# Sloppy Proof in Spivak's Calculus?

1. Jun 11, 2012

### SolsticeFire

So I was reading Spivak's Calculus for fun and I found on the bottom of page 25, 4th edition:
Suppose that $\sqrt{2}$ were rational; that is, suppose there were natural numbers p and q such that $\frac{p^{2}}{q^{2}} = 2$.
We can assume that p and q have no common divisor (since all common divisors could be divided out to begin with). Now we have
p$^{2}$ = 2q$^{2}$.
This shows that p$^{2}$ is even, and consequently p must be even; that is p = 2k for some natural number k. Then
p$^{2}$ = 4k$^{2}$=2q$^{2}$,
so 2k$^{2}$ = q$^{2}$.
This shows that q$^{2}$ is even, and consequently q is even. Thus both p and q are even, contradicting the fact that p and q have no common divisor. This contradiction completes the proof.

Am I missing something or is this proof not complete and sloppy?! Spivak didn't even consider the possibility that q is odd and p is even; in which case, we can write p as 2k and q as 2m+1 and THEN check the equivalency. In this case we find that two sides won't be equivalent when we substitute 2k and 2m+1 in the given equation and that would complete the proof.

2. Jun 11, 2012

### Vargo

The line $2k^2=q^2$ implies that $q^2$ is even which in turn implies that $q$ is even. To recap the whole proof, assume $\sqrt{2}=p/q$, a reduced fraction. $p^2/q^2=2$ implies that p must be even which then implies that q must be even which implies that the fraction is not reduced which is a contradiction.

3. Jun 11, 2012

### AlephZero

He did consider that case. After he divides out any common factors, there are 3 possible cases left
p is odd and q is even
p is even and q is odd
p is odd and q is odd

He then proves that actually both p and q must be even, which means they still have a common factor.

You made a different (and correct) proof of one of the three cases, but Spivak's proof handles all three cases at once by showing they are all impossible

4. Jun 11, 2012

### SolsticeFire

Ahh I see how that completes the proof! The line I was looking at was p$^{2}$=2q$^{2}$. This line suggests that p is even but q can be even or odd. If q is even we have contradiction as we assumed that common divisors were taken care of. So q must be odd and then we can move on to show LHS $\neq$ RHS after substitution of p as 2k and q as 2m+1 in the equation p$^{2}$=2q$^{2}$
But writing p as some number 2k is definitely more efficient.

Thanks a lot!

5. Jun 11, 2012

### SolsticeFire

Yes! I see it now. Thanks a lot :)