1
$\begingroup$

I'm trying to understand each part of this completed proof that my professor did, here is my interpretation in parentheses, please advise as necessary.

Proof: Assume that $2^{1/2}$ is rational, then $2^{1/2}$ = $p/q$ for some integer $p$ and $q$ with $q \neq 0$ and $\gcd(p,q) = 1$

$$(p/q)^2 = 2 = p^2 / q^2 $$ $$\text{ (What was the purpose of squaring √2 and p/q?)}$$

$$p^2 = 2q^2$$ $$ \text{ (Is this basically just canceling out the q^2 on left side, and multiplying the right side?)}$$

2 | p^2 => (2 divides p^2 since it is a factor)

2 | p => (thus 2 divides p)

p = 2k for some integer k => (definition of even)

(2k)^2 = 4k^2 = 2q^2 => (Why are we squaring 2k, and how does 4k^2 = 2q^2?)

q^2 = 2k^2 => (Canceled out 2 on one side, and dividing on other)

2 | q^2 => (2 divides q^2)

2 | q => (thus 2 divides q)

2 | gcd(p,q) => (2 divides gcd of p and q)

2 | 1 Which is a Contradiction (no clue how I got here)

$\endgroup$
3
  • $\begingroup$ Please try to use LaTeX formatting. I've done the first half, can you do the rest? $\endgroup$ Sep 24, 2014 at 23:04
  • $\begingroup$ I didn't know I could do that, give me a minute $\endgroup$ Sep 24, 2014 at 23:12
  • $\begingroup$ You squared both sides because talking about divisibility is easiest when you have all whole numbers. You're squaring $2k$ because you just substituted it for $p$ which was being squared. You got the contradiction because it was assumed at the beginning that gcd$(p,q) = 1$ $\endgroup$
    – genisage
    Sep 24, 2014 at 23:13

3 Answers 3

1
$\begingroup$

I don't know what part you got confused with. I will repeat your steps here with a changed notation (hope it helps). Now, assume the $\sqrt 2 = \frac{p}{q}$ where $p$ and $q$ are integers which are prime with respect to each other. Note that we could show any rational with an infinite number of $p$ and $q$s but enforcing the relative primeness gives us a unique representation for each rational. Anyway, since we don't understand $\sqrt 2$, we will square both sides to reach a more familiar number, that is $2$. We get $${p^2} = 2{q^2}$$Now notice that since $p$ is an integer, it should be odd or even. But the multiplication of two odd numbers gives another odd number which is not in correspondence with the above equation. Hence, $p$ is surely an even integer. That is there exists an integer $k$ such that $p = 2k$. Replacing this result into the above equation gives ${q^2} = 2{k^2}$ which by the same reasoning shows that there exists another integer $l$ such that $q = 2l$. Now, these results show that both $p$ and $q$ are divisible by $2$ which is in contradiction to our initial relative primeness assumption. Hence $\sqrt 2$ does not have a $\frac{p}{q}$ representation in which $p$ and $q$ are relatively prime, in other words, it is not a rational number.

$\endgroup$
2
  • $\begingroup$ Okay, so since we proved both p/q have a gcd of 2 it says they are not rational. So do rational numbers always have a gcd of 1? $\endgroup$ Sep 25, 2014 at 0:00
  • 1
    $\begingroup$ Note that we assumed that $\sqrt{2}$ is in simplest form $p/q$. What we did is contradict it. So basically, the proof says that $\sqrt2$ cannot be in simplest form. $\endgroup$ Sep 25, 2014 at 1:24
1
$\begingroup$

Let me run through it again, but I'll do 2 things. I'll answer your specific questions in the early part of the proof, and I'll take a slightly different tact for the latter part to see if it makes more sense to you. Bear with me if I go into details on stuff you already understand.

Proof: Assume that $\sqrt{2}$ is rational, then $\sqrt{2}= {p\over q}$ for some integer $p$ and $q$ with $q≠0$ and gcd($p$,$q$)=$1$.

(2 things to note here: First, the only thing we assumed is that $\sqrt{2}$ is rational. Everything else above follows directly from that assumption and the definition of rational.

Second, gcd = $1$ means no integer greater than 1 divides both $p$ and $q$. More specifically, $2$ cannot divide them both, so they cannot both be even.)

$(p/q)^ 2 =2={{p^ 2}\over {q^2}}$

(What was the purpose of squaring √2 and p/q?) To get rid of the square root.

$p^2=2q^2$

(Is this basically just canceling out the q^2 on left side, and multiplying the right side?) Yes, to get rid of the fractions.

(Now, here's where I veer off.)

By that last equation, $p^2$ is even (because it's $2$ times something). If $p$ itself were an odd number, then $p^2$ would be odd, so we know now that p is Even.

Now, let's divide both sides by $2$.

${{p^2}\over 2} =q^2$, rearranging a bit, we get

${p\over 2}\times p =q^2$

Now, $p$ is even, so ${p\over 2}$ is an integer. And since $p$ is even, that integer times $p$ is even.

So ${p\over 2}\times p$ is even. Which means $q^2$ is even. But by the same logic we used above, if $q$ itself were an odd number, then $q^2$ would be odd, so we know now that q is Even.

But we said at the beginning that this cannot be true, so we have a contradiction. And since we do, it means one of our assumptions is false. But we only made one assumption, so $\sqrt 2$ must not be rational.

(Final note: all that $2k$ stuff in the original proof was one way of demonstrating that there are two $2$'s in the $p^2$, and so when you divide both sides by $2$, what remains on the left is still even. However, sometimes adding new variables throws people, so I went a different way. Hope it helped.)

$\endgroup$
3
  • $\begingroup$ Thanks for the detailed response. I agree, when I first saw p^2 = 2Q^2 I was confused on why we had to dig further when the definition of even is already shown, for P at least. $\endgroup$ Sep 25, 2014 at 16:05
  • $\begingroup$ If we had made more than 1 assumption, would I have to contradict it as well in order to invalidate the hypothesis? $\endgroup$ Sep 25, 2014 at 16:12
  • $\begingroup$ When you have more than one assumption and you reach a contradiction, you need a way to determine which assumption is incorrect, since failure of any of them (or all of them!) could have led to the contradiction. That gets tricky, so we try to avoid it in general. The only good way I know of to disprove an assumption when you have multiple assumptions snarled together is to show that one of the assumptions causes a contradiction no matter how you choose the others. $\endgroup$
    – Zimul8r
    Sep 25, 2014 at 22:42
1
$\begingroup$

I have always felt that these proofs are way to clumsy and inefficient. Take any fraction $\dfrac ab$ in it's simplest form. Then consider $$ \frac{a^2}{b^2} $$ For any prime $p$ that divides $a^2$ is must divide $a$, thus it does not divide $b$ nor $b^2$. Similarly, if $q$ is any prime that divides $b^2$ it must divide $b$ and not $a$, nor $a^2$. So the fraction $\dfrac{a^2}{b^2}$ must be in it's simplest form too.

So we have two cases:

  1. If $b=1$ the original fraction was an integer, and the square is an integer too
  2. If $b\neq 1$ the original fraction was not an integer, and the square is also not an integer

Since 2. proves that non-integer rationals have non-integer squares we see that integers that are not perfect squares cannot have rational square roots. In particular $2$ is not a perfect square and so $\sqrt 2$ must be irrational.


In a very similar way one can argue that $$ \frac{a}{b}\notin\mathbb Z\implies\frac{a^k}{b^k}\notin\mathbb Z $$ To show that any integer that is NOT a perfect $k$-th power of an integer has an irrational $k$-th root.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .