1. 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?

2. Originally Posted by TheRekz
1. $\forall m,n \in Z, if m + n$ is even then either m and n are both even or
they are both odd
By division algorithm we can write $m=2a+r_1$ where $r_1=0,1$ and $n=2b+r_2$ then $n+m = 2(a + b) + (r_1+r_2)$ for this to be even we require that $(r_1,r_2)=(0,0)$ or $(r_1,r_2)=(1,1)$, i.e. both even or both odd.
2. Prove or disprove using contradiction. Every integer > 11 is the sum of two
composite integers.
If $n>11$ is even then $n - 4 > 2$ is even and these are composite, thus, $n=4+(n-4)$. If $n>11$ and is odd then we $n-9>2$ is even and so composite which means we can write $n = 9 + (n-9)$.

3. 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?

4. 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]

(all number refered to are in $\mathbb{Z}$)

Supose there are $n$ and $m$ such that $n$ is odd and $m$ even and $n+m$ is even.

By supposition there exist $k_1$ and $k_2$ such that $n=2k_1+1,\ m=2k_2$.

Then $n+m=2(k_1+k_2)+1$ which is odd, which contradicts our supposition.

Hence if $n+m$ is even either $m$ and $n$ are both even or they are both odd

RonL

5. 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

6. 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

7. 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 $N> 11$ not the sum of two composite integers.

Then $N$ is even or odd.

Case 1: $N$ even, put $n_1=4,\ n_2=N-4$, then both $n_1$ and $n_2$ are even and greater than $2$ and hence composite, but this contradicts our assumption so $N$ cannot be composite.

Case 2: $N$ odd, put $n_1=9,\ n_2=N-9$, then both $n_1$ is composite and $n_2$ is even and greater than $2$ and hence composite, but this contradicts our assumption so $N$ 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

8. Originally Posted by CaptainBlack
ImPerfectHackers proof is quite simple (and neat):

Suppose that there is an integer $N> 11$ not the sum of two composite integers.

Then $N$ is even or odd.

Case 1: $N$ even, put $n_1=4,\ n_2=N-4$, then both $n_1$ and $n_2$ are even and greater than $2$ and hence composite, but this contradicts our assumption so $N$ cannot be composite.

Case 2: $N$ odd, put $n_1=9,\ n_2=N-9$, then both $n_1$ is composite and $n_2$ is even and greater than $2$ and hence composite, but this contradicts our assumption so $N$ 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??

9. 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