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# Use proof by contradiction to prove that log2(3) is irrational. watch

1. ??? Saw this in a C3 Solomon paper. How do I do it?
2. What is log2(3)?
3. ...
4. (Original post by Drederick Tatum)
What is log2(3)?
Sorry, the 2 is the base of the log
5. (Original post by SimonM)
...
Whaaat? Are we supposed to have covered this in C3? I've never come across this type of question before.
6. Expand it out in terms of p/q
7. (Original post by pimpman777)
Whaaat? Are we supposed to have covered this in C3? I've never come across this type of question before.
I'm surprised it's on C3, but I don't think anything I've written is off syllabus
8. Oh OK, I guess I have some more revision to do on logs lol
9. You may as well assume p and q are positive integers (in what simon M has said) since the log(3) > log(2) = 1.
10. An answer, in two stages: stop after first if you think you can see rest:
Stage 1:
Spoiler:
Show

So as SimonM did above, we do this:
(Original post by SimonM)
...
What we are doing here is assuming that is a rational number: i.e. is the rato of two integers p and q. This implies that for some integer p, q, (by rearrangement really).

Stage 2:
Spoiler:
Show

Note that 3 and 2 are mutually prime: i.e. they share no factors (and are in fact both actual prime numbers). But these numbers, and , are meant to be equal and must therefore have the same, unique, prime number expansion. However if they don't share any factors... So p and q do not exist: there are no integers p and q for which the equality is true.

Hence our original assumption led to a contradiction: so is not rational (so it is irrational).

Bingo!
11. (Original post by Salavant)
An answer, in two stages: stop after first if you think you can see rest:
Stage 1:
Spoiler:
Show

So as SimonM did above, we do this:

What we are doing here is assuming that is a rational number: i.e. is the rato of two integers p and q. This implies that for some integer p, q, (by rearrangement really).

Stage 2:
Spoiler:
Show

Note that 3 and 2 are mutually prime: i.e. they share no factors (and are in fact both actual prime numbers). But these numbers, and , are meant to be equal and must therefore have the same, unique, prime number expansion. However if they don't share any factors... So p and q do not exist: there are no integers p and q for which the equality is true.

Hence our original assumption led to a contradiction: so is not rational (so it is irrational).

Bingo!
Thanks for that I'm still very baffled on why it came up on the C3 paper though, we've never been through this
12. (Original post by pimpman777)
Oh OK, I guess I have some more revision to do on logs lol
The solomon papers where madeon the old syllabus, proof by contradiction isn't on it anymore.
13. (Original post by usainlightning)
The solomon papers where madeon the old syllabus, proof by contradiction isn't on it anymore.
Oh right that explains it
14. i stil do not understand
15. (Original post by misstee91)
i stil do not understand
Proof by contradiction is a method of proof whereby you assume the conclusion is false, and then show this assumption leads to something which can't be true (e.g. 1=0 or "2 is odd").

A number is rational if it is in the form , where are integers ().

Piecing this together, we want to show that is irrational; i.e. that it can't be written in the form for any integers . So, we start our proof by assuming that there exist integers p,q (q nonzero) such that .

By the definition of logarithms, this gives , and raising both sides to the power of q gives . This can only happen if , but we can't have so this can't be true, so the assumption can't be true, so it must be false; hence the proposition is true.
16. nuodai, would you be able to just explain the last stage of this? I don’t quite see why there are no integers p and q for which 2^p=3^q
10^4=100^2 but there the bases have common factors. Is that the point?
17. (Original post by Plato's Trousers)
but surely that only proves there is only one solution of 3^x=2^x (ie x=0). But it doesn't imply that there cannot be two different integers (p and q) for which 2^p=3^q does it?

Sorry, if I'm being thick
2^p=3^q=n

If p and q are not zero then n has two different prime factorisations. This is not possible. You might want to check the proof of that statement.

http://en.wikipedia.org/wiki/Fundame..._of_arithmetic
18. (Original post by Get me off the £\?%!^@ computer)
2^p=3^q=n

If p and q are not zero then n has two different prime factorisations. This is not possible. You might want to check the proof of that statement.

http://en.wikipedia.org/wiki/Fundame..._of_arithmetic
aha! Eureka! Thank you
19. Rather than resort to FTA, in this case it is surely simpler to note that 2^p is even, while 3^q is odd.

(Which some will say is just a special case of the FTA, but I think what I've said is much more accessible at C1-C4 level).
20. (Original post by DFranklin)
Rather than resort to FTA, in this case it is surely simpler to note that 2^p is even, while 3^q is odd.
Oh yes! Nice one.

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