You are watching: Is the square root of 6 irrational
$\implies$ $m^2=6n^2$, for this reason $m^2$ is even. Hence, $m$ is even.
Since $m$ is even, $m=2k, k\in \cg-tower.combbZ$.
$\implies$ $n^2$ is even, for this reason $n$ is even. Yet one of $m$ or $n$ have to be odd, so $x\notin \cg-tower.combbQ$.
Therefore, $\sqrt6$ is irrational.
Does whatever look alright here?
real-analysis radicals rationality-testing
edited Nov 23 "15 at 15:40
51.8k1818 yellow badges162162 silver- badges323323 bronze badges
asked Oct 24 "13 in ~ 16:24
5111 silver- badge22 bronze badges
add a comment |
2 answers 2
active earliest Votes
This is fine and in result you show that $\sqrt2a$ is irrational if $a$ is odd.
reply Oct 24 "13 in ~ 16:28
Hagen von EitzenHagen von Eitzen
add a comment |
Yes, it"s right. You deserve to generalize the method; because that a positive integer $z$ and a prime number $p$, signify by $\mu_p(z)$ the preferably exponent $k$ such that $p^k$ divides $z$ ($\mu_p(z)=0$ if $p$ doesn"t divide $z$).
From the unique factorization, it"s clear the $\mu_p(xy)=\mu_p(x)+\mu_p(y)$.
We want to prove the if $\sqrtz$ is rational, climate $\mu_p(z)$ is even, for any type of prime $p$.
See more: What Type Of Consumer Is A Hawk A Producer Consumer Or Decomposer Or Consumer?
Suppose over there exist $m$ and also $n$ positive 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 specific $z$ is a perfect square.
stack Exchange Network
expropriate all cookie Customize settings