Thread: proof by contradiction

If x is an element of the integers and x^2 is even, then x is even.

assume x is odd when x satisfies the hypothesis. When you square an odd you get an odd because odd times odd is odd. so x^2 is odd and x^2 is even which is falser because a number cannot be even and odd

im unsure if this is right

sorry i forgot i had one more question.

if a,b are elements of the integers and ab is odd, then a and b are both odd.

assume for sake of contradiction a and b are not both odd when the hypothesis is satisfied. so a could be even and b odd. an even number times an odd is even so ab is even. ab cannot be even and odd so there cannot be a counterexmple

Originally Posted by leinadwerdna

sorry i forgot i had one more question.

if a,b are elements of the integers and ab is odd, then a and b are both odd.

assume for sake of contradiction a and b are not both odd when the hypothesis is satisfied. so a could be even and b odd. an even number times an odd is even so ab is even. ab cannot be even and odd so there cannot be a counterexmple

Claim 1. Let be odd, then is odd.
Proof of Claim 1. Assume the contrary that are not both odd but is odd. Then there are two possible cases as follows.
(i) is odd and is even. Then for some , we may write and , which yields (even). Hence this case is not possible.
(ii) and are both even. Then for some , we may write and , which yields (even). Hence this case is not possible either.
Therefore, if are odd, then is odd.

Originally Posted by leinadwerdna

If x is an element of the integers and x^2 is even, then x is even.

assume x is odd when x satisfies the hypothesis. When you square an odd you get an odd because odd times odd is odd. so x^2 is odd and x^2 is even which is falser because a number cannot be even and odd

im unsure if this is right

Claim 2. If is even and is an integer, then is even.
Proof of Claim 2. Again assume the contrary that is even but is odd. Then, we have from Claim 1 that is odd.
This contradiction proves Claim 2.

Probably you are requested to prove Claim 2 by using Claim 1.