# Proof that 1 = 2

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• Jan 2nd 2010, 06:20 AM
Defunkt
Quote:

Originally Posted by TheAbstractionist
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)

Quite late (Rofl), but this proof is incorrect simply due to the fact that in the second quadrant $\displaystyle (\frac{\pi}{2} \leq \theta \leq \pi)$, the result is negative (meaning $\displaystyle cosx = -\sqrt{cos2x+sin^2x}$ ), but you used the positive result.
• Jan 6th 2010, 08:54 AM
rainer
Quote:

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.
• Jan 7th 2010, 03:06 PM
Defunkt
Quote:

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.
• Feb 8th 2010, 06:13 PM
Dawson
Quote:

Originally Posted by TheAbstractionist
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)

you forgot that square rootting something is '+' OR '-'.
• Oct 4th 2011, 11:26 PM
mathbyte
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:

Start with:
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
• Oct 5th 2011, 04:53 AM
Archie Meade
Re: Another Proof that 1 = 2
Nice example of what happens if the variable gets partly treated as a constant.
• Oct 5th 2011, 06:15 PM
agentmulder
Re: Another Proof that 1 = 2
Quote:

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:

Start with:
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 ?
• Oct 6th 2011, 02:04 AM
Archie Meade
Re: Another Proof that 1 = 2
Quote:

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).
• Oct 6th 2011, 03:26 AM
agentmulder
Re: Another Proof that 1 = 2
Quote:

Originally Posted by Archie Meade
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
• Oct 7th 2011, 12:08 PM
Deveno
Re: Another Proof that 1 = 2
let's look at the equation:

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

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

now, let's differentiate that same equation:

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

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.
• Aug 16th 2012, 10:25 PM
manoj9585
Re: Proof that 1 = 2
-1 = 1
0 = 1 + 1
0 = 2
so it will zero not 1 =2 .
word problem help
• Aug 17th 2012, 06:07 AM
Wilmer
Re: Proof that 1 = 2
Go away with your commercials...
• Aug 26th 2012, 09:55 AM
smatik
Re: Proof that 1 = 2
Quote:

Originally Posted by TheAbstractionist
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)

you made a mistake here bro:
$\displaystyle \cos x\ =\ \sqrt{\cos2x+\sin^2x}$ (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.
• Nov 27th 2012, 08:39 AM
Stephen347
Re: Proof that 1 = 2
$\displaystyle \cos^2x-\sin^2x\ =\ \cos2x$ (identity)

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

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

And thus there is no contradiction when $\displaystyle x\ =\ \frac{3\pi}4$. This absolute value sign must be included if we are taking the positive square root on the right hand side of the equation.
• Jul 20th 2013, 02:59 PM
ChessTal
Re: Another Proof that 1 = 2
Quote:

Originally Posted by Archie Meade
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).

More simply: If x is natural x·x(x^2) is not a continuous function therefore not differentiable.(Nod)
If x is not a natural number then x^2 = x+x+...+x is not a valid equality.
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