# Question on proof of no rational number whose square is 2

in the proof that there is no rational number whose square is 2:

Suppose x is a rational number whose square is 2.

Then x can be written in lowest terms as $\frac{a}{b}$ , where a is an integer and b is a positive integer.

Since $x^2 = 2$, $\left (\frac{a}{b} \right )^2$, so $\frac{a^2}{b^2}$ But then a is even, so $a = 2n$ for some integer n.

Then $(2n)^2$ = $2b^2$, so $4n^2 = 2b^2$.

Then $2n^2 = b^2$ , so $b^2$ is even, and thus b is even.

Then a and b both have 2 as a common factor, so $\frac{a}{b}$ cannot be in lowest terms, a contradiction. Thus x cannot be rational.

The part I am struggling with is why we can assume that x can be written in lowest terms. Or in other proofs, they say that the gcd(a,b) must be 1, but I do not understand why this is important to the proof. Any help is appreciated

• It's important because otherwise, how do you obtain a contradiction? – Bernard Oct 4 '17 at 1:15
• @Bernard I misspoke when I said that I didn't know the importance, I was more confused on why we can say the gcd of a rational number is 1 – Curious Student Oct 4 '17 at 1:16
• It is practically a canonical law when dealing with rational numbers, taking them in their lowest terms (take, for instance, $\dfrac 23$ instead of $\dfrac {634} {951}$). On the other hand, $\dfrac ab$ in lowest terms is just equivalent to say that $(a,b)=1$. – Piquito Oct 4 '17 at 1:22

You can assume that $\dfrac{a}{b}$ is in lowest terms because any fraction can be written in lowest terms, so we just take $\dfrac{a}{b}$ to be that lowest terms representation to make the proof easier. This is exactly the same as assuming $\gcd(a,b)=1$. This assumption is crucial, as it allows us to go through the proof, and eventually conclude that $2\mid a$ and $2\mid b$, which gives a contradiction because of our construction that $\gcd(a,b)=1$. Since $\gcd(a,b)=1$ there cannot be any common factors of $a$ and $b$ other than $\pm 1$. In other words, if $\dfrac{a}{b}$ is in lowest terms, the numerator and denominator cannot both be even, because then they would both be divisible by two, and thus the fraction is not in lowest terms.

• I think I understand.. So because $\frac{a}{b}$ is a rational number then you can assume that because it's in the lowest form, a,b can have no common factors besides 1? – Curious Student Oct 4 '17 at 1:15
• @CuriousStudent The definition of "lowest terms" is that there is no positive integer $>1$ that divides both the numerator and the denominator, and hence they have no common factors. – Carl Schildkraut Oct 4 '17 at 1:23
• When you construct a fraction $\frac{a}{b}$, you can assume it is in lowest terms, because any fraction can be represented in such a way. And yes, because the fraction is in lowest terms, $a,b$ have to common factors other than $1$ (and $-1$). – Dave Oct 4 '17 at 1:24

You can always divide both $a, b$ by their GCD and the remaining factors will be coprime. To be more specific lets say $x = \dfrac{a}{b}$ whereas $a = m\cdot d, b = n\cdot d$ with $d =$ GCD of $a,b$. Then $m,n$ must be coprime for if they were not, then take a common factor other than $1$ of $m,n$ say $q$ then you have a factor that is greater than GCD $: qd$ which is greater than $d$ a contradiction to $d$ being GCD.

Every rational can be written in its lowest terms because you can divide numerator and denominator with their common factors until they have no more common factors.

When you do that the proof goes in such a way that it shows that both a numerator and denominator are even, but they cannot be both even since we supposed that our fraction is written in lowest terms and if they are both even then a fraction is not written in its lowest terms.

Since assumption that fraction is written in its lowest terms implies that fraction is not written in its lowest terms we are in obvious contradiction.

Since contradiction arose because we assumed that square root of two is rational it follows that it is not rational because it cannot be both rational and irrational at the same time and it must be either rational or irrational.

Is it clear?

It's not even necessary for $p$ and $q$ to be in lowest terms to complete the proof, it's enough for either $p$ or $q$ (or both) to be odd. It's easy to see that if both are even, we can reduce fraction by 2 until at least one of the numbers is odd.

It's not wholly obvious that any rational, $q,$ can be written as $a/b$ with $a$ and $b$ coprime. However, it is true.

One proof is as follows. Suppose $a$ and $b$ are not co-prime then we can write $$a = k_1 a_1, \, b=k_1 b_1$$ with $k_1 > 1$ and an integer, so $q= a_1/b_1.$ If $a_1$ and $b_1$ are not co-prime, we write $$a_1 = k_2 a_2, \, b_1=k_2 b_2$$ so $q= a_2/b_2.$

Clearly, either this process goes on forever or we reach a representation with $q=a_n/b_n$ with $a_n$ and $b_n$ co-prime.

However, it can't go on forever since after $n$ steps we have $$a=k_1 k_2 \dots k_n a_n \geq 2^n a_n\geq 2^n.$$ So it must stop before $2^n$ is greater than $a.$

We can avoid considering lowest terms. Any $n\in \Bbb N$ is equal to $2^{n_1}\cdot n_2$ where $n_1\in \{0\}\cup \Bbb N$ and $n_2$ is an odd member of $\Bbb N.$

Now $n_1, n_2$ are uniquely determined by $n$: If $n=2^{n_1}\cdot n_2=2^{n'_1}\cdot n'_2$ with $n_1, n'_1\in \{0\}\cup \Bbb N$ and with $n_2,n_2'$ both odd members of $\Bbb N$, then

(i). $n_1=n_1'\implies n_2=n_2'.$

(ii). $n_1<n_1'$ implies the odd number $n_2 =2^{(n'_1-n_1)}\cdot n'_2$ is divisible by $2,$ which is absurd.

(iii). $n_1>n_1'$ implies the odd number $n'_2=2^{(n_1-n'_1)}\cdot n_2$ is divisible by $2,$ which is absurd.

Let $n\in E$ iff $n_1$ is even and $n\in O$ iff $n_1$ is odd. Since $n_1$ is uniquely determined by $n,$ we have $E\cap O=\emptyset.$

Now observe that $a^2,b^2 \in E$ and $2b^2\in O.$ So if $a^2/b^2=2$ then $a^2=2b^2\in E\cap O=\emptyset,$ which is absurd.

Another way to see why every rational numbers can be written in lowest term is that there is no infinite descending chain of natural numbers.

First, let $a$, $b \in \mathbb{N}$ with $b$ being nonzero. If $\frac{a}{b}$ has no lowest term, then we can keep removing common factors, i.e. there are $a_1, a_2 \dots$ and $b_1, b_2, \dots$ such that $$\frac{a}{b} = \frac{a_1}{b_1} = \frac{a_2}{b_2} = \dots$$

Now observe that $$\dots < a_2 < a_1 < a$$ and $$\dots < b_2 < b_1 < b$$ since we are always dividing by a natural number greater than $1$.

This is a contradiction, because there is no infinite descending chain of natural numbers. So every rational numbers can be written in lowest term.

Note the above proof only works for nonnegative rational numbers, but it can be easily modified to work for negative rational numbers (by using the fact that there is no infinite ascending chain of negative integers).

You can just start by assuming that $\left(\dfrac AB\right)^2=2$ where $A$ and $B$ are positive integers. (Since $x^2=(-x)^2$, there is no particular advantage to $A$ or $B$ being negative.)

It can be shown that $\gcd(A,B)$ exists. So we can let $g=\gcd(A,B)$. We can also prove that

$\qquad (1.)\quad a = \dfrac Ag$ and $b = \dfrac Bg$ are integers

$\qquad (2.)\quad \gcd(a,b)=1$

$\qquad (1.)\quad \dfrac ab = \dfrac AB$.

So we may as well start by assuming that $\left(\dfrac ab\right)^2=2$ where $a$ and $b$ are positive, relatively prime integers.

You might think of the problem in another way. It just makes the proof finite.

Say you don't consider a & b to be the lowest terms. Follow it through... You will find something along the lines of:
a = 2c
c = 2d
d = 2e
e = 2f
f = 2g
...

This is analogous to numerical methods whereby each iteration increases accuracy but you never get to 100% accurate - wherever you stop you only have an approximation.

This is the nature of irrational numbers compared to rational numbers - they cannot be defined precisely with integers. Resting upon the lowest factors argument is shorthand for saying "this iteration shouldn't occur infinitely for rational numbers".