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## #76 2005-04-26 02:04:11

Mr T
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### Re: Can this be possible?

SURE. a "friendly" chat.

I come back stronger than a powered-up Pac-Man
I bought a large popcorn @ the cinema the other day, it was pretty big...some might even say it was "large
Fatboy Slim is a Legend

## #77 2005-04-26 19:57:33

jam-pot
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### Re: Can this be possible?

dude your called mrT for gods sake you think you are a T.V star  but did you know that mr T went bankrupt and now does not own his gold coloured pots and pans

the allmighty spatula * want a tip* dont eat yellow snow: the meaning of life is a number and that number is 1

## #78 2005-05-04 03:38:25

Mr T
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### Re: Can this be possible?

actually i was originally called WTF but then my name was changed foo'

I come back stronger than a powered-up Pac-Man
I bought a large popcorn @ the cinema the other day, it was pretty big...some might even say it was "large
Fatboy Slim is a Legend

## #79 2005-05-08 08:51:14

Milos
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### Re: Can this be possible?

Let's take this for exapmle:
a=a
a+a-a=a  , now devide it with a+a
1-a/(a+a)=a/(a+a)
1=1 correct
but if you devide it with a-a then you have
(a-a)/(a-a)+a/(a-a)=a/(a-a)        (a-a)/(a-a) > 0/0 THIS IS NOT  EQUAL TO 1 so we do not have the problem 1=0. If we presume that 0/0=0 than equation would be correct, and it is the only way for it to be correct. SO:
0+a/(a-a)=a/(a-a).Is it true that 0/0=0 - my calculator doesn't think so. ????
The same problem occurred when you(administrator) devided (a-b) and (a-b) where a=b. You calculated that it equals 1 but it is incorrect.
But what I can not solve is the first problem where everithing is ok until this:
x^2-1=x-1 - this is also ok because we said that x=1
(x+1)(x-1)=x-1 - this is also ok - my appologies
THE SAME PROBLEM AS BEFORE 0/0 is not 1

Last edited by Milos (2005-05-08 18:07:07)

## #80 2005-05-08 17:11:49

MathsIsFun

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### Re: Can this be possible?

This one, Milos?

#### Roraborealis wrote:

My maths tutor told me this, and I'm very curious about it. Every time I try to follow it, I get confused. It apparently proves that 1=0.
x=1.
Multiply both sides by x and you get
x^2=x
Take away 1 from each side which becomes
x^2-1=x-1
This can also be expressed as
(x+1)+(x-1)=x-1
Divide each side by x-1 and the answer is:
x+1=1
From this, you can see that x=0. But at the beginning I said that x=1. It has therefore been proved that 1=0.

Is this a trick?

Isn't it just because we are dividing by zero again? "Divide each side by x-1", but we stated that x=1?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #81 2005-05-08 18:04:57

Milos
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### Re: Can this be possible?

Yes that is corret. The same mistake as it was before. My calculation in second pard wasn't good. (x+1)(x-1)=x-1  , x=1 so THIS IS ALSO CORRECT, but I realised that this morning - my appologies.

## #82 2005-05-08 18:19:36

Roraborealis
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### Re: Can this be possible?

Are you a mathematician?

School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?

## #83 2005-05-31 21:15:24

Mr T
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### Re: Can this be possible?

no...he is a donkey taken back in time by Dr. Who

I come back stronger than a powered-up Pac-Man
I bought a large popcorn @ the cinema the other day, it was pretty big...some might even say it was "large
Fatboy Slim is a Legend

## #84 2005-07-09 04:03:03

Roraborealis
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### Re: Can this be possible?

Riiiiiiight...........

School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?

## #85 2005-07-09 16:50:28

ganesh
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### Re: Can this be possible?

#### Milos wrote:

Yes that is corret. The same mistake as it was before. My calculation in second pard wasn't good. (x+1)(x-1)=x-1  , x=1 so THIS IS ALSO CORRECT, but I realised that this morning - my appologies.

Choose arbitrary a and b, and let t = a + b. Then
a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b

This fallacy arises as a result of taking the square-root of both sides of an equation.

Character is who you are when no one is looking.

## #86 2005-07-09 18:04:00

mathsyperson
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### Re: Can this be possible?

Why isn't that allowed? It is because of the ± or something?

Why did the vector cross the road?
It wanted to be normal.

## #87 2005-07-09 21:41:43

ganesh
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### Re: Can this be possible?

You are right! -n is not equal to n, if n ≠ 0

Character is who you are when no one is looking.

## #88 2005-07-10 19:42:34

ganesh
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### Re: Can this be possible?

This one I found in a jokes website:-

Theorem: 3=4
Proof:

Suppose:
a + b = c

This can also be written as:
4a - 3a + 4b - 3b = 4c - 3c

After reorganizing:
4a + 4b - 4c = 3a + 3b - 3c

Take the constants out of the brackets:
4 * (a+b-c) = 3 * (a+b-c)

Remove the same term left and right:
4 = 3

PS:- In the penultimate step, we multiply 4 and 3 with (a+b-c), which is actually zero! This leads to the absurd conclusion!!

Character is who you are when no one is looking.