:

Is the number $\sqrt{a} + \sqrt{b}$ always irrational? Given these conditions:

$a \neq b$

$a = 1,2,3.... \;\;\; and \;\;\;b=1,2,3,....$

$a \neq x^m$ for some numbers $x=1,2,3....$ and $m=1,2,3...$

$b \neq y^n$ for some numbers $y=1,2,3....$ and $n=1,2,3..$

There, I think I've covered all the proper conditions. But if there is one I missed, I'm sure you catch the drift of the exact question I am asking, and can probably guess whatever the missing condition is, not that there is a missing condition, just considering the possibility. Anyway, theres the question. Also, I understand the possibility that there is a simple example that shows the number $\sqrt{a} + \sqrt{b}$ to be rational in some obvious case, and if there is such a case, then this question is solved simply. And if your answer is yes, I"d be interested in seeing a proof, or atleast the "gist" of the proof. Thanks in advance
Isn't this covered here?

http://www.mathhelpforum.com/math-he...tml#post540098

(post #3)
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