Does c always divide b in number theory divisibility?

chimath35
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If a|b then ac=b; now does c always divide b as well?
 
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chimath35 said:
If a|b then ac=b; now does c always divide b as well?

Of course it does.
 
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Thank you Dick; I did pass the exam I took by the way lol. I did not really get to study much as I was ill, but still did good enough compared to other class mates. I think I am growing A LOT mathematically. I am just now learning how to do direct proofs in discrete math so hopefully I will do better next exam, even though I am just auditing this course. I am learning how to become more of a mathematician I guess.
 
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...
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