Prove that $2^{1/2}$ is irrational

Question

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)

Answer 1

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.

Answer 2

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.)

Answer 3

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.

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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.