1.2.1 - Irrationality proofs

This is essetially problem 1.2.1 from the text. It's got two parts.

a) Prove that $\sqrt{3}$ is irrational.
b) Where does the proof that $\sqrt{4}$ is irrational break down?

Part a)

We will start with a contradiction, and assume that $\sqrt{3}$ is rational.

Let $\sqrt{3} = {a / b}$ where $a$ and $b$ $\in \mathbb{Z}$

Therefore $3 = {a^2 / b^2}$

and $a^2 = 3b^2$

If $b$ is odd, then $b^2$ must be odd, and per the above equation $a^2$ and $a$ must also be odd. Now if $b$ and $a$ are even, then our original equation $\sqrt{3} = {a / b}$ is not reduced all the way, so we will assume that $b$ and $a$ are odd, which means that:

$a$ = $2p + 1$

$b$ = $2q + 1$

where $p$ and $q$ $\in \mathbb{Z}$

If we substitute these values into our earlier equation $a^2 = 3b^2$ then we get:

$4p^2 + 4p + 1 = 3(4q^2 + 4q + 1)$

Which can also be written as:

$2(p^2 + p) = 6q^2 + 6q + 1$

The value on the left side of this equation is an even integer, while the value on the right side is an odd integer, therefore this equation has no solution. Therefore there is no solution for our assumption that $\sqrt{3} = {a / b}$

Therefore the contrary to that assumption must be true; that there is no rational solution to $\sqrt{3} = {a / b}$, and $\sqrt{3}$ is irrational.

Who wants to do part b?

As far as part b)
I can see the breakdown begin when we reach $$a^2 = 4b^2$$ because $a^2$ being a multiple of 4 does not immediately imply that $a$ is a multiple of 4.

Take $a = 6$ for example.

What I can't see is where to go from here.

Here is a proof that $\sqrt{p}\not\in\mathbb{Q}$, where $p$ is a prime, which relies on Euclid's Lemma (let $m, n \in\mathbb{Z}$ and $p$ be a prime. If $p|mn$ then $p|m$ or $p|n$). A necessary consequence of this proof is that $\sqrt{3}\not\in\mathbb{Q}$.

Prop: Let $p$ be a prime. Then $\sqrt{p}\not\in\mathbb{Q}$.

Proof: Let $p$ be a prime. By Method of Contradiction suppose that $\sqrt{p}\in\mathbb{Q}$. It follows that $$\sqrt{p}=\frac{a}{b},$$ where $a,b \in\mathbb{Z}$ and $b\neq0$. WLOG assume $gcd(a,b) = 1$. Hence $$p=\frac{a^2}{b^2},$$ and $b^{2}p=a^{2}$. Thus $p|a^2$ and, by Euclid's Lemma, $p|a$. So $pr = a$, where $r\in\mathbb{Z}$. Substituting back into our original equation yields: $$p = \frac{(pr)^2}{b^2}.$$ It follows that $$b^{2}p = p^{2}r^{2}.$$ So $$b^2 = pr^2.$$ Thus $p|b^2$ and, by Euclid's Lemma, $p|b$. This contradicts our assumption that $gcd(a,b) = 1$. Therefore $\sqrt{p}\not\in\mathbb{Q}$. $\blacksquare$

I think that you have it. If you set it up to be a proof by contradiction like we've been using for the prime numbers then you end up with $a^2=4b^2$. Thus since 4 is not prime and has other factors then it does not imply that since b*b is divisible by 4 that b itself is divisible by 4.

It seems to me that a proof of the irrationality of $\sqrt{4}$ falls apart when you consider that four is a composite number - and that it may be written as a product of primes $4 = 2^2$ thus we immediately see that finding the square root of four must yield the rational number 2 since squaring a number and then raising it to the one-half power will always result in the number itself.

I agree with your conclusion but I think it has more to do with the fact that $4$ is a perfect square than it has to do with $4$ being composite. Consider the composite, non-square number $6$, and the contradiction method works.

$$6=\frac{p^{2}}{q^{2}} \implies p^{2} = (2)[(3)q^{2}] \implies \text{ p is even}.$$

Also,

$$6=\frac{p^{2}}{q^{2}} \implies q^{2}=\frac{p^{2}}{6}$$

So we have that $6|p^{2}$ and from this we can conclude that,

$$q^{2}=6 r, r \in \mathbb{Z}$$

It follows that,

$$\frac{q^{2}}{6}=r \implies 6|q^{2} \implies 2|q^{2}$$

Thus, $q$ is even and we have our contradiction. In the case of some perfect square, $z^{2}$, with $z \in \mathbb{Z}.$ We end up with,

$$\frac{p^{2}}{q^{2}}=z^{2} \implies p=zq$$

Which yields no contradiction since $\mathbb{Z} \subset \mathbb{Q}$.
This method is easy to generalize for all even, non-square, composite numbers. So I think that now we need something for odd, non-square, composite numbers.

Hey Spiff - Is this a reply to Josh? If so I think you've got a good point but hit the reply button under his answer, rather than the general Reply button. Make sense??

I have still been thinking about this problem and it still doesn't seem like enough.

Following the proof we did in class, when we get to the point that $p^2 = 4q^2$, can't we still say that $p^2$ is even, thus $p$ is even?

Then the problem would be letting $p = 2r$ and substituting back in would give us $4r^2 = 4q^2$ which implies that $r = q$. This then tells us nothing about q being even or odd and we have no contradiction because $p$ and $q$ could still have no common factors.

Correct me if I am wrong, but the exercise was to show where the proof that the square root of four is irrational breaks down... Meaning you can't prove it is because it is not. Why is every one still trying to prove it is?

Perhaps you should read through them again...