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# Slighly not higher but maths none the less

Started by jambo david, Apr 22 2006 08:33 PM

9 replies to this topic

### #1

Posted 22 April 2006 - 08:33 PM

A friend showed me a little series of equations the other day, i can't remember them but someone here might know them. It started off a=b and ended 2=1

edit- found it:D

a = b

a^2 = ab

a^2 − b^2 = ab − b^2

(a − b)(a + b) = b(a − b)

a + b = b

2b = b

2 = 1

I worked it out why it doesn't work quite quickly

edit- found it:D

a = b

a^2 = ab

a^2 − b^2 = ab − b^2

(a − b)(a + b) = b(a − b)

a + b = b

2b = b

2 = 1

I worked it out why it doesn't work quite quickly

### #2

Posted 22 April 2006 - 09:37 PM

Yeah, it comes up in one of the 'Proofs' bit in Advanced Higher.

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### #3

Posted 22 April 2006 - 09:51 PM

known as proof by contridiction. This means that your assumption that a=b is wrong

If i am not here i am somewhere else

### #5

Posted 22 April 2006 - 11:35 PM

QUOTE

known as proof by contridiction. This means that your assumption that a=b is wrong

so in fact, a b ?

but could it not be a = b = 0?

### #6

Posted 23 April 2006 - 12:26 AM

yeh i guess so

However if we were to say a and b were all the positive integers a random set where a and b werent zero then proof by contridiction is what you would use

we did something like this for proof by induction at uni where we were to proof that everyone has red hair. Whats more we proved it!

However if we were to say a and b were all the positive integers a random set where a and b werent zero then proof by contridiction is what you would use

we did something like this for proof by induction at uni where we were to proof that everyone has red hair. Whats more we proved it!

If i am not here i am somewhere else

### #7

Posted 23 April 2006 - 12:42 AM

interesting

### #9

Posted 23 April 2006 - 11:13 AM

known as proof by contridiction. This means that your assumption that a=b is wrong

That's not correct! This is an example of a 'sophism' - a proof which uses faulty reasoning to 'prove' an absurd result.

It is true that a=b here, throughout the proof. However, the reasoning breaks down at some point, and gives an obviously false result.

(a − b)(a + b) = b(a − b)

a + b = b

This is the crucial step, where you are dividing by a - b = a - a = 0. This is basically illegal in maths, and the result you get from this 'proof' shows exactly why division by zero isn't allowed

There are all sorts of fake proofs like that - I think there's one in Scholar of the statement "All cows are cannibals", using ambiguous language.

### #10

Posted 28 April 2006 - 01:30 PM

In an exam you are not supposed to set any type of equation / function equal to an number/ other function to solve as this can result in what just happened in your equation.

The equation must be set equal to zero before solving by factorisation etc....

The equation must be set equal to zero before solving by factorisation etc....

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