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**Mr T****Member**- Registered: 2005-03-30
- Posts: 1,012

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

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**jam-pot****Member**- Registered: 2005-04-20
- Posts: 142

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

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**Mr T****Member**- Registered: 2005-03-30
- Posts: 1,012

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

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**Milos****Member**- Registered: 2005-05-06
- Posts: 44

About the problem 1=0

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

I hope this was helpful

*Last edited by Milos (2005-05-07 20:07:07)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,713

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

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**Milos****Member**- Registered: 2005-05-06
- Posts: 44

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.

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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

Are you a mathematician?

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

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**Mr T****Member**- Registered: 2005-03-30
- Posts: 1,012

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

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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

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

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

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 48,001

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.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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

Why did the vector cross the road?

It wanted to be normal.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 48,001

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

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 48,001

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!!

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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