Suppose $x\in \tennis2007.orgbbQ$ such the $x^2=6$. Due to the fact that $x\in \tennis2007.orgbbQ$, over there exists $m,n \in \tennis2007.orgbbZ$ whereby either $m$ or $n$ is weird such that $x=\fracmn$.

You are watching: Prove that sqrt 6 is irrational

$\implies$ $x^2=(\fracmn)^2=\fracm^2n^2=6$

$\implies$ $m^2=6n^2$, therefore $m^2$ is even. Hence, $m$ is even.

Since $m$ is even, $m=2k, k\in \tennis2007.orgbbZ$.

Then, $m^2=(2k)^2=4k^2=6n^2$

$\implies$ $n^2$ is even, for this reason $n$ is even. Yet one of $m$ or $n$ should be odd, so $x\notin \tennis2007.orgbbQ$.

Therefore, $\sqrt6$ is irrational.

Does everything look alright here?

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edited Nov 23 "15 in ~ 15:40 boy name Sleziak
inquiry Oct 24 "13 in ~ 16:24 GuestGuest
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This is fine and also in effect you show that $\sqrt2a$ is irrational if $a$ is odd.

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reply Oct 24 "13 at 16:28 Hagen von EitzenHagen von Eitzen
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Yes, it"s right. You have the right to generalize the method; because that a optimistic integer $z$ and a prime number $p$, denote by $\mu_p(z)$ the best exponent $k$ such the $p^k$ divides $z$ ($\mu_p(z)=0$ if $p$ doesn"t division $z$).

From the distinct factorization, it"s clear that $\mu_p(xy)=\mu_p(x)+\mu_p(y)$.

We desire to prove the if $\sqrtz$ is rational, then $\mu_p(z)$ is even, for any type of prime $p$.

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Suppose there exist $m$ and $n$ optimistic integers such the $(m/n)^2=z$. Then$$m^2=zn^2$$Let $p$ it is in a prime; then$$\mu_p(m^2)=2\mu_p(m)=\mu_p(zn^2)=\mu_p(z)+2\mu_p(n);$$therefore $\mu_p(z)=2(\mu_p(m)-\mu_p(n))$ is even. In certain $z$ is a perfect square.

Since \$2^2 3

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