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# Why this is a reasonable approximation? Tweet

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1. Re: Why this is a reasonable approximation?
(Original post by TheMagicMan)
Inclusion exclusion is exactly what you did there...it's just the fancy name for that method of alternating subtracting and adding when calculating the size of sets
Ahh right cool, I've read very little about sets and set theory (and so related techniques) to be honest.

Hmm did you realise a pattern the way dmon1Unlimited did it?

6000 - (1/2)(6000) = 3000 to get rid of multiples of 2
3000 - (1/3)(3000) = 2000 to get rid of multiples of 3
2000 - (1/5)(2000) = 1600 to get rid of multiples of 5

I.e. take a 1/2 to get rid of 2s, 1/3 to get rid of 3s, 1/5 to get rid of 5s, take 1/N to get rid of N (probably only work if all N1, N2 etc are coprime - right?)?

Maybe I'm just tired but I can't figure out why that pattern works.
2. Re: Why this is a reasonable approximation?
(Original post by Dog4444)
Yes, thanks. Now I'm wondering how he managed to do everything in his head.

With a bit of practice, you pretty much jump to the last step straight away.(That said, looking at the writing on my nearby paper, I did it a different way anyways)

(Original post by hassi94)
Ahh right cool, I've read very little about sets and set theory (and so related techniques) to be honest.

Hmm did you realise a pattern the way dmon1Unlimited did it?

6000 - (1/2)(6000) = 3000 to get rid of multiples of 2
3000 - (1/3)(3000) = 2000 to get rid of multiples of 3
2000 - (1/5)(2000) = 1600 to get rid of multiples of 5

I.e. take a 1/2 to get rid of 2s, 1/3 to get rid of 3s, 1/5 to get rid of 5s, take 1/N to get rid of N (probably only work if all N1, N2 etc are coprime - right?)?

Maybe I'm just tired but I can't figure out why that pattern works.
Yeah, one in every N numbers is divisible by N, so you take (1/N)(6000) to get rid of the multiples of N(if you meant something different here, sorry!)
3. Re: Why this is a reasonable approximation?
(Original post by Slumpy)
Yeah, one in every N numbers is divisible by N, so you take (1/N)(6000) to get rid of the multiples of N(if you meant something different here, sorry!)
My point is why are a third of odd numbers multiples of 3. And then again 1/5 of numbers not divisible by 2 or 3 are multiples of 5. I think I can see the simple logic behind it but unsure how I'd write down the reasoning.
4. Re: Why this is a reasonable approximation?
(Original post by hassi94)
My point is why are a third of odd numbers multiples of 3. And then again 1/5 of numbers not divisible by 2 or 3 are multiples of 5. I think I can see the simple logic behind it but unsure how I'd write down the reasoning.
I think it comes down to basic probability and independence: whether a number is divisible by 2 makes it no more probable that it is divisible by 5, whereas the fact that a number is divisible by 6 makes it much more likely that it is divisible by 12. In symbols P(5 divides x|2 divides x)=P(5 divides x) as P((5 divides x) n (2 divides x))=P(5 divides x) X P(2 divides x)
Last edited by TheMagicMan; 15-04-2012 at 14:25.
5. Re: Why this is a reasonable approximation?
(Original post by hassi94)
My point is why are a third of odd numbers multiples of 3. And then again 1/5 of numbers not divisible by 2 or 3 are multiples of 5. I think I can see the simple logic behind it but unsure how I'd write down the reasoning.
I would say it's basically because they're coprime. If a and b are coprime, the events a|x and b|x are independent. In fairness, I'm not sure if this really answers your question, because you could just ask why they're independent, but this isn't too hard:
Look in the region {1,...,ab}. There are a multiples of b in there, and b multiples of a(this should be fairly clear). So the probability of a number being divisible by b is a/ab=1/b, and a similar thing for a. Yes?