Math proof

Mathematics and Science; the subtle and ubiquitous arts

Math proof

Post by Jimmy_xor on Sat Apr 26, 2008 5:55 am
([msg=1386]see Math proof[/msg])

I can make some math proofs which are (obviously) wrong. Your mission is to find out why the proofs ain't correct. When someone have submitted a good explanation, I will give you a new one..

Let's start simple.
x = y
Divide by (x-y)
x/(x-y) = y/(x-y)
Subtract y/(x-y)
x/(x-y) - y/(x-y) = 0
Write on same denominator
(x-y)/(x-y) = 0
Simplify
1 = 0
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Re: Math proof

Post by ELorenz on Sat Apr 26, 2008 11:53 am
([msg=1406]see Re: Math proof[/msg])

Well the first issue is the fact that because x = y you can't divide by x - y, because then you are dividing by 0. The next problem I noticed is that ( x - y )/( x - y ) does not equal 1, if it were possible to divide by x - y it would equal 0/0, and therefore zero. so then you would get 0 = 0.
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Re: Math proof

Post by Inferno96 on Sat Apr 26, 2008 3:11 pm
([msg=1420]see Re: Math proof[/msg])

0/0 is indeterminate not 0 or undefined.
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Re: Math proof

Post by Jimmy_xor on Sat Apr 26, 2008 3:57 pm
([msg=1429]see Re: Math proof[/msg])

Very good ELorenz and Inferno96. It's not allowed to divide by 0.
Please tell me if it's too simple..

Here is the next "proof" (please give answers to both versions if you can):
I suppose you know complex numbers. sqrt(a) means square root of a, sqrt(-1)=i.
1 = sqrt(1)*sqrt(1) = sqrt(1*1) = sqrt(1) = sqrt((-1)*(-1)) = sqrt(-1)*sqrt(-1) = i*i = -1

Here is the version if you don't know what a complex number is.
1 = sqrt(1²) = sqrt(1*1) = sqrt(1) = sqrt((-1)*(-1)) = sqrt((-1)²) = -1
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Re: Math proof

Post by ELorenz on Sat Apr 26, 2008 4:28 pm
([msg=1435]see Re: Math proof[/msg])

The main issue for both is that the property sqrt(xy) = sqrt(x)*sprt(y) only holds for positive real numbers.

Here's a proof by induction to see if you can find the error.

for the series of numbers 1 + 2 + 3 +.....+ n = n(n+1)/2 first show it's true for the test case, n=1, 1 * ( 1 + 1) /2 = 1. it holds,
assume that it's true for the nth case, and show it's true for the n + 1 case.

see if you can tell what I did wrong.

n(n+1)/2 + (n+1)[(n+1)+1]/2 = [n(n+1)+(n+1)^2 + (n+1)]/2 = [n^2 + n + n^2 + 2n + 1 + n + 1 ]/2 = [2n^2 + 4n + 2 ]/2 =
= 2[n^2 + 2n + 1 ]/2 = n^2 + 2n + 1 which is not of the form n(n+1)/2
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Re: Math proof

Post by Jimmy_xor on Sat Apr 26, 2008 5:51 pm
([msg=1457]see Re: Math proof[/msg])

"n(n+1)/2 + (n+1)[(n+1)+1]/2" should be "n(n+1)/2 + (n+1)". Then continue:
n(n+1)/2 + (n+1) = (n^2+3n+2)/2 = [(n+1)^2+(n+1)]/2 = (n+1)((n+1)+1)/2
I never liked induction in high school, but it isn't very difficult.

I got inspired by your sum, so here is a sum for you.
You know (from MacLaurin) that ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + x^5/5 - x^6/6 + ...
Then ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...
Rearrange the terms
ln(2) = (1 + 1/3 + 1/5 + ...) - (1/2 + 1/4 + 1/6 + ...)
ln(2) = (1 + 1/3 + 1/5 + ...) + (1/2 + 1/4 + 1/6 + ...) - 2(1/2 + 1/4 + 1/6 + ...)
Combine the two first parentheses and move the 2 inside the third parenthesis.
ln(2) = (1 + 1/2 + 1/3 + 1/4 + ...) - (1 + 1/2 + 1/3 + 1/4 + ...)
Evaluate.
ln(2) = 0
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Re: Math proof

Post by ELorenz on Sat Apr 26, 2008 6:18 pm
([msg=1466]see Re: Math proof[/msg])

I don't know where you came up with the -2( 1/2 + ...) but I'm really not that good at series manipulation. But there is also the issue that that particular Mclaurin series is defined for -1 < x < 1 so ln(2) wouldn't be within the limits of the domain.
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Re: Math proof

Post by ELorenz on Sun Apr 27, 2008 2:42 pm
([msg=1523]see Re: Math proof[/msg])

Here's one from discrete math.

Show that the composition of 1 - 1 and onto functions is 1-1 and onto.

Let A, B, and C be sets. If f(x) is a 1 - 1 and onto mapping from A --> B and g(x) is a 1-1 and onto mapping from B --> C, show
g(f(x)) is 1-1 and onto.

PF: Let x be an element of C. Then we know that since g(x) is 1 - 1 and onto then there is an element, call it y, of B that
f(y) = x. Then we also know that since f(x) is 1 - 1 and onto there exists an element z of A such that g(z) = y. Since we have an element in C that we can show is an element of A operated on by the two functions then g(f(x)) is 1-1 and onto.

There are a lot of issues with this one, see if you can find them all.
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Re: Math proof

Post by Jimmy_xor on Sun Apr 27, 2008 4:08 pm
([msg=1526]see Re: Math proof[/msg])

Here is the corrected version of your proof.
Let A, B, and C be sets. If f(x) is a 1 - 1 and onto mapping from A --> B and g(x) is a 1-1 and onto mapping from B --> C, show
g(f(x)) is 1-1 and onto.
PF: Let x be an element of C. Then we know that since g(x) is 1 - 1 and onto then there is exactly one element, call it y, of B that
g(y) = x. Then we also know that since f(x) is 1 - 1 and onto there exists exactly one element z of A such that f(z) = y. Since we have an element in C that we can show is an element of A operated on by the two functions then g(f(x)) is 1-1 and onto.
Please tell me if I forgot to correct something. The answer to ln(2)=0 was that the series was conditionally convergent, and it's not allowed to rearrange the terms in such series.

My next "proof" is very simple (or it should be at least):
1$=100c=(10c)²=(0.1$)²=0.01$=1c
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Re: Math proof

Post by yourmysin on Sun Apr 27, 2008 5:06 pm
([msg=1537]see Re: Math proof[/msg])

Such easy proofs :)

Give us something a bit harder. im slightly disappointed that I didn't get the chance to respond to the previous challenges.

1$=100c=(10c)²=(0.1$)²=0.01$=1c

100c=(10sqrt(c))²
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