Proof that n is not a power of an integer.

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Homework Help Overview

The discussion revolves around proving that n! for n>1 cannot be a square, cube, or any other power of an integer. The participants explore the implications of prime factorization in relation to factorials and integer powers.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants attempt a proof by contradiction involving the unique prime factorization of integers and the presence of primes between n/2 and n. Others question the clarity of variable usage between n and n! in the proof.

Discussion Status

There appears to be some productive dialogue regarding the proof's structure, with participants offering suggestions for clarity and addressing potential confusion in variable usage. However, there is no explicit consensus on the correctness of the proof or the interpretation of the hint provided.

Contextual Notes

Participants note the importance of distinguishing between n and n! in the context of the problem, as well as the implications of the hint regarding primes in the factorial's range.

cragar
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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 n!=x^r 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 x^r 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
x^r 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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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 n!=x^r 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 x^r 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
x^r 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?
 
ok thanks, just wanted to make sure it worked
 
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
 
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.
 
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.
 
Hi guys, of course it is! I am the one mixing n! and n :)
Cheers...
 

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