Square root of $2$ is irrational

I am studying the proof that $\sqrt 2$ is an irrational number. Now I understand most of the proof, but I lack an understanding of the main idea which is:

We assume $\frac{m^2}{n^2} = 2$. Then both $m$ and $n$ can't be even.

I do not understand, why can't both $m$ and $n$ be even?

Solution 1:

Before you continue the argument you write the fraction $m/n$ in lowest terms. Then the numerator and denominator can't both be even.

Solution 2:

we can assume that $$\gcd(m,n)=1$$ and $$2=\frac{m^2}{n^2}$$ then $$2n^2=m^2$$ thus the left-hand side is even and so $$m^2$$ this is a contradiction, both numbers $m,n$ can not be even

Solution 3:

Because if $n=2n_1$ and $m=2m_1$ then $\frac{m_1^2}{n_1^2}=2$... and we get an infinite series $n>n_1>n_2>...$ of natural numbers, which is impossible.

Solution 4:

Okay, this may be a bit pendantic.

There are a few fundamentals assumed but not stated.

1) Given that $a = k*n; a,k,n \in \mathbb Z$ and $b = k*m; b, m \in Z$ then then rational value of the ratio $\frac ab $ is the same as the ratio $\frac mn$ and we say $\frac ab = \frac nm$.

Why? Well... because. (If someone has a better answer feel free to leave a comment). I feel this is the basic concept of ratio or proportion and ratio of "$n$ units to $m$ units" is axiomatically the same value regardless of the unit. And if the "units" is a measure of $k$ value then "$n$ $k$-s are in proportion to $m$ $k$-s" as "$n$ is in proportion to $m$" is simply an axiom.

If anyone has a better or more correct way of putting this, let me know.

2) Given any two integers $a,b$ there is some common factor $k$ so that if $a\div k =n$ and $b \div k = m$ and $m,n$ have only $1$ as a common (positive integer) factor.

Why? If $a,b $ don't have any common factor other than then $k=1$ is precisely that common factor. If $a, b$ have a common $j$ other than $1$ then $j > 1$ and $a\div j = n$ then $a > n$ and if $b\div j = m$ then $b > n$. If $a$ and $b$ were such that there is no such common factor, then $j$ can not be that factor so $m$ and $n$ have a common factor $j_2 > 1$ and if $m =j_2*m_2$ and $n = j_2*n_2$, we have $a > n > n_2$ and $b > m > m_2$. If there is no such terminating common factor we can do this infinite so we can get an infinite chain of $a > n > n_2> n_3>.....$ and $b > m > m_2 > m_3> .....$.

This is clearly impossible. Why? Because these are integers the difference between $n_i$ and $n_{i+1}$ is at least $1$ and so $a$ and $b$ must be larger than an infinite $1$s. (i.e. $a$ and $b$ are infinite.)

From those two assumptions, we can conclude:

A) If $q = \frac ab$ is any rational number with $a,b$ are integers. Then $a,b$ have a common factor $k$ so that $n = \frac ak; m = \frac bk$ and $n$ and $m$ have no common factor other than $1$. Thus we can state $q =\frac nm$ where $n$ and $m$ have no factors other than $1$ in common.

We can express any rational $q$ in such a way.

.....

So now you can start your proof:

Assume $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac ab$ for integers $a,b$ so that $a$ and $b$ have no common factor other than $1$.

Then...... < < details omitted > >... $2$ is a factor of $a$ and ..... < < details omitted > > .... $2$ is a factor of $b$.

And therefore $a$ and $b$ have a factor of $2$ in common, which contradicts that $a$ and $b$ have no factors in common other than $1$.

So $\sqrt{2}$ is not rational.

Solution 5:

An another way

Be $\sqrt{2}\in \mathbb{Q}$, and $\mathcal{Q}=\bigg\{q\in\mathbb{N}^*,\quad q\sqrt{2}\in\mathbb{N}\bigg\}\implies \mathcal{Q}\neq\varnothing$

Let $q_s:=\min \mathcal{Q} $

Let $p:=q_s\sqrt{2}-q_s$

$p=q_s\sqrt{2}-q_s<2q_s-q_s=q_s\iff \boxed{p<q_s}\quad (1)$

$p=q_s\sqrt{2}-q_s\iff\sqrt{2}p=2q_s-\sqrt{2}q_s \iff \boxed{\sqrt{2}p\in\mathbb{N}}\quad (2)$

$(1)\land(2)\implies$ we have found $p<q_s$ such that $\sqrt{2}p\in\mathbb{N}$ so we have a contradiction thus $\sqrt{2}\notin \mathbb{Q}$