# Math Help - 2 proofs that I can't do

1. ## 2 proofs that I can't do

Prove using contradiction that( n^2 + n + 2) is always even, where n is an integer

Prove that if 7n^2 is odd then n is odd, given that n is an integer. Use both contradiction and contraposition in your proof.

Can somebody please help me? My book doesn't give any similar examples and I am not making progress on these. Thanks

2. Suppose $n^2+n+2$ is not allways even (contraposition): Then ther exists a number $n_0$ such that $n{_0}^2+n_0+2$ is odd.

Now fill in n= 2k+1 (odd number) and n = 2k (even number) to show this can not be the case.
We will see that the outcome is allways even. That's the contradiction.

Do the other one likewise. Assume that the desired result is false (contraposition) and end with a contradiction.

3. Originally Posted by Dinkydoe
Suppose $n^2+n+2$ is not allways even (contraposition): Then ther exists a number $n_0$ such that $n{_0}^2+n_0+2$ is odd.

Now fill in n= 2k+1 (odd number) and n = 2k (even number) to show this can not be the case.
We will see that the outcome is allways even. That's the contradiction.

Do the other one likewise. Assume that the desired result is false (contraposition) and end with a contradiction.
You mixing up contradiction with contraposition.

For contradiction you have to assume that the statement : for all , n : if nεN ,then $n^2+n+2$ is even :

is not true and end up with a contradiction ,while

For contraposition you have to prove that:

For all ,n : if $n^2+n+2$ is not even ,then n does not belong to Natural Nos

4. Originally Posted by alexandros
You mixing up contradiction with contraposition.

For contradiction you have to assume that the statement : for all , n : if nεN ,then $n^2+n+2$ is even :

is not true and end up with a contradiction ,while

For contraposition you have to prove that:

For all ,n : if $n^2+n+2$ is not even ,then n does not belong to Natural Nos
No for proof by contradiction you assume that there is some $\dispaystyle n \in \mathbb{Z}$ such that $n^2+n+2$ is not even and from this derive a contradiction (which in this case is that $\dispaystyle n^2+n+2$ is even)

CB

5. Originally Posted by CaptainBlack
No for proof by contradiction you assume that there is some $\dispaystyle n \in \mathbb{Z}$ such that $n^2+n+2$ is not even and from this derive a contradiction (which in this case is that $\dispaystyle n$ is odd)

CB
We are talking the same thing Captain>

If : for all ,n [ $n\in N\Longrightarrow n^2+n+2 is even$] is not true ,as i pointed out in my post:

Then : There exists an n,such that :nεN and n^2+n+2 is not even.

Now since nεN ,n=2k+1 or n=2k ,where kεN

And examining each case separately we end up with n^2+n+2 even

Hence a contradiction, since we assumed n^2+n+2 not even

6. Originally Posted by babbagandu
Prove using contradiction that( n^2 + n + 2) is always even, where n is an integer

Prove that if 7n^2 is odd then n is odd, given that n is an integer. Use both contradiction and contraposition in your proof.

Can somebody please help me? My book doesn't give any similar examples and I am not making progress on these. Thanks
I'll help with the second part:

Premise $7n^2$ is odd implies n is odd.
Suppose n is even and $7n^2$ is odd (contraposition)
Then $7n^2$ can be written as $7(2k)^2$ where 2k=n.
This gives $7(4k^2)$ which can be rewritten as $4(7k^2)$ with the commutative property.
Now by factorization this gives $2(2(7k^2))$. Let m= $2(7k^2)$. Now you have $2(2(7k^2))$=2m which by definition is even. This is a contradiction so therefore, if $7n^2$ is odd, then n is odd. QED

7. Originally Posted by babbagandu
Prove using contradiction that( n^2 + n + 2) is always even, where n is an integer
$n^2+n+2=n(n+1)+2$

Now $n(n+1)$ is always even since one of $n$ or $n+1$ will always be even.

8. Originally Posted by oldguynewstudent
I'll help with the second part:

Premise $7n^2$ is odd implies n is odd.
Suppose n is even and $7n^2$ is odd (contraposition)
Then $7n^2$ can be written as $7(2k)^2$ where 2k=n.
This gives $7(4k^2)$ which can be rewritten as $4(7k^2)$ with the commutative property.
Now by factorization this gives $2(2(7k^2))$. Let m= $2(7k^2)$. Now you have $2(2(7k^2))$=2m which by definition is even. This is a contradiction so therefore, if $7n^2$ is odd, then n is odd. QED
You also mixing up contradiction with contrapositive proof:

$7n^2$ odd, as a given premise and we must assume n as even .

This will lead us to the contradiction: $7n^2$ being odd and even .

For contrapositive proof we have to prove:

n even implies $7n^2$ even

n even $\Longrightarrow 7n^2$ even

9. "proof by contradiction" includes "proving the contrapositive".

In proving contrapostive, to prove "if A then B" we show that "not B" implies "not A".

In proof by contradiction, we assume both "A" and "not B" and show that that leads to a contradiction. of course, if we could start from "not B" and show "not A" (contrapositive) we would have a contradiction- both "A" and "not A".

They are not exactly the same because it is possible to start from "A" and "not B" and show that we can, from them, arrive at two completely different statements, "C" and "C*", which contradict one another.

10. Originally Posted by HallsofIvy
"proof by contradiction" includes "proving the contrapositive".

I think,this is wrong.

We call contrapositive the whole law ,[ $(A\Longrightarrow B)\Longleftrightarrow(\neg B\Longrightarrow\neg A)$ ] and not ,~A or ~B ,which are the negations of A or B

, if we could start from "not B" and show "not A" (contrapositive) we would have a contradiction- both "A" and "not A".
Wrong ,there is no contradiction because we have not assumed A in the 1st place.