I"m make the efforts to perform this evidence by contradiction. I understand I need to use a lemma to develop that if $x$ is divisible through $3$, then $x^2$ is divisible by $3$. The lemma is the easy part. Any kind of thoughts? how should I extend the proof for this to the square root of $6$?

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

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So $3 = \fraca^2b^2$ and also $3b^2 = a^2$. Currently $a^2$ should be divisible by $3$, however then so need to $a $ (fundamental to organize of arithmetic). Therefore we have actually $3b^2 = (3k)^2$ and also $3b^2 = 9k^2$ or also $b^2 = 3k^2 $ and also now we have actually a contradiction.

What is the contradiction?

The-Duderino by the way, the proof for $ \sqrt6 $ follows in the exact same steps nearly exactly. $\endgroup$

suppose $\sqrt3$ is rational, climate $\sqrt3=\fracab $ for some $(a,b)$suppose we have actually $a/b$ in most basic form.\beginalign\sqrt3&=\fracab\\a^2&=3b^2\endalignif $b$ is even, climate a is likewise even in which situation $a/b$ is not in simplest form.if $b$ is odd then $a$ is also odd.Therefore:\beginaligna&=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)+1\endalignSince $(n^2+n)$ is one integer, the left hand side is even. Due to the fact that $(3m^2+3m)$ is an integer, the best hand side is odd and we have discovered a contradiction, as such our hypothesis is false.

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edited Mar 2 in ~ 10:38

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answer Sep 14 "14 at 4:24

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A an alleged equation $m^2=3n^2$ is a straight contradiction to the fundamental Theorem that Arithmetic, due to the fact that when the left-hand next is expressed as the product the primes, there are evenly plenty of $3$’s there, when there are oddly plenty of on the right.

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answered Sep 14 "14 in ~ 5:05

LubinLubin

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The number $\sqrt3$ is

**irrational**,it cannot be expressed together a proportion of integers a and also b. To prove that this declare is true, let us Assume the it is

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

So the presumptions states that :

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

Where a and b room 2 integers

**Now because we desire to refuse our assumption in stimulate to acquire our preferred result, us must show that there room 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 since $a^2=3b^2$ (3 times an weird number squared is odd) and also Ofcourse a is weird too, due to the fact that $\sqrtodd number$ is also odd.

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

$a=2x+1$

$b=2y+1$

Where x and also y should be integer values, otherwise clearly a and b no be integer.

Substituting this equations to $3b^2=a^2$ provides :

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

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

Then simplying and using algebra us get:

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

You should understand 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 side is an even number. Why? (Similar Reason)

$2x^2+2x$ is **even** Always, and there is NO +1 like there remained 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 b which satisfy the connection = $\fracab$ **cannot** it is in found.