# Proof by Contradiction

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• Oct 7th 2009, 11:05 AM
kallivis
Proof by Contradiction
I'm on one my last Math problems of the assignment, and I can't seem to figure out where to start or where to go with it. The problem is:
-Use Proof by contradition to prove the following:
Assume p is rational and q is irrational. Prove if p =/= 0, then p*q is irrational.

Any help would be appreciated, thanks (Happy)
• Oct 7th 2009, 11:08 AM
Matt Westwood
Quote:

Originally Posted by kallivis
I'm on one my last Math problems of the assignment, and I can't seem to figure out where to start or where to go with it. The problem is:
-Use Proof by contradition to prove the following:
Assume p is rational and q is irrational. Prove if p =/= 0, then p*q is irrational.

Any help would be appreciated, thanks (Happy)

The way I'd look at it is:

If pq were rational it could be expressed in the form s/t where s and t are integers.

Now express p in the same form (m/n, say) and multiply pq by 1/p.

Continue from there.
• Oct 7th 2009, 11:29 AM
kallivis
I appreciate the assistance, so I put in what you said and now have
q= s/tq which should mean p*q= s^2/t^2q, but wouldn't that mean since that is a ratio that p*q is a rational nymber would be true, therefore not giving a contradition? Or am I missing something still?
• Oct 7th 2009, 12:12 PM
Matt Westwood
Quote:

Originally Posted by kallivis
I appreciate the assistance, so I put in what you said and now have
q= s/tq which should mean p*q= s^2/t^2q, but wouldn't that mean since that is a ratio that p*q is a rational nymber would be true, therefore not giving a contradition? Or am I missing something still?

I don't think you read my post correctly.

I said: "... express $pq$ in the form $s/t$" not "express $p$ in the form $s/t$".

Look, it goes a bit like this.

You want to show that pq is irrational if p is rational and q irrational. So what we're going to do is say: if pq were rational, then we would show that q also has to be rational, which contradicts our premise that q is irrational, hence demonstrating a contradiction.

So, suppose $pq = s/t, p = m/n$.

Then $q = 1/p (s/t) = (n/m)(s/t) = (ns) / (mt)$.

But $m,n,s,t$ are all integers so $ns$ and $mt$ are integers ...

... which means ...?
• Oct 7th 2009, 01:08 PM
HallsofIvy
Have you already proved that the rational numbers are closed under multiplication? If so, you can use a proof by contradiction. Suppose, with p rational, not equal to 0 and q irrational, pq = r is a rational number, then p= r(1/q) is a product of rational numbers.