# Math Help - Proof By Contradiction

I have so many questions about what is really going on for proofs by contradiction.

first, what is the original questions? can that be determined from viewing this proof?

second, what is the goal of a contradiction? to make a complete reverse of the original question?

third, from reading the proof, it seems its said that that there are integers that are even for both n, then its said that n is not even, then it says because n is not even, then both n^2 and n are both not even. Is this correct?

is that what were trying to proof, that there both not even? and then must i supply work to support this.

heres the proof..

proof: We will argue by contradiction.

Assume it is not the case that for every integer n, if n^2 is even then n is even

Therefore, There is an integer n s.t. n^2 is even and n is not even.

Since n is not even, n is odd by n previous Theorem. By the given result, n^2 is odd. By previous theorem
n^2 is not even --- contradiction.

--------- work goes here??? ------------

By the Method of arguing by contradiction,

For all n that exists in Z, if n^2 is even Then n is even, <end of proof>

2. Originally Posted by rcmango
I have so many questions about what is really going on for proofs by contradiction.

first, what is the original questions? can that be determined from viewing this proof?

second, what is the goal of a contradiction? to make a complete reverse of the original question?

third, from reading the proof, it seems its said that that there are integers that are even for both n, then its said that n is not even, then it says because n is not even, then both n^2 and n are both not even. Is this correct?

is that what were trying to proof, that there both not even? and then must i supply work to support this.

heres the proof..

proof: We will argue by contradiction.

Assume it is not the case that for every integer n, if n^2 is even then n is even

Therefore, There is an integer n s.t. n^2 is even and n is not even.

Since n is not even, n is odd by n previous Theorem. By the given result, n^2 is odd. By previous theorem
n^2 is not even --- contradiction.

--------- work goes here??? ------------

By the Method of arguing by contradiction,

For all n that exists in Z, if n^2 is even Then n is even, <end of proof>

first,
as i read it, the question was (i think) is that to show that "if n^2 is even, then n is even".

second, no..
the goal of a contradiction is to show the statement "If P is true, then Q is true" to be true by assuming "P to be true but Q is false"; and since you can work with "Q" and as you work with "Q" and find contradiction/s to some of your arguments and usually, you found out that "P is also "false" for a false Q", you can conclude that Q must be true hence you were able to show "If P is true, then Q is true".

third, what you are trying to prove is the one i wrote in first.
using proof by contradiction, you assume that n^2 is even but n is odd.. using your previous theorems, you were able to show that if n is odd, then n^2 is also odd, which is a contradiction now to your assumption that n^2 is even.. so you can conclude now the n must be even and not odd..

3. yes, after reading this through careful several times, i can now understand what the goal is. Maybe someone could supply an example worked out. Or a solution to this problem. thanks alot.

4. Lets assume that $n^2$ is even if $n$ is odd.

Therefore $n=2k+1$ for where k is some integer.

So $n^2 = (2k+1)^2$

$n^2 = 4k^2 +4k +1$

$n^2 = 2(2k^2 + 2k) +1$

Since $n^2$ is even, then $2(2k^2 + 2k) +1$ is even.

But $2(2k^2 + 2k) +1$ is odd since $2k^2 + 2k$ is an integer. Therefore we have a contradiction. So we conclude that our assumption is false, and $n^2$ is even if n is even.