Thread: Proving and Disproving

1. [tex] \forall m,n \in Z, if m + n is even then either m and n are both even or
they are both odd [\math]

2. Prove or disprove using contradiction. Every integer > 11 is the sum of two
composite integers.

For the first number, is it the same if I prove that if m and n are both even then m +n is even and if m and n are both odd then m + n is even?

Originally Posted by TheRekz

1. is even then either m and n are both even or
they are both odd

By division algorithm we can write where and then for this to be even we require that or , i.e. both even or both odd.

2. Prove or disprove using contradiction. Every integer > 11 is the sum of two
composite integers.

If is even then is even and these are composite, thus, . If and is odd then we is even and so composite which means we can write .

is there any other way to proof this?? your explanation seems complicated although it makes sense. I don't quite understand where did you get the m = 2a + r from?? so you mean r can be either 0 or 1 here depends whether m is odd or even? if m is even then r is 0?? r can therefore also be 2 right?

Originally Posted by TheRekz

1. [tex] \forall m,n \in Z, if m + n is even then either m and n are both even or
they are both odd [\math]

By contradiction:

(all number refered to are in )

Supose there are and such that is odd and even and is even.

By supposition there exist and such that .

Then which is odd, which contradicts our supposition.

Hence if is even either and are both even or they are both odd

RonL

Originally Posted by TheRekz

1
For the first number, is it the same if I prove that if m and n are both even then m +n is even and if m and n are both odd then m + n is even?

No you have to show that if n+m is even both n and m are even or both are odd, and that you will not have done.

RonL

Originally Posted by CaptainBlack

No you have to show that if n+m is even both n and m are even or both are odd, and that you will not have done.

RonL

can you help me to do number 2?? Cause I don't really get it on the post no.2 answer

Originally Posted by TheRekz

can you help me to do number 2?? Cause I don't really get it on the post no.2 answer

ImPerfectHackers proof is quite simple (and neat):

Suppose that there is an integer not the sum of two composite integers.

Then is even or odd.

Case 1: even, put , then both and are even and greater than and hence composite, but this contradicts our assumption so cannot be composite.

Case 2: odd, put , then both is composite and is even and greater than and hence composite, but this contradicts our assumption so cannot be composite.

Case 1 and Case 2 together contradict the original assumption and so the theorem: Every integer > 11 is the sum of two composite integers; is proven by contradiction.

RonL

Originally Posted by CaptainBlack

ImPerfectHackers proof is quite simple (and neat):

Suppose that there is an integer not the sum of two composite integers.

Then is even or odd.

Case 1: even, put , then both and are even and greater than and hence composite, but this contradicts our assumption so cannot be composite.

Case 2: odd, put , then both is composite and is even and greater than and hence composite, but this contradicts our assumption so cannot be composite.

Case 1 and Case 2 together contradict the original assumption and so the theorem: Every integer > 11 is the sum of two composite integers; is proven by contradiction.

RonL

just one more question, how do we know that n2 is even here?? thanks??

Originally Posted by TheRekz

just one more question, how do we know that n2 is even here?? thanks??

In Case 1:By hypothesis N is even, and > 11, so N-4 is even (and >7)

In Case 2:By hypothesis N is odd, and > 11, so N-9 is even (and >2)

(even-even is even, and odd-odd is also even)

RonL