Can the Product of n Consecutive Positive Integers Be Divisible by n!?

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Homework Help Overview

The discussion revolves around proving that the product of n consecutive positive integers is always divisible by n!. The original poster is exploring the use of a minimal counterexample technique to establish this claim.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the concept of a least counterexample and the implications of its existence. There are attempts to set up algebraic expressions to relate the product of consecutive integers to factorials. Questions arise regarding the meaning of certain variables in the expressions and the nature of the proof being employed.

Discussion Status

The discussion is ongoing, with participants providing insights and clarifications about the setup of the proof. There is an acknowledgment of potential complexities in the problem, and some participants are questioning the definitions and assumptions involved in the proof strategy.

Contextual Notes

There are indications of confusion regarding the proof method, particularly with respect to the use of inductive reasoning versus the least counterexample approach. Participants are also clarifying definitions related to divisibility.

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Homework Statement


Using the minimal counterexample technique prove that the product of n consecutive positive integers is always divisible by n!

The Attempt at a Solution



Suppose that the statement is not true and the product of n consecutive positive integers is not always divisible by n!...

I'm stuck here. How can I continue this to get a contradiction and thus show that the product of is always divisible by n! ? Any guidance is greatly appreciated.
 
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Then there is a least counterexample k for which the proposition does not hold. Since this is the least counterexample, the proposition holds for k-1. Now you need to do some algebraic tricks to show that the proposition holding for k-1 implies that it holds for k. Your set up should look something like
c*(k-1)!= a * (a+1) *...*(n+a-1), for some a,c>0
 
JThompson said:
Then there is a least counterexample k for which the proposition does not hold. Since this is the least counterexample, the proposition holds for k-1. Now you need to do some algebraic tricks to show that the proposition holding for k-1 implies that it holds for k. Your set up should look something like
c*(k-1)!= a * (a+1) *...*(n+a-1), for some a,c>0

What does that "c" mean? Could you please explain a bit more how you got this expression?
 
Note: Sorry- the last term in the equation I posted had an error- the last term should have been a+k-2.

Definition of "divides": For integers x and y, where x does not equal 0, we say x divides y if there exists an integer c so that y=xc.

So in the equation c*(k-1)!= a * (a+1) *...*(a+k-2), the terms a, a+1, a+2 ... a+k-2 are the consecutive integers (note that there are a+k-2 - a +1 = k-1 of them), and c is the integer from the definition of integer divisibility above. This is the inductive assumption.

Edit: Forget the part about the inductive assumption- different type of proof. Forgot this one used least counterexample.
 

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