Can you see a counter example that i can't, divisibility problem

mr_coffee
Messages
1,613
Reaction score
1
HEllo everyone. I'm trying to find a counter example that will prove this false. But it may be true but I'm hoping it isn't :)

For all integers a and b, if a|10b then a|10 or a|b. I said false, a = 3, b = 5. 3 is not divisible by 50. 3 is also not divisble by 10 nor 3 divisible by 5. But then i saw, p has to be true for q to be true. I have to prove q to be false, then p is also false. Its looking like this has to be true, can anyone spot a counter example? I"m not looking for an answer, but it would motivate me to keep looking for one.

THanks!
 
Physics news on Phys.org
Try having a > 10 and a > b
 
So we're starting with

prove or disprove that for all a,b if a|10b then a|10 or a|b.

A disproof does require one counter example. However, why did you pick a=3 and b=5, since we need to have a|10b and a doesn't divide 10 or b for a counter example, and 3 does not divide 50.

Hint: p is *prime* if and only if for all m,n p|mn implies p|m or p|n. (so of course a counter example exists, and can be found just by thinking about prime factorizations.
 
mr_coffee said:
HEllo everyone. I'm trying to find a counter example that will prove this false. But it may be true but I'm hoping it isn't :)

For all integers a and b, if a|10b then a|10 or a|b. I said false, a = 3, b = 5. 3 is not divisible by 50. 3 is also not divisble by 10 nor 3 divisible by 5.
The question would not be whether 3 is divisible by 50 but whether 50 is divisible by 3. a| 10b means "a divides 10b". I.e. "10b is divisible by a". In any case, the statement says "If a|10b" so you cannot choose an example in which that is not true.
But then i saw, p has to be true for q to be true. I have to prove q to be false, then p is also false.
Well, I don't know because you didn't tell us what p and q are!
Its looking like this has to be true, can anyone spot a counter example? I"m not looking for an answer, but it would motivate me to keep looking for one.

THanks!
I hope you mean you are not asking for an answer. You certainly should be looking for one! Try this: pick b to be anything you like and pick a to be the biggest number you can find that will divide 10b.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top