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Shroomerited


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Quote:
Celestial Traveler said: Or maybe
a*b = (2c)*b = 2cb --> 2cb/2 = c*b (integer product is integer) therefore (a*b) is even.
You don't even need that last part. When we say that a number is even, we mean that it equals 2 times some integer.
But yes, that's it.
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JMcDoogle
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Shroomerited


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Re: How to learn advanced math? [Re: JMcDoogle]
#19163144 - 11/20/13 12:52 AM (10 years, 3 months ago) |
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Now prove n even implies n^2 even. After that, prove n^2 odd implies n is odd.
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Celestial Traveler
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Re: How to learn advanced math? [Re: Shroomerited]
#19163162 - 11/20/13 12:57 AM (10 years, 3 months ago) |
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n = 2k (n,k) E N [i.e. n and k are natural numbers].
n^2 = 4k^2 = 2*(2*k*k) ---> 2*k*k E N because (2,k,k) E N (i.e. natural numbers)
n = 2k - 1 , n,k positive integers
n^2 = 4k^2 - 4k + 1 = (4k^2 - 4k + 2) - 1
= 2*(2k^2 - 2k + 1) - 1
[2k^2 - 2k + 1 is an integer because product of integers, and addition or subtraction of integers is an integer)
n^2 is odd.
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Shroomerited


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Quote:
Celestial Traveler said: n = 2k (n,k) E N [i.e. n and k are natural numbers].
n^2 = 4k^2 = 2*(2*k*k) ---> 2*k*k E N because (2,k,k) E N (i.e. natural numbers)
n = 2k - 1 , n,k positive integers
n^2 = 4k^2 - 4k + 1 = (4k^2 - 4k + 2) - 1
= 2*(2k^2 - 2k + 1) - 1
[2k^2 - 2k + 1 is an integer because product of integers, and addition or subtraction of integers is an integer)
n^2 is odd.
First one is good.
But for the second one, you proved that n odd implies n^2 odd. I asked you to prove that n^2 odd implies n is odd.
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Celestial Traveler
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Re: How to learn advanced math? [Re: Shroomerited]
#19163173 - 11/20/13 01:01 AM (10 years, 3 months ago) |
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Ah I see...ok...
Workin on it again then...although technically by proving n^2 must be even if n is even, I have already proved that n^2 is odd implies n is odd.
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Shroomerited


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I'm being a bit of a smartass here though.
Quote:
Celestial Traveler said: n = 2k (n,k) E N [i.e. n and k are natural numbers].
n^2 = 4k^2 = 2*(2*k*k) ---> 2*k*k E N because (2,k,k) E N (i.e. natural numbers)
Therefore n even implies n^2 even. Also equivalently n^2 odd implies n odd.
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Shroomerited


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Quote:
Celestial Traveler said: Ah I see...ok...
Workin on it again then...although technically by proving n^2 must be even if n is even, I have already proved that n^2 is odd implies n is odd.
Yup! That was my point.
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Celestial Traveler
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Working backwards then...
n^2 = 2t - 1 (n and t are integers), n^2 is odd.
Let t = 2k^2 - 2k + 1 , with k an integer.
then
n^2 = 4k^2 - 4k + 1
n = 2k - 1
n is odd.
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Celestial Traveler
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Shroomerited


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Quote:
Celestial Traveler said: Working backwards then...
n^2 = 2t - 1 (n and t are integers), n^2 is odd.
Let t = 2k^2 - 2k + 1 , with k an integer.
then
n^2 = 4k^2 - 4k + 1
n = 2k - 1
n is odd.
Approaching it that way, you need to prove that t is of the form _____.
But the main idea was just that you already proved it!
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lessismore
Registered: 02/10/13
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Re: How to learn advanced math? [Re: skatealex2]
#19163203 - 11/20/13 01:09 AM (10 years, 3 months ago) |
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Quote:
skatealex2 said: Anyone have any ideas how to learn advanced math?
Also if you can do advanced math did it come easy to you, did you learn in high school or later in college, etc.?
I'm thinking of trying math and see if I'm capable of going advancd with it. In the past I've sucked at math but I still feel like it could ultimately be a matter of studying and learning it even though some of it looks complicated as #%^*. I also realize some prestigious careers involve math and calculus so there's that too.

The same way you learn anything else
through hard work for years, or decades
if you want to learn advanced maths, good luck if you never had luck with it before 
not impossible, but you will have to work for (at least) 10 years before you get good at it likely/get a degree that is long enough for a job in the field
do what you love instead
and if that is math, keep on working on it... anything is possible if you dedicate enough time
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Celestial Traveler
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Re: How to learn advanced math? [Re: Shroomerited]
#19163204 - 11/20/13 01:09 AM (10 years, 3 months ago) |
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Do I win a million dollars?
Btw have you heard about the Beal Conjecture, and the millionaire (or billionaire) who has offered a large sum as a reward to whomever can prove it?
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Celestial Traveler
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I think it'd be really cool, but I've looked at it and I just have no clue how I would tackle it.
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Acidic_Sloth
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Re: How to learn advanced math? [Re: skatealex2]
#19163208 - 11/20/13 01:10 AM (10 years, 3 months ago) |
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practice. khan academy is a good place to start. start with the lower level math to reinforce the really important foundational concepts which crop up all the time in advanced math. a good foundation goes a long fucking way. get comfortable with fractions.
anyone can be good at math with enough practice. i can't say that enough in my tutoring sessions.
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Shroomerited


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Quote:
Celestial Traveler said: Do I win a million dollars?
Btw have you heard about the Beal Conjecture, and the millionaire (or billionaire) who has offered a large sum as a reward to whomever can prove it?
Mhmm. It would be interesting to see if it's true or not.
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Celestial Traveler
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Re: How to learn advanced math? [Re: Shroomerited]
#19163235 - 11/20/13 01:19 AM (10 years, 3 months ago) |
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It is true (both the conjecture and the fact that there is a prize offered for solving it) but it is just very hard to prove.
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Celestial Traveler
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Shroomerited


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If you're up for a challenge, prove the Archimedean Property of the natural numbers. That is, prove the set of natural numbers has no upper bound. Proof is straight forward, but needs to be done.
Hint: every set of real numbers that has an upper bound, has a least upper bound. ie the lub of (0,1) is 1. Or the lub of the set {0,2,34} is 34. An upper bound is any number greater than the set. Like 2 or 4 or 35344 are upper bounds for (0,1) but 1 is the least.
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stimpson
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just watch sum PE exam prep videos on youtube and do cocaine.
i thought all MIT math whizzes luved cocaine.
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