Assignment where i have to make conjectures.

[choob]

I am in the process of completing an assignment where i have to make conjectures. As of now I have made conjectures that for function P(n)=(n^x)-n is divisible by x for all x that are prime numbers,

And that for x choose r=k, k belonging to the natural numbers, that k is a multiple of x when r=2.

Now I have to make a converse, I have absolutely no idea what a converse is. Through googling "converse conjecture" I have an idea that a converse is a counter-argument to my arguments, but that would mean I would have to contradict myself?

Help and input much appreciated.

 

[HallsofIvy]

The converse to "if a then b" is "if b then a". You just switch the "hypothesis" (a) and the conclusion (b). No, a "converse" is NOT counter-argument. I don't know how you got that impression.

Notice that the converse of "if b then a" is "if a then b" because we have just swapped the hypothesis and conclusion back again.

For example, if the statement is "if n is divisible by 6 then it is divisible by 2", then its converse would be "if n is divisible by 2 then it is divisible by 6". Notice that, in this case, the first statement is true but the second statement is false.

Another example would be "if n is divisible by both 2 and 3 then it is divisible by 6" which has converse "if n is divisible by 6 then it is divisible by both 2 and 3". In this case, both the statement and the converse are true.

Knowing the original statement is true tells you nothing about whether the converse is true or false (and "conversely").

 

[choob]

Thanks.
So for my first conjecture n^x-n is divisible by x that are prime numbers then the converse would be if n^x-n and x is a prime number than n^x-n is divisible by x?

 

[Dick]

Try and state the conjecture clearly in an if-then form. Your conjecture is "IF p is prime THEN n^p-n is divisible by p for all n". Now it's easy to state the converse. Just reverse the IF and THEN parts. Try it. Both are interesting questions and have names and history.

 

[choob]

If x is a prime number than n^x-n is divisible by x
Converse: If n^x-n is divisible by n then x is a prime number

If r is 2 then x choose r is divisible by x
Converse: If x choose r is divisible by x then r=2

Is that it?:D

Also any chance you could give me the name to this problem? I've tried googling it to no avail.

 

[Dick]

You are leaving out the "for all n" part. But, ok. And in the converse I hope you meant n^x-n is divisible by x, there's no reason to reverse n and x. The conjecture is Fermat's little theorem. It's true. The converse is the problem of pseudoprimes. It's false.

 

[choob]

Thanks Dick

 

[choob]

When I'm doing an assignment for math, is there a specific way to make references? For example, I am copying a passage from the late Dijkstra's personal website.