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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: 7,664

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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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,209

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 is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,209

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

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,209

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 is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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

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