# Falsity of assumptions question.

1. Aug 10, 2011

### tylerc1991

In the course of proving that $\sqrt{3}$ is irrational, I had another question pop up. To prove that $\sqrt{3}$ is irrational, I first assumed 2 things: $\sqrt{3}$ is rational, and the rational form of $\sqrt{3}$ is in it's lowest form. I then broke the proof up into cases and showed that none of these cases could occur.

My question boils down to: did I actually show that $\sqrt{3}$ is irrational?

From a purely logical standpoint, let's say that the 2 assumptions I made were named A and B. I successfully showed that A $\wedge$ B is false. However, this doesn't mean that BOTH A and B are false. More specifically, A could be true and B could be false, and I would still arrive at A $\wedge$ B being false.

On the other hand, the second assumption that was made (the rational form of $\sqrt{3}$ is in it's lowest form) shouldn't (doesn't?) change the problem.

Could someone give me solace and explain this little technicality I have? Thank you very much!

2. Aug 10, 2011

### Hurkyl

Staff Emeritus
If you are merely assuming that, then you do have a problem, and you have merely proved:
If sqrt(3) is rational, then it cannot be expressed as a fraction in lowest terms​
(or something equivalent)

But you don't have to merely assume that a rational number can be written in lowest terms, do you?

3. Aug 10, 2011

### tylerc1991

So this isn't really an assumption, per se? This is more of a 'without loss of generality' statement?

4. Aug 10, 2011

### Hurkyl

Staff Emeritus
Right, although I wouldn't have chosen that phrasing.