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

In summary, the problem at hand is to use the minimal counterexample technique to prove that the product of n consecutive positive integers is always divisible by n!. After setting up a least counterexample k, it is necessary to use algebraic tricks to show that the proposition holds for k-1, which will then imply that it holds for k. The equation c*(k-1)!= a*(a+1)*...*(a+k-2) is used to represent the consecutive integers, where c is the integer from the definition of integer divisibility. However, this problem may have a more complicated solution."
  • #1
roam
1,271
12

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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  • #2
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
 
  • #3
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?
 
  • #4
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.
 
  • #5
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Related to Can the Product of n Consecutive Positive Integers Be Divisible by n!?

1. What is a minimal counterexample?

A minimal counterexample is the smallest possible example that disproves a given statement or hypothesis.

2. How is a minimal counterexample different from a regular counterexample?

A regular counterexample is any example that disproves a statement or hypothesis, while a minimal counterexample is the smallest possible example that still disproves the statement or hypothesis.

3. Why are minimal counterexamples important in scientific research?

Minimal counterexamples are important because they help us identify the exact conditions under which a statement or hypothesis is false. By finding the smallest counterexample, we can gain a better understanding of the limitations or exceptions to a general rule or theory.

4. How can one find a minimal counterexample?

To find a minimal counterexample, one must systematically test different scenarios or conditions until the smallest one that disproves the statement or hypothesis is found. This often involves conducting experiments or gathering data.

5. Can a minimal counterexample ever be proven wrong?

No, a minimal counterexample cannot be proven wrong. By definition, it is the smallest example that disproves a statement or hypothesis. However, as new evidence and research emerge, the conditions or assumptions that led to the minimal counterexample may be revised or updated.

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