1 = 0

TripleAce · 10-12-2008 03:29PM #1

a=b

therefore

0=b-a

Now we divide both by b-a:

0/(b-a) =(b-a)/(b-a)

and so:

0 = 1

Why?

henbane · 10-12-2008 03:44PM

You're dividing by zero which doesn't work

seamus · 10-12-2008 03:48PM

0/0 = undefined != 1

MathsManiac · 11-12-2008 10:09PM

You've made the puzzle for too transparent! You need to try to bury the division by 0 more, so that it's a bit harder to spot. Try this:

let a=b=1
multiply both sides by b: ab = b^2
subtract a^2 from both sides: ab - a^2 = b^2 - a^2
Factorise: a(b-a) = (b-a)(b+a)
Cancel the common factor: a = b+a
But we already know that a=b=1: 1 = 1+1
That is: 1 = 2.

TripleAce · 12-12-2008 12:12PM

MathsManiac wrote: »

You've made the puzzle for too transparent! You need to try to bury the division by 0 more, so that it's a bit harder to spot. Try this:

let a=b=1
multiply both sides by b: ab = b^2
subtract a^2 from both sides: ab - a^2 = b^2 - a^2
Factorise: a(b-a) = (b-a)(b+a)
Cancel the common factor: a = b+a
But we already know that a=b=1: 1 = 1+1
That is: 1 = 2.

COOL

McSeamus ORiley · 13-12-2008 02:52AM

For all n, n/n = 1

at n=0 0/0 = 1

so 0/0 + 0/0 = 2

but 0/0 + 0/0 = (0+0)/0 = 0/0 =1

giving 1=2

add 1 to both sides 2=3

but 2=1 so 1=3

likewise 1=4 1=5 1=6 .....

Therefore for all n, n=1

:pac:

LeixlipRed · 13-12-2008 11:52AM

But that breaks down straight away when you say for all n, n/n=1. Then you say 0/0=1

McSeamus ORiley · 13-12-2008 12:50PM

Dumbass

jhegarty · 13-12-2008 12:53PM

TripleAce wrote: »

(b-a)/(b-a)

I am surprised boards didn't crash with a divide by zero error

Thomas_S_Hunterson · 13-12-2008 12:55PM

McSeamus ORiley wrote: »

For all n, n/n = 1

at n=0 0/0 = 1

so 0/0 + 0/0 = 2

but 0/0 + 0/0 = (0+0)/0 = 0/0 =1

giving 1=2

add 1 to both sides 2=3

but 2=1 so 1=3

likewise 1=4 1=5 1=6 .....

Therefore for all n, n=1

:pac:

1^1 = 1
1^0 = 1
=> 1=0
:pac::p

LeixlipRed · 13-12-2008 01:04PM

McSeamus ORiley wrote: »

Dumbass

The whole point of a fallacy is that you don't explicitly state the mistake in the second line of the "proof".

LeixlipRed · 13-12-2008 01:14PM

Sean_K wrote: »

1^1 = 1
1^0 = 1
=> 1=0
:pac::p

Much more subtle this time

Here's another one that doesn't involve division by 0.

4 - 6 = 1 - 3
4 - 6 + 9/4 = 1 - 3 + 9/4
(2 - 3/2)^2 = (1 - 3/2)^2
2 - 3/2 = 1 - 3/2
2 = 1

McSeamus ORiley · 13-12-2008 01:14PM

LeixlipRed wrote: »

The whole point of a fallacy is that you don't explicitly state the mistake in the second line of the "proof"

PROTIP: It's just a joke. :rolleyes:

LeixlipRed · 13-12-2008 01:17PM

Here's another one I'd forgotten about as well. The derivative of x^2 is 2x using the Power Rule. But x^2=x+x+....+x (x times) e.g 4^2=16 or 4+4+4+4=16. The derivative of x is one so the derivative of x^2=x+x+...+x (x times) is 1 added up x times or just x!

There are two things wrong with this "proof" by the way.

McSeamus ORiley · 13-12-2008 01:23PM

Rectangles and triangles.

Thomas_S_Hunterson · 13-12-2008 01:29PM

Conjecture:
McSeamus ORiley==e05bf05a

LeixlipRed · 13-12-2008 01:35PM

Sean_K wrote: »

Conjecture:
McSeamus ORiley==e05bf05a

Well they're banned now but maybe if they come back we might investigate that

Michael Collins · 13-12-2008 06:50PM

LeixlipRed wrote: »

Much more subtle this time

Here's another one that doesn't involve division by 0.

4 - 6 = 1 - 3
4 - 6 + 9/4 = 1 - 3 + 9/4
(2 - 3/2)^2 = (1 - 3/2)^2
2 - 3/2 = 1 - 3/2
2 = 1

Too obvious! But better than the divide by zero one...

Spoiler

It's the fourth line here that has the false step - need to take account of the possiblity of the root being negative...which it turns to be in this case

LeixlipRed wrote: »

Here's another one I'd forgotten about as well. The derivative of x^2 is 2x using the Power Rule. But x^2=x+x+....+x (x times) e.g 4^2=16 or 4+4+4+4=16. The derivative of x is one so the derivative of x^2=x+x+...+x (x times) is 1 added up x times or just x!

There are two things wrong with this "proof" by the way.

Spoiler

Well x is a variable, so the amount of x's on the RHS will be a function of x, so you cannot just assume it's constant. Not sure about the second error, is it to with the fact derivatives only operate on functions, and the RHS isn't a proper function?

How about this one:

1 = sqrt (-1 * -1)
1 = sqrt(-1)*sqrt(-1)
1 = i * i
1 = -1

LeixlipRed · 13-12-2008 07:09PM

Michael

Spoiler

You can only right x^2 as a sum like that if x is a positive integer. Then obviously you can't differentiate. Suppose that's not two seperate things wrong with the proof really. One follows from the other.

dyl10 · 14-12-2008 12:34AM

TripleAce wrote: »

a=b

therefore

0=b-a

Now we divide both by b-a:

0/(b-a) =(b-a)/(b-a)

and so:

0 = 1

Why?

When I read that at a glance, I thought it was me being stupid, then I realised the truth..... :P

e05bf05a · 22-12-2008 05:15PM

Michael Collins wrote: »

Too obvious! But better than the divide by zero one...
Spoiler
It's the fourth line here that has the false step - need to take account of the possiblity of the root being negative...which it turns to be in this case

Spoiler
Well x is a variable, so the amount of x's on the RHS will be a function of x, so you cannot just assume it's constant. Not sure about the second error, is it to with the fact derivatives only operate on functions, and the RHS isn't a proper function?

How about this one:

1 = sqrt (-1 * -1)
1 = sqrt(-1)*sqrt(-1)
1 = i * i
1 = -1

i got banned because he didnt cop onto this

LeixlipRed · 22-12-2008 05:32PM

You'll get banned again and if you don't start acting with a bit of cop on.

token56 · 23-12-2008 02:36PM

Michael Collins wrote: »

How about this one:

1 = sqrt (-1 * -1)
1 = sqrt(-1)*sqrt(-1)
1 = i * i
1 = -1

Spoiler

the sqrt(-1* -1) has two roots, -1 and +1 and in this case we are assuming the root is the positive one, the first line should really be -1 or +1 = sqrt(-1*-1) for the equation to be true

1 = 0

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