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

1. Dec 1, 2004

### Astronamus

I am deleting

Last edited: Dec 1, 2004
2. Dec 1, 2004

### Tide

Does this help?

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

Last edited: Dec 1, 2004
3. Dec 1, 2004

### 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.

4. Dec 2, 2004

### 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.

5. Jan 4, 2005

### 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?

6. Jan 4, 2005

### 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?

Last edited: Jan 4, 2005
7. Jan 4, 2005

### daster

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

8. Jan 4, 2005

### Rogerio

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

9. Jan 4, 2005

### 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.

10. Jan 4, 2005

### 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

11. Jan 4, 2005

### daster

Wow... That's brilliant.

Well done!

12. Jan 4, 2005

### 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})?

13. Jan 4, 2005

### 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?

14. Jan 4, 2005

### 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.

Last edited by a moderator: Jan 4, 2005
15. Jan 4, 2005

### 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?

16. Jan 4, 2005

### 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)

17. Jan 4, 2005

### daster

Yeah, naturally.

Thanks again.

18. Jan 4, 2005

### Rogerio

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.

19. Jan 4, 2005

### daster

Yes, sorry, that's what I meant. :rofl:

20. Jan 5, 2005

### daster

matt,
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.

21. Jan 6, 2005

### 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?

Last edited: Jan 6, 2005
22. Jan 6, 2005

### 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.

23. Jan 6, 2005

### matt grime

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