Proof by contradiction

**nikie1o2** · April 9th 2009, 03:07 PM

I need to prove that SqRt of 3 is irrational using a proof my contradiction and the lemma: an interger n is a multiple of 3 iff n^2 is a multiple of 3

I know to assume the SqRt of 3 is rational. Then it equals a/b, where a,b are integers and b doesnt equal to 0..not too sure where to go after that

**Plato** · April 9th 2009, 03:39 PM

Let’s prove a more general result: If n is a non-square positive integer then $\sqrt{n}$ is irrational.
So suppose for the purpose of proof by contradiction that $\sqrt{n}$ is rational.
Using the well order of the positive integers, there is a first positive integer $K$ such that $K\sqrt n \in \mathbb{Z}^ +$ .
Using the floor function: $0 < \sqrt n - \left\lfloor {\sqrt n } \right\rfloor < 1$ .
Now $0 < K\sqrt n - K\left\lfloor {\sqrt n } \right\rfloor < K$ and $\left[ {K\sqrt n - K\left\lfloor {\sqrt n } \right\rfloor } \right] \in \mathbb{Z}^ +$ . (WHY?)
However, $\left[ {K\sqrt n - K\left\lfloor {\sqrt n } \right\rfloor } \right]\left( {\sqrt n } \right) \in \mathbb{Z}^ +$ which is a clear contradiction to K being minimum.

**xalk** · April 11th 2009, 04:12 PM

Originally Posted by nikie1o2

I need to prove that SqRt of 3 is irrational using a proof my contradiction and the lemma: an interger n is a multiple of 3 iff n^2 is a multiple of 3

I know to assume the SqRt of 3 is rational. Then it equals a/b, where a,b are integers and b doesnt equal to 0..not too sure where to go after that

Let $\sqrt{3}= \frac{a}{b}$ where a,b integers and $b\neq 0$ and in their lowest terms with no common factor.

Now square both sides of the equation and we get :

$3b^2 =a^2$ .That indicates that a^2 is multiply of 3 ,since it is equal to 3b^2,hence according to the above lemma a is a multiple of 3,thus a=3c.

Now substitute that to the above equation and we have:

$3b^2=9c^2$ or $b^2=3c^2$ .

Therefor we see that also b is a multiple of 3,since b^2 is a multiple of 3.

Thus b=3r ,contrary to our assumption that a,and b have no common factor.

Hence $\sqrt{3}$ is not rational

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