http://imgur.com/fuoLUOT
In proofs of contradiction isn't the whole point to have >P be false and have a false statement Q, which makes the proof true? Why in the proof, do we assume >P is true? kind of confuses me..
http://imgur.com/fuoLUOT
In proofs of contradiction isn't the whole point to have >P be false and have a false statement Q, which makes the proof true? Why in the proof, do we assume >P is true? kind of confuses me..
I must have wrote this down wrong..or perhaps he said something about irrationality that I forgot. He also left us to do another problem to do at home.
exercise: Prove by method of contradiction that square root (3) is not a rational number.
basically what's going on is you determine a polynomial who's roots are $\sqrt{2}$ or $\sqrt{3}$ or whatever.
Then you assume these are rational roots (to be contradicted), and use the rational root theorem on the polynomial to determine what the possible rational roots are.
You then show that none of these possible rational roots are in fact roots and thus it must be that the roots are irrational.
What you posted is the classical proof for the $\sqrt2$ being irrational.
Assume $\sqrt2$ is rational. That means there are two relatively prime integers $p~\&~q$ such that:
$\sqrt2=\dfrac{p}{q}$. So $2=\dfrac{p^2}{q^2}$ squaring both sides.
But that leads us to $2q^2=p^2$ which implies $p=2k$ i.e. $p$ is even.
So we get $2q^2=4k^2$ or $q$ is also even. How does that contradict that $p~\&~q$ are relatively prime?