Proof that n is not a power of an integer.

In summary, the conversation discusses a proof that n! for n>1 cannot be a square, cube, or any other power of an integer. The proof involves assuming n!=x^r where x and r are natural numbers and using the fact that there is always a prime between n/2 and n if n>3. The conversation also mentions that the use of "n" and "n!" may be confusing and suggests rephrasing to avoid confusion.
  • #1
cragar
2,552
3

Homework Statement


Prove that n! for n>1 cannot be a square or cube or any other power of an integer.
Hint: There is always a prime between n/2 and n if n>3

The Attempt at a Solution


Lets assume for contradiction that [itex] n!=x^r [/itex] where x and r are natural numbers
and n>3 , so there is some prime p that is between n/2 and n. well this occurs once in prime factorization of n!. Now x has some prime factorization and in order for [itex] x^r [/itex] to be equal to n! the prime p must be in x's prime factorization, but if r>1 then p will have r copies in
[itex] x^r [/itex] so this is a contradiction because numbers have unique prime factorizations so
n! is never a power of an integer for r>1.
 
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  • #2
cragar said:

Homework Statement


Prove that n! for n>1 cannot be a square or cube or any other power of an integer.
Hint: There is always a prime between n/2 and n if n>3

The Attempt at a Solution


Lets assume for contradiction that [itex] n!=x^r [/itex] where x and r are natural numbers
and n>3 , so there is some prime p that is between n/2 and n. well this occurs once in prime factorization of n!. Now x has some prime factorization and in order for [itex] x^r [/itex] to be equal to n! the prime p must be in x's prime factorization, but if r>1 then p will have r copies in
[itex] x^r [/itex] so this is a contradiction because numbers have unique prime factorizations so
n! is never a power of an integer for r>1.

Sounds just fine to me. Any question?
 
  • #3
ok thanks, just wanted to make sure it worked
 
  • #4
Hi cragar, yes it looks good, but you are mixing n and n! here and there so it is a bit confusing, you might want to rephrase it.
The original statement is probably at the origin of it, but when they say
"Hint: There is always a prime between n/2 and n if n>3", it's not the same n that was used at the beginning of the problem, and your proof, although correct in the spirit (I think) would benefit from using different variable names or to be more explicit about when you talk about n or n!
they are freely intermixed and it could be considered as just wrong
 
  • #5
oli4 said:
you are mixing n and n! here and there so it is a bit confusing, you might want to rephrase it.
The original statement is probably at the origin of it, but when they say
"Hint: There is always a prime between n/2 and n if n>3", it's not the same n that was used at the beginning of the problem,
Looks like the same n to me.
 
  • #6
haruspex said:
Looks like the same n to me.

Sure it's the same n. Knowing there's a prime between n!/2 and n! is pretty useless.
 
  • #7
Hi guys, of course it is! I am the one mixing n! and n :)
Cheers...
 

FAQ: Proof that n is not a power of an integer.

1. What is the definition of a power of an integer?

A power of an integer is a number that can be expressed as the product of the integer multiplied by itself a certain number of times. For example, 32 = 9 is a power of 3.

2. How can you prove that a given number is not a power of an integer?

One way to prove that a number is not a power of an integer is to find a prime factorization of the number and check if any of the prime factors have a fractional exponent. If there is a fractional exponent, then the number is not a power of an integer.

3. Can a negative number be a power of an integer?

Yes, a negative number can be a power of an integer. For example, (-2)3 = -8 is a power of -2.

4. How do you determine if a number is not a power of an integer using logarithms?

If a number is not a power of an integer, then its logarithm will not be a whole number. Therefore, if we take the logarithm of the number and it is not a whole number, we can conclude that the number is not a power of an integer.

5. Can a decimal number be a power of an integer?

No, a decimal number cannot be a power of an integer because a power of an integer is always a whole number. If a decimal number is expressed as a power of an integer, then the exponent must also be a decimal, which goes against the definition of a power of an integer.

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