# [SOLVED] Please check proof

• Mar 22nd 2009, 01:15 PM
meg0529
[SOLVED] Please check proof
So I need to prove that$\displaystyle n^2 -1$ is divisible by 8 iff n is odd
I wrote n= 2P+1 where P = 2k

then $\displaystyle (2P+1)^2 - 1 = 4P^2+4P + 1-1$ subing 2k back in I get

16k^2+8k = 8(2k^2+k)

Is this okay?
• Mar 22nd 2009, 02:27 PM
Reckoner
Quote:

Originally Posted by meg0529
So I need to prove that$\displaystyle n^2 -1$ is divisible by 8 iff n is odd
I wrote n= 2P+1 where P = 2k

then $\displaystyle (2P+1)^2 - 1 = 4P^2+4P + 1-1$ subing 2k back in I get

16k^2+8k = 8(2k^2+k)

Is this okay?

No. By setting $\displaystyle n=4k+1,$ you have only proven the statement for every other odd number: $\displaystyle \dots,\,-7,\,-3,\,1,\,5,\,9,\,13,\dots$.

Fortunately, you can fix it by handling the case for odd $\displaystyle P$ separately. Then you get $\displaystyle n=4k+3,$ and

$\displaystyle n^2-1=16k^2+24k+9-1=8(2k^2+3k+1).$
• Mar 22nd 2009, 02:40 PM
Soroban
Hello, meg0529!

Quote:

I need to prove that $\displaystyle X \:=\:n^2 -1$ is divisible by 8 iff $\displaystyle n$ is odd
I wrote $\displaystyle n\:=\: 2p+1$, where $\displaystyle p = 2k$ . . . . no

$\displaystyle p$ does not have to be even.
$\displaystyle p$ can be any integer, and $\displaystyle n$ will be odd.

So we have: .$\displaystyle X \:=\:n^2-1 \:=\:(2p+1)^2-1 \:=\:4p^2 + 4p \:=\:4p(p+1)$

We see that $\displaystyle X$ is divisible by 4.

$\displaystyle \text{Note that: }\:X \:=\:4\cdot\!\!\!\!\!\!\!\!\underbrace{p(p+1)}_{\t ext{consecutive integers}}$

Since $\displaystyle p(p+1)$ is the product of two consecutive integers,
. . one of them is even, the other is odd.
Hence: .$\displaystyle p(p+1)$ is divisible by 2.

Therefore, $\displaystyle X$ is divisible by 8.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

It said if and only if, so we must prove it "the other way."

. . If $\displaystyle n^2-1$ is divisible by 8, then $\displaystyle n$ is odd.

We have: .$\displaystyle n^2-1$ is divisible by 8.

. . Then: .$\displaystyle n^2-1 \:=\:8k$, for some integer $\displaystyle k.$

And we have: .$\displaystyle n^2 \:=\:8k + 1$

We see that $\displaystyle 8k$ is even . . . hence: $\displaystyle 8k+1$ is odd.

. . So we have: .$\displaystyle n^2 \:=\:\text{odd}$

It can be shown that:
. . If the square of an integer is odd, the integer is odd.

The proof is quite simple, but omitted here.

Therefore: .$\displaystyle n$ is odd.

• Mar 22nd 2009, 02:45 PM
meg0529
Reckoner,

Ok I see your point, So you are saying I need to do this is two different steps? Is there a way for me to do this in one shot?

Thanks
• Mar 22nd 2009, 02:48 PM
Reckoner
Quote:

Originally Posted by meg0529
Reckoner,

Ok I see your point, So you are saying I need to do this is two different steps? Is there a way for me to do this in one shot?

Certainly there is, and Soroban has provided it.

Take care.

Edit: Oh, and be sure you understand what is meant by "if and only if" (abbreviated iff). I misread your post initially and thought it only went one direction; there are basically two proofs that need to be made. Again, Soroban has this covered.
• Mar 22nd 2009, 03:44 PM
meg0529
Got it. Thanks you both very much
• Mar 22nd 2009, 04:22 PM
Reckoner
I see you have edited your post, but I will post my response anyway since it is already written.

Quote:

Originally Posted by meg0529
So If my sentence contains an if and only if in it, I should set it up as an If then statement?

No. An if-then statement means the implication goes in only one direction.

The statement "If $\displaystyle p,$ then $\displaystyle q,$" which can also be written as $\displaystyle \text{}p$ only if $\displaystyle q,\text{''}$ means that if $\displaystyle p$ is true, then $\displaystyle q$ must also be true (but it might be possible for $\displaystyle q$ to be true even though $\displaystyle p$ is false).

The statement $\displaystyle \text{}p$ if $\displaystyle q,\text{''}$ means that if $\displaystyle q$ is true, then $\displaystyle p$ must also be true (but it might be possible for $\displaystyle p$ to be true even though $\displaystyle q$ is false).

The statement $\displaystyle \text{}p,$ if and only if $\displaystyle q\text{''}$ means that if $\displaystyle p$ is true, then $\displaystyle q$ must also be true, and if $\displaystyle q$ is true, then $\displaystyle p$ must also be true. Do you see how the implication goes both ways?

When proving a statement like $\displaystyle p\text{ if{f} }q,$ you need to prove both of these statements:
1. If $\displaystyle p,$ then $\displaystyle q$
2. If $\displaystyle q,$ then $\displaystyle p$
• Mar 22nd 2009, 06:16 PM
meg0529
Ya haha I re-read the chapter, thanks!