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ProofMaster23

Member

Registered: 2011-02-24

Posts: 2

What is Mathematically wrong with my proof?

I did this proof but my teacher says there are some mathematical problems? I have tried forever, any help?

Proposition: For any integer n, the integer n is even if and only if   is even.

Proof: Let n be an integer so that n^2   is even.  By definition of even, we know that n^2 = 2k . Assume  n = 2m and so we have that  n^2 = (2m)^2 = 2^2 * m^2 = 2(2m^2) .  Since m   is an integer, then 2m^2   is an integer, since the integers are closed under multiplication.  Note that n  is even.

Hope that makes sense. Thanks!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: What is Mathematically wrong with my proof?

Hi;

Your proposition is missing something, is it not?

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

ProofMaster23

Member

Registered: 2011-02-24

Posts: 2

Re: What is Mathematically wrong with my proof?

ahh yes sorry about that its supposed to say:

Proposition: For any integer n, the integer n is even if and only if n^2  is even.

Proof: Let n be an integer so that n^2   is even.  By definition of even, we know that n^2 = 2k . Assume  n = 2m and so we have that  n^2 = (2m)^2 = 2^2 * m^2 = 2(2m^2) .  Since m   is an integer, then 2m^2   is an integer, since the integers are closed under multiplication.  Note that n  is even.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: What is Mathematically wrong with my proof?

Maybe you can try this on him/her.

If n^2 is even then it is n x n = (2m)(2m) ( Why? ),

then

n is even when n^2 is even.

If n^2 is odd then it is n x n = (2m+1)(2m+1) ( Why? )

then

if n^2 is odd then n is odd.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: What is Mathematically wrong with my proof?

hi ProofMaster23

If you are asked to prove tha (A is true) if and only if (B is true), then you have to show that

(A is true) => (Bis true) AND (B is true) => (A is true)

Your proof in post #1 does the first of these but not the second.

You still have to show that (n^2 is even) => (n is even).

ie start with n^2 = 2m and somehow get to n = 2p for some integer p.

In practice it is probably easier to show (A is not true) => (B is not true) which is logically equivalent to  (B is true) => (A is true)

So I would start with 'suppose that n is odd, ie n = 2m + 1 for some integer m'

You can probably see how to finish this off. (ie. get to therefore, n^2 is odd)

Bob

Last edited by Bob (2011-02-24 20:31:13)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob