Proof by induction: multiplication of two finite sets.

whyme1010
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Homework Statement



prove by induction that if A and B are finite sets, A with n elements and B with m elements, then AxB has mn elements



Homework Equations


AxB is the Cartesian product. AxB={(a,b) such that a is an element of A and b is an element of B}


The Attempt at a Solution


normally, proof by induction involves one variable. here it includes two. I guess I could say that mn=nm and therefore proving it for n+1 is the same thing as proving it for m+1. Then any increase in m or n after that is just the same thing. But somehow I still feel this isn't really justifiable.
 
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whyme1010 said:

Homework Statement



prove by induction that if A and B are finite sets, A with n elements and B with m elements, then AxB has mn elements



Homework Equations


AxB is the Cartesian product. AxB={(a,b) such that a is an element of A and b is an element of B}


The Attempt at a Solution


normally, proof by induction involves one variable. here it includes two. I guess I could say that mn=nm and therefore proving it for n+1 is the same thing as proving it for m+1. Then any increase in m or n after that is just the same thing. But somehow I still feel this isn't really justifiable.

So induction has three steps. Where you assume the case n = 1, the induction assumption and then the proof for the n+1 case.

Case : n = 1 = m

Assume that A and B have one element in each corresponding set. Say A = {(a0,b0)} and B = {(c0,d0)} and go from here.
 
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yeah, but how do I prove that if it is true for the n+1 case, it is also true for the m+1 case. and what if m+1 and n+1 are occurring at the same time?

I guess what I'm asking is, if I prove it true for the n+1 case (which is pretty easy), then how do I show that this means I can add one element to A and B as desired and the result mn would still be valid.
 
whyme1010 said:
yeah, but how do I prove that if it is true for the n+1 case, it is also true for the m+1 case. and what if m+1 and n+1 are occurring at the same time?

I guess what I'm asking is, if I prove it true for the n+1 case (which is pretty easy), then how do I show that this means I can add one element to A and B as desired and the result mn would still be valid.

That's the beauty of induction. Imagine if you held the amount of elements in B constant the whole time. Would it change your outcome?

Remember that proving something n+1 times is sufficient when thinking about this.
 
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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