False proof 1=2

**winner45** · December 27th 2008, 07:38 AM

Step 1: a and b > 0
Step 2: a = b
Step 3: a2 = ab
Step 4: a2 - b2 = ab - b2
Step 5: (a + b)(a - b) = b(a - b)
Step 6: (a + b) = b
Step 7: b + b = b
Step 8: 2b = b
Step 9: 2 = 1

This is a flase proof but I wanted to see how many people can find the false step in this.

**Aliquantus** · December 27th 2008, 07:40 AM

Since a=b, a-b=0 and division by zero is not allowed.

**masters** · December 27th 2008, 07:50 AM

Originally Posted by winner45

Step 1: a and b > 0
Step 2: a = b
Step 3: a2 = ab
Step 4: a2 - b2 = ab - b2
Step 5: (a + b)(a - b) = b(a - b)
Step 6: (a + b) = b
Step 7: b + b = b
Step 8: 2b = b
Step 9: 2 = 1

This is a flase proof but I wanted to see how many people can find the false step in this.

Originally Posted by Aliquantus

Since a=b, a-b=0 and division by zero is not allowed.

What he said.

**TKHunny** · December 27th 2008, 08:23 AM

This is one of the weaker such "proofs".

"Step 1: a and b > 0" -- This step is entirely pointless.

"Step 2: a = b" -- The whole thing is really quite pointless after this.

Substitute this expression into the rest of the "proof".

"Step 3: b^2 = b*b" -- So? Hardly interesting.

"Step 4: b^2 - b^2 = b*b - b^2" -- So? Hardly interesting.

"Step 5: (b + b)(b - b) = b(b - b)" -- or (2b)(0) = b(0) So? Hardly interesting.

Why would you even consider the division step after that?

**winner45** · December 27th 2008, 02:19 PM

Ok good the people on here are smarter than the ones in my town there was only 2 people other than me that found out what was wrong with this proof and sadly they were all kids.

**TKHunny** · December 27th 2008, 04:32 PM

What did you conclude was wrong with it?

1) Just the division-by-zero problem?
2) The fact that it contains useless and distracting information?
3) The necessity to write the equations in a specific visual way in order for it to be deceiving?
4) That you, a confessed non-mathematician, even for a moment, wondered if people who actually pose as mathematicians might be stumped by such foolishness?

There are quite a few things wrong with it, not just one.

**Jhevon** · December 27th 2008, 04:51 PM

there are a lot better false proofs here

**bobak** · December 27th 2008, 05:00 PM

Okay were going down this path now......

Statement: All horses are the same colour.

Proof (by induction): For one horse the statement is true. assume the statement is true for any set of k horses, then consider a set of k+1 horses. $( h_1 , h_2 , ... , h_k , h_{k+1} )$ then take any two distinct k element subsets $( h_1 , h_2 , ... , h_k ,)$ and $( h_2 , ... , h_k , h_{k+1} )$ for example. we know that all the horses in the k element subsets are the same colour, so $h_1 = h_2 = ... = h_k$ and $h_2 = ... = h_k = h_{k+1}$ . this gives $h_1 = h_2 = ... = h_k = h_{k+1}$ . so this prove that for any set of k+1 horse they are the same colour. so by induction all horses are the same colour.

Bobak

**Jhevon** · December 27th 2008, 05:03 PM

Originally Posted by bobak

Okay were going down this path now......

Statement: All horses are the same colour.

Proof (by induction): For one horse the statement is true. assume the statement is true for any set of k horses, then consider a set of k+1 horses. $( h_1 , h_2 , ... , h_k , h_{k+1} )$ then take any two distinct k element subsets $( h_1 , h_2 , ... , h_k ,)$ and $( h_2 , ... , h_k , h_{k+1} )$ for example. we know that all the horses in the k element subsets are the same colour, so $h_1 = h_2 = ... = h_k$ and $h_2 = ... = h_k = h_{k+1}$ . this gives $h_1 = h_2 = ... = h_k = h_{k+1}$ . so this prove that for any set of k+1 horse they are the same colour. so by induction all horses are the same colour.

Bobak

haha, i've never seen this one before!

**bobak** · December 27th 2008, 05:05 PM

Originally Posted by Jhevon

haha, i've never seen this one before!

have you seen the error ?

Bobak

**bobak** · December 27th 2008, 05:28 PM

The problem with the proof is that I have shown that being true for k implies also true for k+1 however this proof only works if the k element subsets of a k+1 element set overlap, namely k > 1, so I have shown that if true for 2 also true for 3 ect. but the statement holding for 1 horse does not imply it is true for 2 horses so we have a gap, I must show by other means that the statement is true for 2 horses to complete my proof, but that is impossible.

Bobak

**winner45** · December 28th 2008, 10:14 AM

Originally Posted by TKHunny

What did you conclude was wrong with it?

1) Just the division-by-zero problem?
2) The fact that it contains useless and distracting information?
3) The necessity to write the equations in a specific visual way in order for it to be deceiving?
4) That you, a confessed non-mathematician, even for a moment, wondered if people who actually pose as mathematicians might be stumped by such foolishness?

There are quite a few things wrong with it, not just one.

Sorry if that insulted anyone but my math teacher couldnt figure it out so
Ive been using it to see how good someone is at math.

**Jhevon** · December 28th 2008, 01:49 PM

Originally Posted by winner45

... but my math teacher couldnt figure it out so

i find that very hard to believe. but ok. it's all good fun though, the first few times you see it. so no worries.

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