A math puzzle for you!

ntg

ntg

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Joined
24 Nov 2008
Posts
2,499
alright,


Let's say n=m,

Then nm=n^2

Then we just choose to subtract m^2 from both sides (it doesn't affect the equation) nm-m^2=n^2-m^2

Then we got m(n-m)=(n-m)(n+m)

Let's go on and divide with n-m and we have

m(n-m)/(n-m)=(n-m)(n+m)/(n-m) which gives us

m = n+m,

since n=m this means that m=2m

If we divide by m both sides then 1=2.

So what's wrong with this???
 
m(n-m)/(n-m)=(n-m)(n+m)/(n-m) which gives us


at this part change every m to an n because n = m
n(n-n)/(n-n) = (n-n)(n+n)/(n-n)

so you have
0/0 = 0/0

You've just rearranged everything into nothing. Why would you want to do this anyway?
n = m
nm = n^2
so nn = n^2
n^2 = n^2

I don't see why you're rearranging stuff that just equals itself.
 
1 = 3/3

1/3 = 0.3 recurring
0.3 recurring * 3 = 0.9 recurring

Therefore 1 = 0.9 recurring

My maths teacher always used to pull this out, quite silly really.
 
*Insert overused divide by zero picture*

Ok

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2aep9gx.jpg

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1 = 3/3

1/3 = 0.3 recurring
0.3 recurring * 3 = 0.9 recurring

Therefore 1 = 0.9 recurring

My maths teacher always used to pull this out, quite silly really.

So? 0.9999recurring IS actually equal to 1, it's not a fallacy or inequality.

The proof is simple really as let x = 0.3333333
10x = 3.333333
subtract x
10x-x = 3.333333-0.333333
9x=3
3x=1, so 3*0.333333333 = 1

0.x recurring is accepted in maths that it is equal to an integer.
 
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