Proof that 1/2 +2/3 + 3/4 does not equal a whole number

[Astronamus]

I am deleting

 

[Tide]

Does this help?

\sum_{j = 1}^{N} \frac {j}{j + 1} = \sum_{j = 1}^{N} \left( 1 - \frac {1}{j + 1}\right) = N - \sum_{j = 1}^{N} \frac {1}{j + 1}

 

[shmoe]

Minor quibble-Tide you lost a negative sign.

Tide has reduced this nicely to showing any the Harmonic series never hits an integer. This might be tough to prove so I'll kick out a hint.

Let 2^k be the largest power of 2 less than or equal to N, so 2^k<=N<2^k+1. Assume the sum is an integer. Clear the denominators carefully so every term becomes an integer, except the 1/2^k term. This gives a contradiction.

I'll leave it to you how to clear them 'carefully', but seeing how to do this for the first few values of N might help.

 

[mathwonk]

if you clear denominators you see that n! must divide the top. It follows for example that n must divide (n-1)! which is impossible say, if n is prime. If not, and p is the largest prime number less than n, it follows again that p must divide some number between p+1 and n. I.e. n must be at least 2p. But this contradicts the fact that there must be another prime between p and 2p, hence another between p and n.

 

[daster]

This is kind of related, is:

\sum_{r=1}^{n} r!

ever a perfect square? I know this is true for n=1 and n=3, but what about in general?

I'd like hints only, please.

 

[matt grime]

I'd probably pick some useful number and look at things mod that number. for it is divisible by 3 for n>1, so it must be WHAT if it is a square?

 

[daster]

It must be in the form 3(n^2) or in the form 3^(2n+1)?

 

[Rogerio]

This is kind of related, is:

\sum_{r=1}^{n} r!

ever a perfect square? I know this is true for n=1 and n=3, but what about in general?

I'd like hints only, please.

For n>4, the last digit is always '3', so it will never be a perfect square.

 

[daster]

I swear I was just thinking the same thing!

m! where m>4 always ends in zero. So when n>4, we have 10p+(4!+3!+2!+1!)=10p+30+3=10q+3, but I really didn't know that if a number ends in 3 it's a perfect square.

 

[matt grime]

just work out the squares of the residues mod 10, which is waht i meant by considering a suitable mod thing

and if 3 divides N and N is a square then 9 divides N is what I meant

 

[daster]

just work out the squares of the residues mod 10, which is waht i meant by considering a suitable mod thing

and if 3 divides N and N is a square then 9 divides N is what I meant

Wow... That's brilliant.

Well done!

 

[daster]

For the sake of curiosity...

What you're saying is:
if N>3 and N = 0 (mod 3), then N = 0 (mod 9).
Right?

Can we generalize this to:
if N>3^n and N = 0 (mod 3), then N = 0 (mod 3^{n+1})?

 

[matt grime]

This is realiant on N being a square, you realize, and nothing to do with N being greater than some power of three, nor for that matter is it to do with 3.

Let N be the n'th power of q where q is an integer. Write q as the product of primes, now, what is the decomposition of N relative to q?


For example, let n=3 so we#re considering cubes. What are some cubes? 8, 27, 216. How may times do 2 and 3 occur each decomposition? it's always a multiple of n isn't it?

 

[daster]

I know this might sound naive, but why is that?

Wait, nevermind... I figured it out.

So, to sum up, if q divides p^n then so does q^n?

Thanks for all your help matt.

 

[matt grime]

Did you write out q as a product of primes? Raise it to the power n? What do you get?

for instance if q=12=2^2.3, then q^3= 2^^.3^3 doesn't it?

 

[matt grime]

in response to your last corrected qusetion, only if the divisor is a prime, it is one of the equivalent criteria for being prime (in the integers)

 

[daster]

Yeah, naturally.

Thanks again.

 

[Rogerio]

...but I really didn't know that if a number ends in 3 it's a perfect square.

And it is not!
As I had said, if a number ends in 3, it is not a perfect square.

There is no n>4 which satisfies your equation.

 

[daster]

And it is not!
As I had said, if a number ends in 3, it is not a perfect square.

Yes, sorry, that's what I meant.

 

[daster]

matt,
I have one more question about your proof.
S = 33 (mod 10), but does S really need to be divisible by 3? 43 = 33 (mod 10), but 43 isn't divisible by 3.

 

[matt grime]

what proof? I didn't say that I'd proved it, I offered two things for you to think about that might prove it. One of them does more easily than the other. I've no idea if the sum of the first r!'s is not divisible by 9 or not, but that seems a reasonable thing to consider. Divisbility by 3 and the modulo 10 bit are completely unrelated. The mod 10 bit proves it straightforwardly.


And I've no idea what's puzzling you in your question. Of course there's no reason for two numbers in the equivalence class of 3 mod 10 to both be divisible by 3, there's no reason for either of them to be divisible by 10. IF they are both divisible by 3, so is their difference, which, by definition, is also divislbe by 10, and hence by 30. And?

 

[daster]

Oh, sorry. I misread what you posted. I know that the mod 10 bit proves (since 33 isn't a square mod 10), but I thought you'd approached it from a different angle.

 

[matt grime]

I had. A little check, which I intended you to do demonstrates the div by 9 thing doesn't actually help.