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I read the proof of the irrationality of $\sqrt{3}$ in my textbook (Richard Hammack's Book of Proof), and I was wondering if my proof is legitimate as well.

Prop: $\sqrt{3}$ is irrational.

Suppose by way of contradiction that $\sqrt{3}$ is rational. Hence, $\sqrt{3} = \frac{m}{n}, m \in \mathbb{Z}, n \in \mathbb{N}$. Furthermore, suppose both $m,n$ are not even, so the fraction is reduced. Then, $3n^2=m^2$. Suppose $n$ is even, so $n = 2a, a \in \mathbb{Z}$. Therefore, $3 \cdot 4a^2=m^2 \longrightarrow m^2=12a^2 \longrightarrow m^2=2(6a^2)$ and $m^2$ is even, hence $m$ is even. This contradicts our assumption that both $m,n$ were not even, and hence $\sqrt{3}$ must be irrational. $\blacksquare$

Also, I'm wondering if I also need to show the case where $m$ is even?

Edit: Thank you for all the help. I realize that I was essentially trying to make the same argument as the classic proof of the irrationality of $\sqrt{2}$, but it doesn't quite work the same. I've done some research about the prime factorization theorem and I agree that $3n^2=m^2$ being a contradiction is definitely a more elegant proof that a case-by-case approach.

Julian L
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  • This looks good. But what if both m and n are odd? – Adam Howard Dec 09 '19 at 19:49
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    See this question. – Dietrich Burde Dec 09 '19 at 19:50
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    Even means "divisible by two". This is useful when you have the equation $m^2=2n^2$; less so for $m^2=3n^2$. – Angina Seng Dec 09 '19 at 19:51
  • @AdamHoward Good point. So the way I've written this proof I would have to have cases in which $m$ is even, $n$ is even, or neither $m,n$ is even, which definitely removes the succinctness I was trying to accomplish. – Julian L Dec 09 '19 at 19:59
  • @JulianL You need to remove the mention of “even” altogether. – egreg Dec 09 '19 at 20:11
  • You've proven $\sqrt{3}$ doesn't have an even (reduced) numerator. That's a far cry from proving is irrational. After all, your proof will work equally well for $\sqrt{25}$ or $\sqrt {\frac {49}{121}}$. Your comment that not both even means reduced, and your assumption that reduced functions will have an even numerator is very strange. – fleablood Dec 09 '19 at 21:49
  • I hate to be heavy handed but even and odd have no relevance. You will end up proving that neither $m$ nor $n$ are even which doesn't mean anything. There are LOTS of reduced rationals that do no have even numeratiors or denominators. Proving a numerator/denominator is not anything won't get us anywhere. There is always some rational that is NOT what we want. What DOES matter is if the numerator/denominator is divisible by $3$. In that case you will see that the numerator IS be divisible by $3$ and then denominator IS too, and that is the contradiction. – fleablood Dec 09 '19 at 21:57

3 Answers3

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There are at least two flaws:

  • you say $m,n$ are both not even, which in fact means neither $m$ nor $n$ are even.

  • you say "so the fraction is reduced", but is $\frac{15}{25}$ reduced ?

Allowing only irreducible fractions, $$\sqrt3=\frac pq\iff p^2=3q^2.$$

So $p^2$ is a multiple of $3$, and so must $p$ be. Then $p^2$ is a multiple of $9$ and $q^2$ is a multiple of $3$. And so must $q$ be !

  • Thank you. Maybe I shall change it to "such that $m,n$ do not have the same parity and the fraction is in reduced form". If I did that and considered the case where where $m$ is even would the proof be complete? – Julian L Dec 09 '19 at 20:01
  • @JulianL: "do not have the same parity" is wrong: odd/odd is possible. Stick to "in reduced form" and forget the parity. –  Dec 09 '19 at 20:02
  • Ok. That explains why the original proof is much longer than I thought it needed to be, because it considers all cases. – Julian L Dec 09 '19 at 20:04
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    @JulianL: parity is irrelevant to the question. If you wanted you could also handle the mutliples of $42$ separately, but this is extraneous. –  Dec 09 '19 at 20:05
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It's incomplete. You showed that $n$ cannot be even. But $n$ can still be odd and then your proof does not say that there is a contradiction (which there should be).

Here is a shorter proof:

Assume $3= m^2/n^2$ where $m,n \in \mathbb{Z}$, $n \neq 0$. Then

$$3n^2 = m^2$$

Comparing the unique prime factorizations of both sides we find that the prime $3$ occurs to odd power on the left side, but even power on the right side, contradicting the uniqueness of prime factorizations.

This same proof generalises to $\sqrt{p}$ is irrational for every prime $p$.

Bill Dubuque
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J. De Ro
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  • Your claim that "the LHS always has one more prime factor than the RHS" requires proof and it is not clear what proof you intend. Clearer: use parity: prime 3 occurs to odd power on LHS, but even power on RHS, contra FTA (existence an uniqueness of prime factorizations). – Bill Dubuque Dec 09 '19 at 20:14
  • You are right. I meant that, but it came out in a wrong way. – J. De Ro Dec 09 '19 at 20:15
  • I edited accordingly. Thanks. – J. De Ro Dec 09 '19 at 20:16
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    I edited to make it clearer that existence and uniqueness is used in many places. – Bill Dubuque Dec 09 '19 at 20:21
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The $nth$ root of a number is rational precisely when $N$ is the $nth$ power of the rational number $\frac{a}{b}$.

$$\sqrt[n] N=\frac{a}{b} \iff N=\left (\frac{a}{b} \right )^n$$

Suppose that $\sqrt[n] N$ is not. There is a positive number $r$ subtracting $N$ so that their root is rational. We just need to define what $r$ is.

$$\sqrt[n] {N-r}= \frac{a}{b} \iff N=\left (\frac{a}{b} \right )^n+r, r>0$$

If $(\frac{a}{b})^n$ is the first term of a binomial expansion, then because we're adding something to it, $r$ is defined by its following terms, otherwise known for this summation starting at $1$.

$$r=\sum_{k=1}^n {n \choose k} \left (\frac{a}{b} \right )^{n-k} \left (\frac{c}{d} \right )^k$$

And when you replace and expand this in the expression, you end up with the following binomial, whose purpose is to show that N is a $nth$ power.

$$N=\left (\frac{a}{b}+\frac{c}{d} \right )^n$$

However, does it have a rational base, when it's expressed by the sum of two numbers? The only way to find out is to make it a linear combination.

$$N=\left (\frac{a}{b}+\frac{c}{d} \right )^n \iff \sqrt[n] N=\frac{a}{b}+\frac{c}{d} \\ \frac{a}{b}=\lambda\sqrt[n] N\land\frac{c}{d}=\left (1-\lambda\right)\sqrt[n] N\implies1\sqrt[n] N=\lambda\sqrt[n] N+\left(1-\lambda\right)\sqrt[n] N$$

Therefore, using our assumption that $\sqrt[n] N$ is irrational, $\frac{a}{b}$ and $\frac{c}{d}$ must be too for some rational $\lambda$. Should the former be rational, the latter two are as well. The number $r$ that we defined is not necessarily irrational due to its products in the expansion.

Let's use the proposition in the opening question: $\sqrt 3$ is irrational.

$$\sqrt {3-r}=\frac{a}{b} \iff 3=\left (\frac{a}{b}\right )^2+r,r>0$$

We consider $r$ to be the binomial expansion using $n=2$.

$$r=2\left (\frac{a}{b}\right )\left (\frac{c}{d}\right )+\left (\frac{c}{d}\right )^2 \implies 3=\left (\frac{a}{b}+\frac{c}{d} \right )^2$$

Then find two numbers for the linear combination to hold.

$$\frac{a}{b}=\frac{\sqrt 3}{2}\land\frac{c}{d}=\frac{\sqrt 3}{2}\implies\sqrt 3=\frac{a}{b}+\frac{c}{d}$$

Therefore, since our assumption was that $\sqrt 3$ was irrational, those two numbers are as well. The number $r$ is rational:

$$r=2\left (\frac{\sqrt 3}{2}\right)\left (\frac{\sqrt 3}{2}\right )+\left (\frac{\sqrt 3}{2}\right )^2=\frac{9}{4}$$

Should this exercise be the square root of $4$, and because we can find two integers whose square is equal to it, the expression would unfold differently and you would find that the numbers in that linear combination are equal to $1$. It means that you can find a lower square by following that process even after the assumption that it has integer roots.

Marco_O
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