Proof by contradiction:

assume

This contradicts the assumption, therefore E_1 + E_2 = E_3

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- March 29th 2009, 07:58 AMbmp05Proove: the sum of two even numbers is even

Proof by contradiction:

assume

This contradicts the assumption, therefore E_1 + E_2 = E_3 - March 29th 2009, 12:51 PMHallsofIvy
Isn't 2m+ 2n= 2(m+n) simpler?

- March 29th 2009, 12:53 PMGrandad
Hello bmp05Are you required to prove this by contradiction? The direct proof is very straightforward. We define an even number in the obvious way: is even .

Then is even and is even

is even

If you're going to use a 'contradiction' method, I think you have to be a little more careful with your use of notation. The definition of your sets should be

Then your hypothesis would be:

Assume

Then

. Contradiction.

Grandad - March 29th 2009, 01:21 PMbmp05
Thanks grandad! I was trying to follow your last example as well, btw. I read that you should be careful to not neglect the direct proof, because it is easier if possible, but then I have hardly ever seen it used.(Clapping)

- March 29th 2009, 02:05 PMbmp05
And now to prove uneven.even = even

is even .

is uneven .

Then is even and is uneven

is even

Is this a trivial case, because you prove that but using the definition of an even number? (But that's the same for adding two even numbers as well...) - March 29th 2009, 11:31 PMGrandad
Hello bmp05Well done. You've got this pretty well complete, except for the bit I've marked in red, where you've just gone back to the original expression.

I agree, it does look more or less trivial, but what you should have said is

which is now in the format

So that you can now say:

is even

using the part of the definition that is even

Do you understand this subtle, but important, point?

Grandad - March 30th 2009, 01:23 AMbmp05
Yes- thanks again, I realised that I needed to show something analogue to the definition of the even number, but I can see that it is much clearer to substitute of the natural numbers as opposed to . But the most important thing missing from the proof is this substitution of an

__equivalent natural number p__.

This reminds me of another probelm that I'm having proving the distributive laws in set theory- it is somehow simple, but hard to express in the right way!

Proove

__Part1__:

__Part2__:

__It follows that__,

Unfortunately, it's not very clear to read. - March 30th 2009, 06:30 AMGrandad
Hello bmp05

Quote:

__Part1__:

Quote:

__Part2__:

You need to use the Fundamental Theorem of Arithmetic and say that if , where the are primes, then . So , for some , and therefore .

Grandad - March 31st 2009, 01:26 PMbmp05
Hello Grandad,

I'm not sure about the Fundamental Theorem of Arithmetic, I had a look at the wikipedia page and it looks interesting, but the course is moving very quickly, unfortunately and although I can't write very good proofs yet, we have already moved onto fields (help!)- anyway, I thought of a good idea for this problem, but I'm not sure how to proove it.

The idea is that

This would mean that I can jump straight into the proof, because can only ever equal one of the 3 possibilities above. But the proof becomes something like:

, so I would like to say something like:

Proof by contraditiction, suppose

[Errr... well, I- this seems to be true, but I don't know how to...?]

[... do the same for the other or]

__This is a contradiction__, therefore

Well, this is a terrible proof, but on the good side, if done correctly this would actually be an "if and only if" proof! (?) - April 1st 2009, 01:54 AMGrandad
Hello bmp05

Well done. I think this is pretty well OK. I should just change the last part a bit (the part where you weren't so sure) so that instead of:I think I'd factorise the right-hand sides, and say:

Contradiction, since and both leave remainder when divided by .

Grandad - April 1st 2009, 12:33 PMMoo
Mathematics, or how to complicate simple things (Giggle)