# Truth value of a quantifier statement

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• February 10th 2010, 07:05 PM
Nostalgia
Truth value of a quantifier statement
Hello,

I need to determine whether the following statement is true of false:

$\forall x(x > 1\to x^2 > x)$ Domain: All reals

I think the statement is true since I cannot find a value which would make the first part true while the second part false. However, I don't know how to prove the statement true without plugging in various numbers, but doing so would not prove that the statement is true in general, so I was wondering how one would start this question.

Thanks
• February 10th 2010, 08:14 PM
Danneedshelp
Quote:

Originally Posted by Nostalgia
Hello,

I need to determine whether the following statement is true of false:

$\forall x(x > 1\to x^2 > x)$ Domain: All reals

I think the statement is true since I cannot find a value which would make the first part true while the second part false. However, I don't know how to prove the statement true without plugging in various numbers, but doing so would not prove that the statement is true in general, so I was wondering how one would start this question.

Thanks

Here are some thoughts

Notice, $x^{2}=xx>x$ $\Leftrightarrow$ $x>\frac{x}{x}=1$.
So, clearly a contradiction will arise if we assume $x\leq\\1$ for $\forall\\x$.
• February 11th 2010, 05:58 AM
Grandad
Hello Nostalgia
Quote:

Originally Posted by Nostalgia
Hello,

I need to determine whether the following statement is true of false:

$\forall x(x > 1\to x^2 > x)$ Domain: All reals

I think the statement is true since I cannot find a value which would make the first part true while the second part false. However, I don't know how to prove the statement true without plugging in various numbers, but doing so would not prove that the statement is true in general, so I was wondering how one would start this question.

Thanks

The proof is quite simple, provided we are given that we may multiply both sides of an inequality by a positive number; i.e. provided we know that:
$a > b$ and $c > 0 \Rightarrow ac > bc$
For we simply multiply both sides by $x$, noting that $x>1 \Rightarrow x >0$. So:
$x>1 \Rightarrow xx > 1x \Rightarrow x^2>x$
Grandad
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