Proof by induction: multiplication of two finite sets.

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The discussion focuses on proving by induction that the Cartesian product of two finite sets A and B, with n and m elements respectively, contains mn elements. Participants express concern about applying induction with two variables, suggesting that proving the case for n+1 could be analogous to proving it for m+1. They discuss the necessity of establishing a base case, particularly when both n and m increase simultaneously. The importance of holding one variable constant during the proof is highlighted as a potential simplification. Ultimately, the conversation emphasizes the inductive reasoning process and its application to this two-variable scenario.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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