In the proof of irrationality of $\sqrt{2}$ or $\sqrt{7}$, why do the numerator and denominator of a rational number have to be in their lowest term?

Question

I have been confused by this problem for a very long period of time, and I think I am personally opposed to this concept and refused to agree with it in my introduction to mathematical proof course.

It's pretty much the same thing, the problem arises in the proof that shows square root of 2 or square root of 7 is an irrational number.

I agree with the beginning that assumes these two numbers are rational numbers for the sake of applying contradiction, but as the solution in example proceeds, I notice that they all assume that the numerator and denominator must be in their lowest term, which I think they are relatively prime?

I mean seriously, why do they have to be in the lowest term? I mean there are lots of rational numbers that don't satisfy this condition, like 4/2, 10/2 right?

Can someone explain to me why this in lowest term must be satisfied in order to show a contradiction?

Appreciated.

Answer 1

What they all assume (correctly) is that every rational number $q>0$ can be written (in one and only one way, by the way) as $q=\frac mn$, where $m$ and $n$ are co-prime natural numbers.

Concerning your examples, we have:

• $\dfrac42=\dfrac21$ and $2$ and $1$ are co-prime;
• $\dfrac{10}2=\dfrac51$ and $5$ and $1$ are co-prime.

right?!

Answer 2

The point is that all rational numbers can be expressed in lowest terms. They don't have to be. Yes, $\frac 24$ is a perfectly good rational number, but the same number can be expressed as $\frac 12$. When we prove $\sqrt 2$ is irrational, we assume it is rational, then say to express that rational in lowest terms as $\frac ab$. The reason to do that is that we then show both $a$ and $b$ are even, so $\frac ab$ is not in lowest terms. If we had not assumed it was in lowest terms we would not have the contradiction we are seeking.

Answer 3

It arises when we try to prove an irrational rational that $a$ and $b$ are not relatively prime. Although whether a number is irrational does not depend on how much the two numbers are scaled, in order for the statement to be a contradiction, we must assume $\frac{a}{b}$ is in lowest terms, meaning $a$ and $b$ must be relatively prime.

Answer 4

The question has been answered, but I think it's worth pointing out that the proof can be slightly modified to avoid supposing the fraction is reduced to lowest terms. This modified version may give some additional insight into why that supposition is made in the original proof.

So let $\sqrt{2}=a_0/b_0$, where $a_0$ and $b_0$ are integers, not necessarily in lowest terms. This is equivalent to $2b_0^2=a_0^2$, which implies $a_0$ is even. Write $a_0=2a_1$, so that $2a_1^2=b_0^2$, implying $b_0$ is even. Write $b_0=2b_1$, so that $2b_1^2=a_1^2$. Notice that starting from $\sqrt{2}=a_0/b_0$ we have obtained a new way of expressing $\sqrt{2}$, namely as $a_1/b_1$, where $a_1$ and $b_1$ are integers half the size of the original $a_0$ and $b_0$.

Nothing prevents the argument from being repeated as many times as we like. So we get $\sqrt{2}=a_2/b_2$, where $a_2$ and $b_2$ are integers half the size of $a_1$ and $b_1$, or a quarter the size of $a_0$ and $b_0$. Continuing 98 more times, we come to $\sqrt{2}=a_{100}/b_{100}$ where $a_{100}$ and $b_{100}$ are integers a factor of $2^{100}$ smaller than $a_0$ and $b_0$. And so on.

In other words, if they exist, the integers $a_0$ and $b_0$ have the remarkable property that we can keep pulling out factors of $2$ indefinitely. We now see that there can be no such integers. (More formally, $n\mapsto2^n$ is an increasing integer-valued function of $n$, so there is an $n$ such that $2^n>b_0$, and we therefore can't divide $b_0$ by $2_n$ without leaving a remainder.)

This "infinite descent" proof is a close relative of your proof, but the contradiction is obtained by a slightly different strategy.

Answer 5

In the standard proof of the irrationality of $\sqrt{2}$, we assume for the sake of contradiction that $\sqrt{2} = \frac{a}{b}$. Then, we say something to the effect that "without loss of generality we may assume that $\frac{a}{b}$" is in lowest terms, for if it is not, we always can put it in lowest terms.

Now, that "without loss of generality" is necessary because that is how we end up obtaining our contraction. Without the assumption of the fraction being in lowest terms, no contradiction is obtained:

$$\sqrt{2}=\frac{a}{b} \longrightarrow 2 = \frac{a^{2}}{b^{2}} \longrightarrow 2b^{2} = a^{2} \longrightarrow a = 2k \rightarrow a^{2} = 4k^{2} $$

And so,

$$ 2b^{2} = 4k^{2} \longrightarrow b^{2} = 2k^{2} \longrightarrow b = 2s. $$

Therefore, our fraction $\frac{a}{b}$, which is presumed to be in lowest terms becomes

$$ \frac{a}{b} = \frac{2k}{2s},$$

which is not in lowest terms, and we have arrived at the desired contradiction---which we could not have had we not required (without loss of generality) that $\frac{a}{b}$ be in lowest terms.

Answer 6

In fact, they don't have to. Let $a$ and $b$ be any positive numbers with $(a/b)^2=2$. In that case it's easy to show that also $$\left(\frac{2b-a}{a-b}\right)^2=2$$ and $0<2b-a<a$ and $0<a-b<b$. Now that leads immediately to a contradiction if $a$ and $b$ were integers.