I"m make the efforts to execute this evidence by contradiction. I understand I need to use a lemma to establish that if $x$ is divisible by $3$, then $x^2$ is divisible by $3$. The lemma is the straightforward part. Any type of thoughts? just how should I expand the proof because that this to the square root of $6$?

Say $ sqrt3 $ is rational. Then $sqrt3$ can be stood for as $fracab$, where a and also b have no common factors.

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So $3 = fraca^2b^2$ and $3b^2 = a^2$. Now $a^2$ have to be divisible by $3$, yet then so have to $a $ (fundamental organize of arithmetic). For this reason we have actually $3b^2 = (3k)^2$ and $3b^2 = 9k^2$ or even $b^2 = 3k^2 $ and also now we have a contradiction.

What is the contradiction?

The-Duderino through the way, the proof because that $ sqrt6 $ adheres to in the very same steps practically exactly. $endgroup$

suppose $sqrt3$ is rational, then $sqrt3=fracab $ for some $(a,b)$suppose we have actually $a/b$ in easiest form.eginalignsqrt3&=fracab\a^2&=3b^2endalignif $b$ is even, then a is also even in which instance $a/b$ is not in simplest form.if $b$ is odd then $a$ is additionally odd.Therefore:eginaligna&=2n+1\b&=2m+1\(2n+1)^2&=3(2m+1)^2\4n^2+4n+1&=12m^2+12m+3\4n^2+4n&=12m^2+12m+2\2n^2+2n&=6m^2+6m+1\2(n^2+n)&=2(3m^2+3m)+1endalignSince $(n^2+n)$ is an integer, the left hand side is even. Due to the fact that $(3m^2+3m)$ is one integer, the right hand side is odd and we have found a contradiction, because of this our hypothesis is false.

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A claimed equation $m^2=3n^2$ is a straight contradiction to the basic Theorem the Arithmetic, because when the left-hand next is expressed as the product of primes, there are evenly numerous $3$’s there, when there room oddly plenty of on the right.

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answered Sep 14 "14 at 5:05

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The number $sqrt3$ is

**irrational**,it can not be expressed as a ratio of integers a and also b. To prove that this statement is true, let united state Assume the it is

**rational**and then prove it isn"t (Contradiction).

So the presumptions states that :

**(1)** $sqrt3=fracab$

Where a and also b room 2 integers

**Now due to the fact that we desire to refuse our presumption in order to acquire our desired result, us must show that there space no such 2 integers.**

Squaring both sides give :

$3=fraca^2b^2$

$3b^2=a^2$

(**Note** : If $b$ is **odd** then $b^2$ is Odd, climate $a^2$ is odd because $a^2=3b^2$ (3 time an weird number squared is odd) and Ofcourse a is odd too, because $sqrtodd number$ is also odd.

With **a** and also **b** odd, we deserve to say that :

$a=2x+1$

$b=2y+1$

Where x and also y need to be essence values, otherwise clearly a and b wont be integer.

Substituting these equations come $3b^2=a^2$ offers :

$3(2y+1)^2=(2x+1)^2$

$3(4y^2 + 4y + 1) = 4x^2 + 4x + 1$

Then simplying and also using algebra us get:

$6y^2 + 6y + 1 = 2x^2 + 2x$

You should know that the LHS is an odd number. Why?

$6y^2+6y$ is **even** Always, therefore +1 come an even number gives an strange number.

The RHS next is an even number. Why? (Similar Reason)

$2x^2+2x$ is **even** Always, and there is NO +1 favor there was in the LHS to do it ODD.

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There room no options to the equation due to the fact that of this.

Therefore, integer worths of a and also b which meet the partnership = $fracab$ **cannot** it is in found.