# Prove Number Theory Proofs: Sum Irrational, (m+dk) mod d, x^2=x, n^2 mod 3

• and1bball4mk
So for n=1 mod 3, n^2=1 mod 3 and for n=2 mod 3, n^2=4 mod 3 which is equivalent to 1 mod 3. Therefore, n^2 mod 3=1. In summary, for any integer n not divisible by 3, n^2 mod 3=1.

#### and1bball4mk

1. For any positive integer n, if 7n+4 is even, then n is even.
2.Sum of any two positive irrational numbers is irrational.
3. If m, d, and k are nonnegative integers with d=/=0 then (m+dk) mod d = m mod
4. For all real x, if x^2=x and x=/=1 then x=0
5. If n is an integer not divisible by 3, then n^2 mod 3=1

2. Basically I have to prove (or disprove) all of those and I'm stuck. Any advice and feed back would be appreciated.

3. Here are my attempts at solutions
1. By contraposition, if n is odd then 7n+4 is odd. If N is odd, then n=2k+1 for some integer k. so 7(2k+1)+4, and by algebra we get 2(7k+5)+1. 7k+5 is an integer, so it must be odd.

Is this right? I think it is but I'm never sure because I'm terrible at number theory.

2. For this one I found a counterexample and I think that is sufficient to get the problem right but I want to know why it's a false statement..anyone have any insight?

3. This is confusing. Basically after half an hour of writing stuff I haven't reached any conclusions. I'm sure it has something to do with the quotient remainder theorem because the form of (m+dk) is very similar to QRT where given any interger n and positive integer d, there exists unique integers q and r such that n=dq+r. Any help would be appreciated.

4. This one is really bothering me. I know that it's true but I'm trouble proving it. I keep going back to the method of exhaustion but obviously that won't work. I have a feeling that this is really simple and I'm just overlooking something.

5. This is another that I know is true but I don't know how to prove it. I think part of the problem I'm having here is defining n as a integer not divisible by 3.

Any help on any of those would be appreciated.

Thanks

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Actually I might of made some headway on number 4.

If x^2=x then x=x/x. So x/x will always reduce to 1 unless x=0.

Is this sufficient for a proof?

You're likely to get more responses if you put each problem in itw own thread, rather than posting a whole slew of them all at once.

and1bball4mk said:
Actually I might of made some headway on number 4.

If x^2=x then x=x/x. So x/x will always reduce to 1 unless x=0.

Is this sufficient for a proof?

and1bball4mk said:
4. For all real x, if x^2=0 and x=/=1 then x=0
Which is it? x^2 = 0 or x^2 = x?

Mark44 said:
Which is it? x^2 = 0 or x^2 = x?

It is x^2=x

For 5, if n is an integer not divisible by 3, then its remainder when divided by 3 has to be either 1 or 2. I.e., n = 1 mod 3 or n = 2 mod 3.

Investigate each case to say something about n^2.

Mark44 said:
For 5, if n is an integer not divisible by 3, then its remainder when divided by 3 has to be either 1 or 2. I.e., n = 1 mod 3 or n = 2 mod 3.

Investigate each case to say something about n^2.

Thanks.