# Math Help - Proof that 1 = 2

1. ## 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)

2. Division by zero $(b-a)$.

3. Well done.

I can't believe it took me that long to figure it out!

4. 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.

5. Originally Posted by paulrb
Are there any 1=2 "proofs" that do not involve dividing by zero?
Here is one.

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

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

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

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

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

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

$1\ =\ 2$ (adding $\frac32$ to both sides)

6. Originally Posted by TheAbstractionist

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

$1\ =\ 2$ (adding 1 to both sides)
The last step should lead to $0=2$...

7. Originally Posted by Chris L T521
The last step should lead to $0=2$...
^ How?

The Abstractionist is correct!

8. Originally Posted by blueirony
^ How?

The Abstractionist is correct!
Read what I quoted. Its correct now since he edited it.

9. 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. (To Chris L T521: Thanks! )

By the way, did you spot the fallacy in the “proof”?

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

11. Here's what could be the simplest -1 = 1 proof.
$1 = \sqrt{1} = -1$

So 1 = -1.

Is it that we can also write $\sqrt{0+\frac{1}{2}} = \frac{-1}{\sqrt{2}}$ and hence $-\frac{1}{2} = -\frac{1}{2}$ i.e. -1 = -1...
It’s along those lines, but I wouldn’t write $\sqrt{0+\frac{1}{2}} = \frac{-1}{\sqrt{2}}.$ The LHS is generally taken to mean the positive square root, so the equation is technically incorrect.

Hint:
Spoiler:
If $a^2=b^2$ (where $a$ and $b$ are real) then $a=\ldots?$

13. 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.

14. A is racing against B; A wears sweater#1, B wears sweater#2.
Thet hit "finish line" at exactly same time.....WELL?

15. 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.

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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