Elementry number theory question... please help all you math people..

[chiwawa626]

I can't seem to figure this out:

Lets say you have a|(n+m) and a|(n-m). Does it follow that a|n, if it is also known that:
1. a is even
2. a is odd


(the notation | means "divides")

 

[Fiveohhh]

should use / for division iirc | is absolute value

 

[upsciLLion]

I am obviously incapable of reading.

ups

 

[upsciLLion]

Originally posted by: Fiveohhh
should use / for division iirc | is absolute value

a|b means a divides evenly into b, or more specifically it implies b=ac where c is an integer.

 

[chiwawa626]

well not divides, saying a|z means z is divisible by a... | is the notation my professor and my book uses, ive seen it elsewhere as well. |a| is absolute value or magnitude

 

[Fiveohhh]

Originally posted by: chiwawa626
well not divides, saying a|z means z is divisible by a... | is the notation my professor and my book uses, ive seen it elsewhere as well. |a| is absolute value or magnitude

ahh been a few yrs since math...

 

[chiwawa626]

Originally posted by: upsciLLion
No. 5|(2+3) where a=2 and 5|(3+2) where a=3, but in neither case does 5|a.

ups

btw, a|(n+m) and a|(n-m) is given so from ur example that dosent make sense then? 5|(2+3) but 5 (not) |(2-3)

 

[gopunk]

Originally posted by: upsciLLion
No. 5|(2+3) where a=2 and 5|(3+2) where a=3, but in neither case does 5|a.

ups

i don't think that's what he was asking...

 

[gopunk]

Originally posted by: chiwawa626
I can't seem to figure this out:

Lets say you have a|(n+m) and a|(n-m). Does it follow that a|n, if it is also known that:
1. a is even
2. a is odd


(the notation | means "divides")

well if n is odd and m is 1, then a|n is not true if a is even. note that 2m has to be a multiple of a... i can thnk of a case where it doesn't hold...

 

[chiwawa626]

Yeah the given situation is the numbers a, n, m satisfy a|(n+m) and a|(n-m).

So does it follow that a|n if a is odd or even?

 

[Muzzan]

If a | (n + m) and a | (n - m), we must have:

{ (n + m) / a = p
{ (n - m) / a = o

for some natural numbers p and o. Suppose now that a is even (i.e a = 2k for some natural k):

{ (n + m) / (2k) = p
{ (n - m) / (2k) = o

Or equivalently:

{ n + m = (2k)p
{ n - m = (2k)o

Now, add these two equations to get:

n + m + n - m = (2k)p + (2k)o
2n = (2k)p + (2k)o
(2n) / (2k) = p + o
n/k = p + o
n / (a/2) = p + o
n/a * 2 = p + o

If one could somehow motivate that p + o is even, we would be done... Numerical evidence suggests that p is odd and o is odd, but I have no proof.

 

[cchen]

Originally posted by: Muzzan
If a | (n + m) and a | (n - m), we must have:

{ (n + m) / a = p
{ (n - m) / a = o

for some natural numbers p and o. Suppose now that a is even (i.e a = 2k for some natural k):

{ (n + m) / (2k) = p
{ (n - m) / (2k) = o

Or equivalently:

{ n + m = (2k)p
{ n - m = (2k)o

Now, add these two equations to get:

n + m + n - m = (2k)p + (2k)o
2n = (2k)p + (2k)o
(2n) / (2k) = p + o
n/k = p + o
n / (a/2) = p + o
n/a * 2 = p + o

If one could somehow motivate that p + o is even, we would be done... Numerical evidence suggests that p is odd and o is odd, but I have no proof.

*claps* well done

 

[fizmeister]

cchen:

How's the operations research program at Columbia? I'll have my applied math degree next year (with a slew of graduate couress) and I'm still trying to figure out where to apply (besides the top 5).

 

[Amorphus]

a|n only follows if a|m, also, I think.

 

[cchen]

Originally posted by: fizmeister
cchen:

How's the operations research program at Columbia? I'll have my applied math degree next year (with a slew of graduate couress) and I'm still trying to figure out where to apply (besides the top 5).

depends on what you want to specialize in. mainly here because its in nyc the OR program focuses on finance, financial engineering, options pricing, derivatives, etc. stuff like that. also pretty good in optimization and stochastics.

I'm probably gonna do a master's in OR after I finish my undergrad here since I won't have to take any of the OR core courses

 

[gopunk]

Originally posted by: Muzzan
If a | (n + m) and a | (n - m), we must have:

{ (n + m) / a = p
{ (n - m) / a = o

for some natural numbers p and o. Suppose now that a is even (i.e a = 2k for some natural k):

{ (n + m) / (2k) = p
{ (n - m) / (2k) = o

Or equivalently:

{ n + m = (2k)p
{ n - m = (2k)o

Now, add these two equations to get:

n + m + n - m = (2k)p + (2k)o
2n = (2k)p + (2k)o
(2n) / (2k) = p + o
n/k = p + o
n / (a/2) = p + o
n/a * 2 = p + o

If one could somehow motivate that p + o is even, we would be done... Numerical evidence suggests that p is odd and o is odd, but I have no proof.

nice i think you can just do it by contradiction for evens though...

if a is even:
choose a = 2, n = 3, m = 1. 2 | (3 + 1) and 2 | (3 - 1), but it is not true that 2 | 3. therefore we know that it does not follow in the first case.

 

[gopunk]

and for odds:

if a is odd, then let a = 2k + 1 for k >= 0

then

n + m / 2k + 1 = p
n - m / 2k + 1 = o

so

n + m = p(2k + 1)
n - m = o(2k + 1)

sum up

2n = (2k + 1) (p + o)
2n / (2k + 1) = p + o

p and o are integers by definition, so their sum must also be an integer.

2n must be even since n is an integer, and (2k + 1) is odd since k is an integer.

thus, 2n must be an even multiple of 2k + 1... that is 2n = 2j * (2k + 1), for some integer j >= 0, since an odd multiple of an odd number would result in an odd number.

2k + 1 = a, thus 2n must be an even multiple of a. if 2n is an even multiple of a, then 2n = 2ja for some integer j >= 0.

n = ja -> n / a = j, j is an integer by definition, thus the hypothesis holds for odds.

 

[huesmann]

I'd advise you to brush up on your verbal skills too in addition to your mathematical skills.

 

[Muzzan]

nice i think you can just do it by contradiction for evens though...

if a is even:
choose a = 2, n = 3, m = 1. 2 | (3 + 1) and 2 | (3 - 1), but it is not true that 2 | 3. therefore we know that it does not follow in the first case.

Doh, that's a pretty simple counterexample... Can't believe I overlooked it