# Thread: Proof that 1 = 2

1. Originally Posted by TheAbstractionist
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)
Quite late , but this proof is incorrect simply due to the fact that in the second quadrant $(\frac{\pi}{2} \leq \theta \leq \pi)$, the result is negative (meaning $cosx = -\sqrt{cos2x+sin^2x}$ ), but you used the positive result.

2. Originally Posted by 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.
Irrational number - Wikipedia, the free encyclopedia

I saw this the other day. Look under the "History" section. Hippasus' proof that an odd number is even. It's not exactly the same as a 1=2 "proof," but is at the very least relevant to what you are touching upon here.

It's worth mentioning that--if I've understood correctly--a major difference is that Hippasus's proof is valid. No fallacies in logic can be pointed out as in the examples here. I therefore speculate that it is conceivable that a valid 1=2 proof can exist, and that it would merely prove, as Hippasus' proof does, that irrational numbers exist.

My speculation is probably totally hare-brained though, as I am not a real mathematician.

3. Originally Posted by rainer
Irrational number - Wikipedia, the free encyclopedia

I saw this the other day. Look under the "History" section. Hippasus' proof that an odd number is even. It's not exactly the same as a 1=2 "proof," but is at the very least relevant to what you are touching upon here.

It's worth mentioning that--if I've understood correctly--a major difference is that Hippasus's proof is valid. No fallacies in logic can be pointed out as in the examples here. I therefore speculate that it is conceivable that a valid 1=2 proof can exist, and that it would merely prove, as Hippasus' proof does, that irrational numbers exist.

My speculation is probably totally hare-brained though, as I am not a real mathematician.
Actually, in his proof he assumes (in contradiction), that there are no irrational numbers. He then reaches a cotradiction - that the number b, which he specified, must be both even and odd, however no such number exists. therefore, the claim that there are no irratioal numbers is false.

4. Originally Posted by TheAbstractionist
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)

you forgot that square rootting something is '+' OR '-'.

5. ## Another Proof that 1 = 2

I'm surprised no one brought this one up, since it's been quite a number of years since I first heard of it.
Here is a 1=2 proof using differentiation:

x^2 = x^2

x^2 is x*x, so we can represent one side by addition:
x + x + x + ... + x = x^2
where there are x x's.

Now, we differentiate both sides:
1 + 1 + 1 + ... + 1 = 2x

There are x 1's, so we can sum all the 1's to get
x = 2x

And dividing by x we get
1=2

6. ## Re: Another Proof that 1 = 2

Nice example of what happens if the variable gets partly treated as a constant.

7. ## Re: Another Proof that 1 = 2

Originally Posted by mathbyte
I'm surprised no one brought this one up, since it's been quite a number of years since I first heard of it.
Here is a 1=2 proof using differentiation:

x^2 = x^2

x^2 is x*x, so we can represent one side by addition:
x + x + x + ... + x = x^2
Theres got to be something fishy about that representation

x + x + x + ... + x = x^2

if x = -1, how many x do you have on the left of your equation? How about if x = 1/3 ?

8. ## Re: Another Proof that 1 = 2

Originally Posted by agentmulder
Theres got to be something fishy about that representation

x + x + x + ... + x = x^2

if x = -1, how many x do you have on the left of your equation? How about if x = 1/3 ?
Even if "x" is natural,
the number of x's summed is itself a function of x
and must be taken into account when calculating the derivative (product rule).

9. ## Re: Another Proof that 1 = 2

Even if "x" is natural,
the number of x's summed is itself a function of x
and must be taken into account when calculating the derivative (product rule).
I agree. To me it looks fishy even before the derivative is taken, x^2 = x^2 holds for any x, not so sure about the other representation.

Yes, x*x = x^2 doesn't pose any problems i can think of

10. ## Re: Another Proof that 1 = 2

let's look at the equation:

$\sum_{i=1}^kx = x^2$

when we solve for k, we get k = x.

now, let's differentiate that same equation:

$\sum_{i=1}^k 1 = 2x$.

when we solve for k, we get k = x/2.

evidently, x = x/2, so 2x = x, so x = 0.

therefore, the flaw in the proof is the very last step...we have divided by 0.

11. ## Re: Proof that 1 = 2

-1 = 1
0 = 1 + 1
0 = 2
so it will zero not 1 =2 .
word problem help

13. ## Re: Proof that 1 = 2

Originally Posted by TheAbstractionist
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)
you made a mistake here bro:
$\cos x\ =\ \sqrt{\cos2x+\sin^2x}$ (taking square root)
there should absolute function over cosx. since sqrt(x^2)=abs(x). Its the most common mistake that students makes while learning Pre-calculus.But we soon realize the importance of absolute function when we study limits.

14. ## Re: Proof that 1 = 2

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

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

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

And thus there is no contradiction when $x\ =\ \frac{3\pi}4$. This absolute value sign must be included if we are taking the positive square root on the right hand side of the equation.
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