# Proof that 1 = 2

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• May 5th 2009, 11:56 PM
blueirony
Proof that 1 = 2
This is a silly proof that my thirteen-year old cousin showed me the other day. I'm ashamed to admit that it took me a full ten minutes to figure it out.

Let a = 1
Let b = 1

therefore:
a = b

multiply both sides by b:
ab = b^2

subtract a^2:
ab - a^2 = b^2 - a^2

factorise:
a(b - a) = (b - a)(b + a)

divide by (b - a):
[a(b - a)]/(b -a) = [(b - a)(b + a)]/(b -a)

cancel:
a = b + a

returning to our initial property:
a = 1
b = 1

therefore:
a = b + a
1 = 1 + 1
1 = 2

What is wrong with the above proof? (It's not difficult, but it's a good one to pull out to confuse people)
• May 6th 2009, 12:26 AM
vemrygh
Division by zero $\displaystyle (b-a)$.
• May 6th 2009, 04:34 AM
blueirony
Well done.

I can't believe it took me that long to figure it out!
• Jun 3rd 2009, 10:19 AM
paulrb
Are there any 1=2 "proofs" that do not involve dividing by zero? I am trying to think of other mathematical rules that can be broken to derive such a proof.
• Jun 3rd 2009, 10:59 AM
TheAbstractionist
Quote:

Originally Posted by paulrb
Are there any 1=2 "proofs" that do not involve dividing by zero?

Here is one.

$\displaystyle \cos^2x-\sin^2x\ =\ \cos2x$ (identity)

$\displaystyle \cos^2x\ =\ \cos2x+\sin^2x$ (rearramging)

$\displaystyle \cos x\ =\ \sqrt{\cos2x+\sin^2x}$ (taking square root)

$\displaystyle \cos\frac{3\pi}4\ =\ \sqrt{\cos\frac{3\pi}2+\sin^2\frac{3\pi}4}$ (substituing $\displaystyle x=\frac{3\pi}4)$

$\displaystyle -\frac1{\sqrt2}\ =\ \sqrt{0+\frac12}=\frac1{\sqrt2}$ (evaluating the substitution)

$\displaystyle -\frac12\ =\ \frac12$ (multiplying both sides by $\displaystyle \frac1{\sqrt2})$

$\displaystyle 1\ =\ 2$ (adding $\displaystyle \frac32$ to both sides)
• Jun 3rd 2009, 11:41 AM
Chris L T521
Quote:

Originally Posted by TheAbstractionist

$\displaystyle -1\ =\ 1$ (multiplying both sides by $\displaystyle \sqrt2)$

$\displaystyle 1\ =\ 2$ (adding 1 to both sides)

The last step should lead to $\displaystyle 0=2$...
• Jun 3rd 2009, 08:07 PM
blueirony
Quote:

Originally Posted by Chris L T521
The last step should lead to $\displaystyle 0=2$...

^ How?

The Abstractionist is correct!
• Jun 3rd 2009, 08:39 PM
Chris L T521
Quote:

Originally Posted by blueirony
^ How?

The Abstractionist is correct!

Read what I quoted. Its correct now since he edited it.
• Jun 4th 2009, 01:45 AM
TheAbstractionist
Quote:

Originally Posted by blueirony
^ How?

The Abstractionist is correct!

I made a mistake in my original post, which was spotted by Chris L T521. I do this quite often, and when someone spots my mistakes, I normally thank the person who spotted them and just go back and edit my post. (Smile) (To Chris L T521: Thanks! (Happy))

By the way, did you spot the fallacy in the “proof”?
• Jun 4th 2009, 03:27 AM
Is it that we can also write $\displaystyle \sqrt{0+\frac{1}{2}} = \frac{-1}{\sqrt{2}}$ and hence $\displaystyle -\frac{1}{2} = -\frac{1}{2}$ i.e. -1 = -1...
• Jun 4th 2009, 03:35 AM
Here's what could be the simplest -1 = 1 proof.
$\displaystyle 1 = \sqrt{1} = -1$

So 1 = -1.
• Jun 4th 2009, 03:39 AM
TheAbstractionist
Quote:

Is it that we can also write $\displaystyle \sqrt{0+\frac{1}{2}} = \frac{-1}{\sqrt{2}}$ and hence $\displaystyle -\frac{1}{2} = -\frac{1}{2}$ i.e. -1 = -1...

It’s along those lines, but I wouldn’t write $\displaystyle \sqrt{0+\frac{1}{2}} = \frac{-1}{\sqrt{2}}.$ The LHS is generally taken to mean the positive square root, so the equation is technically incorrect.

Hint:
Spoiler:
If $\displaystyle a^2=b^2$ (where $\displaystyle a$ and $\displaystyle b$ are real) then $\displaystyle a=\ldots?$
• Jan 1st 2010, 06:07 PM
joll
This is logical 'proof' rather than mathematical....

Consider the following statement;
" If this statement is true, then 1= 2. " (let's call this A)

First, let's assume A to be true. on this supposition;
As statement A is true, "If statement A is true, then 1=2" is true.
and again, as statement A is true and
"if statement A is true, then 1=2" is true, we obtain 1=2.

Now we have proved that "If this statement is true, then 1=2 " is true.

Then, as 'this statement' is true, 1=2 (q.e.d.)

By using the same trick, anything can be proved.
• Jan 2nd 2010, 03:28 AM
Wilmer
A is racing against B; A wears sweater#1, B wears sweater#2.
Thet hit "finish line" at exactly same time.....WELL? (Giggle)
• Jan 2nd 2010, 04:17 AM
One empty bottle and two empty bottles contain the same amount of liquid.
If x empty bottles contain zero liquid in total, x is not immediately determineable.

Quote:

Originally Posted by blueirony

Let a = 1
Let b = 1

therefore:
a = b

multiply both sides by b:
ab = b^2

subtract a^2:
ab - a^2 = b^2 - a^2
this is now 0=0.

factorise:
a(b - a) = (b - a)(b + a)
this is now 1(0)=(0)2, only non-zero values can be factorised.

What is wrong with the above proof? (It's not difficult, but it's a good one to pull out to confuse people)

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