Number Theory: nth root of n is irrational

[barylwires]

1. For n ≥2, n^(1/n) is irrational.
Hint provided: Use the fact that 2^n > n2. This is probably familiar to many.
By contradiction, n = a^n/b^n
--> a^n = n(b^n)
--> n|a^n
--> n|a
Am I trying to force the same contradiction as with 2^1/2 is rational, that is, that a/b are not in lowest terms? Or do I need prime factorization to do something? Guidance is appreciated.

 

[meiji1]

The fact that 2^n > n means that n cannot have n or more prime factors. Supposing that a and b are coprime also helps.

 

[barylwires]

The fact that 2^n > n means that n cannot have n or more prime factors. Supposing that a and b are coprime also helps.

I don't think I'm using the hint in the following way; is it correct?
n=a^n/b^n --> a^n=n•b^n --> b^n|a^n --> b|a --> n^(1/n) is an integer,
but by hyp., n≥ 2 --> 2^(1/2) is an integer --><-- (relies, of course, on sqrt(2) known to be irrational)

 

[meiji1]

That's not quite a contradiction. You've assumed the statement is true in the subcase that $$n \ge 3$$, in which the statement "2^(1/2) is an integer" has no effect.

Maybe go up to a^n=n•b^n, and look at primes dividing n on the RHS.

 

[barylwires]

That's not quite a contradiction. You've assumed the statement is true in the subcase that $$n \ge 3$$, in which the statement "2^(1/2) is an integer" has no effect.

Maybe go up to a^n=n•b^n, and look at primes dividing n on the RHS.

Why are primes dividing n on the RHS? Do you mean I should consider the division of n by primes? As is , the problem implies two divisions: n|a^n and b^n|a^n.

 

[meiji1]

I mean that you should consider division of n by primes, yes.

 

[barylwires]

I mean that you should consider division of n by primes, yes.

I see that n has r<n prime factors, and that n|a^n.
And a^n/n = b^n.
I can't see the connection between this feature, the fact that a and b are co-prime, and a contradiction.

 

[Dick]

I see that n has r<n prime factors, and that n|a^n.
And a^n/n = b^n.
I can't see the connection between this feature, the fact that a and b are co-prime, and a contradiction.

There is no way that (a/b)^n with a/b in lowest terms can be an integer unless b=1. Can you show that?

 

[barylwires]

There is no way that (a/b)^n with a/b in lowest terms can be an integer unless b=1. Can you show that?

Thanks.
My initial contradictory supposition was that n^(1/n) was rational. My biggest problem here (I think) is seeing where rational a/b must be integer a/1. Is that what the hint is supposed to help me see? That no integer x can have x or more factors, together with gcd(a, b)= gcd(a^n, b^n) = 1. ?

 

[Dick]

Thanks.
My initial contradictory supposition was that n^(1/n) was rational. My biggest problem here (I think) is seeing where rational a/b must be integer a/1. Is that what the hint is supposed to help me see? That no integer x can have x or more factors, together with gcd(a, b)= gcd(a^n, b^n) = 1. ?

If (a/b)^n=n and a/b is in lowest terms, and b is not 1, then there is a prime p that divides b but not a. So you can write b=pc. So a^n=(pc)^n*n. Conclusion? It's really just like the sqrt(2) is irrational proof.

 

[barylwires]

If (a/b)^n=n and a/b is in lowest terms, and b is not 1, then there is a prime p that divides b but not a. So you can write b=pc. So a^n=(pc)^n*n. Conclusion? It's really just like the sqrt(2) is irrational proof.

Many thanks for all the good pushing. What I have is clunky at best, and probably flawed:

n = (a/b)^n
$$\Rightarrow$$b^n|a^n
$$\Rightarrow$$gcd(a^n, b^n) = b^n = gcd(a, b) = 1
$$\Rightarrow$$b=1
$$\Rightarrow$$n=a^n.
Since n≥2, a>1, but
n≤2^n
$$\Rightarrow\Leftarrow$$

 

[Dick]

Many thanks for all the good pushing. What I have is clunky at best, and probably flawed:

n = (a/b)^n
$$\Rightarrow$$b^n|a^n
$$\Rightarrow$$gcd(a^n, b^n) = b^n = gcd(a, b) = 1
$$\Rightarrow$$b=1
$$\Rightarrow$$n=a^n.
Since n≥2, a>1, but
n≤2^n
$$\Rightarrow\Leftarrow$$

Keep most of that proof. But it's not generally true that gcd(a^n,b^n)=gcd(a,b), right? gcd(a^n,b^n)=gcd(a,b)^n. Do you have that theorem already proved? If so then it's fine. Otherwise you have to prove it. Or you could prove if gcd(a,b)=1 then gcd(a^n,b^n)=1. Either way there is a step in the proof that you don't state a reason for, so I'm not sure how you are filling it in.

 

[barylwires]

Keep most of that proof. But it's not generally true that gcd(a^n,b^n)=gcd(a,b), right? gcd(a^n,b^n)=gcd(a,b)^n. Do you have that theorem already proved? If so then it's fine. Otherwise you have to prove it. Or you could prove if gcd(a,b)=1 then gcd(a^n,b^n)=1. Either way there is a step in the proof that you don't state a reason for, so I'm not sure how you are filling it in.

Groan. Must have been thinking a^n|b^n $$\Rightarrow$$a|b, but writing something entirely different. I suppose sleep is everything they say it is.

 

[Dick]

Groan. Must have been thinking a^n|b^n $$\Rightarrow$$a|b, but writing something entirely different. I suppose sleep is everything they say it is.

a^n|b^n does imply a|b. And nothing you said is wrong. I'm just asking if you have to prove it. Or is that already a proven theorem, so you can just use it?

 

[barylwires]

a^n|b^n does imply a|b. And nothing you said is wrong. I'm just asking if you have to prove it. Or is that already a proven theorem, so you can just use it?

I should have written that, by hyp., gcd(a, b)=1 $$\Rightarrow$$ gcd(a^n, b^n) = 1, then gcd(a^n, b^n) = b^n = 1. Does that do it? I'm referring specifically to these, particular, relatively prime a and b. Clearly, gcd(4^2, 6^2) = 4 ≠ 2 = gcd(4, 6) is a counter example of gcd(a^n, b^n) = gcd (a, b), in general, right?

 

[Dick]

I should have written that, by hyp., gcd(a, b)=1 $$\Rightarrow$$ gcd(a^n, b^n) = 1, then gcd(a^n, b^n) = b^n = 1. Does that do it? I'm referring specifically to these, particular, relatively prime a and b. Clearly, gcd(4^2, 6^2) = 4 ≠ 2 = gcd(4, 6) is a counter example of gcd(a^n, b^n) = gcd (a, b), in general, right?

It sure is. You are saying nothing but true statements. I'm just asking which of them you have a theorem to back up, and which you might have to find a proof for. If you already know gcd(a,b)=1 => gcd(a^n,b^n)=1 then you are home free.