Why will this function always be an integer?

In summary, the conversation discusses the definition of f(n) as 11x22x33...xnn and the question of whether f(n)/(f(r).f(n-r)) is always an integer for 0 < r < n. It is suggested that this can be proven using the fact that ##{n \choose c} = \frac{n!}{c! (n-c)!}## is always an integer, and the concept of a "hyperfactorial" ##H(n) = \prod_{i=1}^{n} i^i ## is introduced. Other members are encouraged to contribute to finding an explicit proof for this question.
  • #1
Avichal
295
0
f(n) is defined as 11x22x33...xnn

Then it seems as if f(n)/(f(r).f(n-r)) is always an integer for 0 < r < n.
I tried a few cases. Its true for them. Is it always true? I cannot seem to figure out any ways to prove it.
 
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  • #2
I presume r and n are integers?
 
  • #3
x means multiplication? * would be clearer.

You can prove it with the fact that ##{n \choose c} = \frac{n!}{c! (n-c)!}## is always an integer.
 
  • #4
It is trivially true since the set of all integers is closed under the operation of multiplication (exponentiation being a special case of multiplication).
 
  • #6
mfb said:
There is a division...
I think I might have missread the f(n)/(f(r).f(n-r)) statement in the OP by reading to quickly. When I look at it now it makes no sense at all unless someone defines the "."-operator.
 
  • #7
I assumed it to be a multiplication, too.
 
  • #8
This is the "hyperfactorial", ##H(n)## (assuming the x is multiplication). It is defined as
##H(n) = \prod_{i=1}^{n} i^i ##.
Just by looking at it, I don't see any reason to believe it wouldn't be an integer at any point. I don't know if there's an explicit proof of this out there, but generally, an integer multiplied any number of times by itself is an integer, and when added to another integer that has undergone the same process, it should still be an integer. Of course, ##n## can be really large, so I can't say with any certainty what happens as the product has more and more (perhaps infinite) terms.
Interesting question--I'm curious now!
I'd love for other members to contribute to this.
 
  • #9
There is a division! It makes the question not trivial. Here it is in red:
Avichal said:
f(n)/(f(r).f(n-r))
$$\frac{f(n)}{f(r)f(n-r)}$$

Still always an integer, but you have to prove it.
 

1. What is an integer?

An integer is a whole number, meaning it does not have a fractional or decimal component. Examples of integers include -3, 0, and 5.

2. Why is it important for a function to always return an integer?

Functions are mathematical operations that are used to solve problems and make calculations. In some cases, the output of a function needs to be a whole number, making it necessary for the function to always return an integer.

3. Can a function ever not return an integer?

Yes, a function can sometimes return a non-integer value, such as a decimal or fraction. This depends on the input values and the specific function being used.

4. How can we ensure that a function always returns an integer?

One way to ensure that a function always returns an integer is to use specific mathematical operations, such as division and multiplication, that will result in a whole number. Another way is to use conditional statements to check the output and round it to the nearest integer if needed.

5. Are there any real-life applications where a function must always return an integer?

Yes, there are many real-life applications where it is necessary for a function to return an integer. For example, when calculating the number of items in a store inventory, the result must be a whole number. In financial calculations, such as interest rates, the output must also be an integer.

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