I've proved that 1 = 2.

[EternalUniverse]

Actually, I haven't. I encountered a fun math puzzle that purportedly proves that 1 = 2. What is wrong with the proof? Assuming there is one

1. Let
x = 1

2. It follows that
x = x

3. Square both sides to get
x² = x²

4. Subtract x² from both sides
x² - x² = x² - x²

5. Factor both sides to get
x(x-x) = (x+x)(x-x)

6. Divide both sides by (x-x)
x=(x+x)

7. Since x = 1
1=(1+1)

8. So
1 = 2


edited to add: oops, I've posted on the wrong forum

 

[(S)]

Division by zero. This is pointed out in every highschool mathbook -- both the 'proof' and the 'flaw' -- since the beginning of time.

If you assume 1 = 2, then I am the Pope.
[Proof:
Let S be the set consisting of me and the Pope. The order of S is 2, since S contains two elements. But 2 = 1, so the order of S = 1 and S really only contains 1 element, which must mean that I am the Pope.
[Thank you, Bertrand Russel]]

 

[scribble]

Wow!

Two great threads arguing for EVERYONE TO LEARN CALCULUS (and do a better job of it than I did!!).

 

[Brown]

JREF trivia: The very first puzzle in the "puzzles" section posted a variation of this "proof," with the goal being to explain what was wrong with it.

The first two responses were incorrect.

 

[Virgil]

you are wrong because I am the pope.

please send money

Virgil

 

[phildonnia]

Actually a huge flaw, often missed by math teachers that should know better, is that the first line of this proof is not an axiom or theorem. If we treat it as a hypothetical, then this becomes a proof by contradiction that x cannot equal 1.

 

[Agammamon]

I think step one can be eliminated.

Step 2 says that x=x and if you follow the rest of the steps you end up with x=2x.

so you can say that 1=2
2=4
3=6
4=8 etc.


And I think the real flaw here is that in step 6 you're dividing 0 by 0 which give my calculator headaches.

 

[Brown]

I think step one can be eliminated.

Agreed. To make the problem more bufuddling, I recommend removing the step that says "Divide both sides by (x-x)" and replacing it with "Cancel (x-x) from both sides." This way, it looks as though there is no dividing going on.

 

[Cecil]

Here's an alternative proof of this using calculus:

x^2 = x + x + ... + x (x times)
d(x^2)/dx = d(x + x + ... + x)/dx
2x = 1 + 1 + ... + 1 (x times)
2x = x
2 = 1

And one using complex numbers:
-1/1 = 1/-1
sqrt(-1/1) = sqrt(1/-1)
sqrt(-1)/sqrt(1) = sqrt(1)/sqrt(-1)
i/1 = 1/i
i/2 = 1/2i
i/2 + 3/2i = 1/2i + 3/2i
i(i/2 + 3/2i) = i(1/2i + 3/2i)
i^2/2 + 3i/2i = i/2i + 3i/2i
-1/2 + 3/2 = 1/2 + 3/2
1 = 2

 

[phildonnia]

x^2 = x + x + ... + x (x times)
d(x^2)/dx = d(x + x + ... + x)/dx
2x = 1 + 1 + ... + 1 (x times)
2x = x
2 = 1

That's pretty clever! It took me a while to find the flaw.

 

[TeaBag420]

Actually, I haven't. I encountered a fun math puzzle that purportedly proves that 1 = 2. What is wrong with the proof? Assuming there is one

1. Let
x = 1



2. It follows that
x = x

It doesn't "follow", it's AXIOMATIC that x=x

3. Square both sides to get
x² = x²

You've already said x = 1, so now you're saying that 1=1

4. Subtract x² from both sides
x² - x² = x² - x²

So now you're saying that 0=0

5. Factor both sides to get
x(x-x) = (x+x)(x-x)

6. Divide both sides by (x-x)
x=(x+x)

Wrong, since division by zero is undefined in algebra and calculus.

7. Since x = 1
1=(1+1)

Wrong. See above.

8. So
1 = 2

[/QUOTE]

 

[Crossbow]

Actually, I haven't. I encountered a fun math puzzle that purportedly proves that 1 = 2. What is wrong with the proof? Assuming there is one

1. Let
x = 1

2. It follows that
x = x

3. Square both sides to get
x² = x²

4. Subtract x² from both sides
x² - x² = x² - x²

5. Factor both sides to get
x(x-x) = (x+x)(x-x)

6. Divide both sides by (x-x)
x=(x+x)

7. Since x = 1
1=(1+1)

8. So
1 = 2


edited to add: oops, I've posted on the wrong forum

The error is in the factoring on the right side of Step 5.

(x+x)(x-x) = 2x² - 2x²

Or put another way:

x² - x² = (x+x)(x-x)/2

Soooo, the proof should look thus:

1. Let
x = 1

2. It follows that
x = x

3. Square both sides to get
x² = x²

4. Subtract x² from both sides
x² - x² = x² - x²

5. Factor both sides to get
x(x-x) = (x+x)(x-x)/2

6. Divide both sides by (x-x)
x=(x+x)/2

7. Since x = 1
1=(1+1)/2

8. So
1 = 1

How is that?

 

[zenith-nadir]

Meanwhile, as the math argument continues, next door at zenith-nadir's dorm room....

 

[Cecil]

That's pretty clever! It took me a while to find the flaw.

I must admit it's not my creation, but I found it fascinating enough that it stuck around in my brain.

 

[Skeptoid]

Re: Re: I've proved that 1 = 2.

How is that?

You're still dividing by zero in step 6.

 

[Yahweh]

The solution to 1 = 2 will be resolved in approximately 200 more posts. Carry on.

 

[Cecil]

The solution to 1 = 2 will be resolved in approximately 200 more posts. Carry on.

I have proven above that 1=2 beyond all doubt. All who disagree with me therefore do not understand math.

3 proofs that 1 = 2 have been shown so far in this thread. You skeptics have no proofs that 1 != 2. I will continue to repeat these proofs approximately every 10 posts until the unbelievers surrender.

 

[(S)]

Apply Pope proof. It shows that, in a set containing Cecil and me, there is actually only one element. Therefore, I am Cecil, and I concede that 1 != 2.

 

[scribble]

I have proven above that 1=2 beyond all doubt. All who disagree with me therefore do not understand math.

3 proofs that 1 = 2 have been shown so far in this thread. You skeptics have no proofs that 1 != 2. I will continue to repeat these proofs approximately every 10 posts until the unbelievers surrender.

Now that's a good post!

 

[Crossbow]

Re: Re: Re: I've proved that 1 = 2.

You're still dividing by zero in step 6.

I have read what other people have been writing, and I do not think that business about dividing by zero is correct; because what is happening in this problem is that it is a case of dividing zero into zero which should produce a quotient of one.

And even if that is not the case, it still does not change the fact that the expansion offered in the original proof is incorrect, therefore the proof is still incorrect.