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Maths is broken?

Col Man

OK obviously maths isn't broken.... can you tell me why not?

(From UCL Mathematical Fallacies)

Suppose that u is a solution of u = 1 + u^2. Clearly u =/= 0 (does not equal), so we can divide by u. Hence 1 = 1/u + u. Therefore u = 1 + u^2 = (1/u + u) + u^2 [substituting 1/u + u for 1]. Hence u = 1/u + u + u^2 and so 0 = 1/u + u^2. Therefore 0 = 1 + u^3, so that u^3 = -1 and so u = -1.

Hence u = 1 + u^2 becomes -1 = 1 + (-1)^2 = 2

Therefore 3 = 0. Good times.

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Ass

if u have a plane and it is on a treadmill and it is going forward on the treadmill will it take off?

[Deleted User]

Is it possible for u to be a solution of u = 1 + u^2? I can't see how it would be.

I'm going to take a stab and say the flaw lies in assuming u is a solution to the equation.

Col Man

No that's not quite it...

[Deleted User]

The only other thing I can see is that you're subbing the value of 1 that you got back into the equation which gave you that value. That isn't logical, so my guess is there's some flaw there. Still, I can't see any division by 0 or the like. I'll get some pen and paper and have another look now. Good riddle.

Also, the solutions would be complex, wouldn't they?

Col Man

Complex indeed!

Ha, this is a good riddle all right. Good ol' UCL. They have another few, I'll post them up later

red_fox

LaTeX won't work under spoilers

By dividing by u, then adding that to the original equation (cancelling the 1) and then multiplying by u you essentially multiply the whole equation by (u+1) leading to the solution of u=-1, which is not a solution of the origional equation.

The solutions are the two complex third roots of unity.

Nice, these usually divide by zero.

Col Man

Nice work sir

[Deleted User]

Any more Col Man?

Col Man

Realising the difficulty in posting integration questions, I'll settle for this one from somewhere online:

In any group that consists of just one person, everybody in the group has the same age, because after all there is only one person.

Therefore, statement S(1) is true.

The next stage in the induction argument is to prove that, whenever S(n) is true for one number (say n=k), it is also true for the next number (that is, n = k+1).

We can do this by (1) assuming that, in every group of k people, everyone has the same age; then (2) deducing from it that, in every group of k+1 people, everyone has the same age.

Let G be an arbitrary group of k+1 people; we just need to show that every member of G has the same age.

To do this, we just need to show that, if P and Q are any members of G, then they have the same age.

Consider everybody in G except P. These people form a group of k people, so they must all have the same age (since we are assuming that, in any group of k people, everyone has the same age).

Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age.

Let R be someone else in G other than P or Q.

Since Q and R each belong to the group considered in step 7, they are the same age.

Since P and R each belong to the group considered in step 8, they are the same age.

Since Q and R are the same age, and P and R are the same age, it follows that P and Q are the same age.

We have now seen that, if we consider any two people P and Q in G, they have the same age. It follows that everyone in G has the same age.

The proof is now complete: we have shown that the statement is true for n=1, and we have shown that whenever it is true for n=k it is also true for n=k+1, so by induction it is true for all n.

Meh, might entertain some of you for a while. Or maybe everyone does have the same age? Who knows.

MathsManiac

This is the faulty step:

Let R be someone else in G other than P or Q.

I won't spell out why!

Col Man

Indeed! Are you not spelling out why so as to leave it as a challenge to someone else?

hivizman

When I studied mathematical induction a long time ago, my teacher gave a piece of practical advice, which was always to check that the formula or result worked for n=2 as well as n=1 before going to the actual induction step. This is a lovely example of why that advice was useful, and MathsManiac has correctly identified the step in the argument that introduces a hidden assumption that invalidates the "proof" for n=2, and hence for all subsequent values of n.

Col Man

Easy one this:

-2 = -2
4 - 6 = 1 - 3
4 - 6 + 9/4 = 1 - 3 + 9/4
(2 - 3/2)2 = (1 - 3/2)2
2 - 3/2 = 1 - 3/2
2 = 1

[Deleted User]

I assume the 2 outside the bracket is actually ^2?

In the 3rd line you completed the square, and in the 5th line you've only taken the positive value of the square root, either one could be the negative value, too.

Col Man

Lol whoops.... yeah supposed to be ^2. And yeah that's basically the answer, ha. Nice

[Deleted User]

This is an absolutely brilliant one, takes a bit of time but it's worth it.

http://www.manbottle.com/trivia/einstein_s_riddle

(Not so much a riddle, more of a puzzle).

[Deleted User]

how can 1 = 1/u +1 when u just said that u=/= 0? U can't sub in that if its not true. 1 cannot = 1/u +1 no matter what value u take for u

Col Man

Oh sorry, I substituted correctly in the actual sum, just in the [explanation inside these brackets] I put the wrong thing. Lol oops

Nappy

Just got that einstein riddle there! took an age tho, am extremely self satisfied and thot i should share with everyone!

Fremen

Not bad babbage, but I would say they're not quite in the same league as Col Man's. It's clear that you can't sum a divergent series, and that the Complex Log function has a branch cut.

Here's a nice one. It's actually "true" for some suitably warped value of "true".

There is a sense in which

(1 + 1 + 1 + 1 + .... ) = -1/2.

Short wikipedia article here

Fremen

Reading further, apparently Zeta regularisation applies to the last two examples given by babbage too.
Weird.

rjt

Babbage wrote: »

Travel from the bottom of a triangle to the top in a series of infinitesimal steps like this:

All the horizontal steps add up to a, and all the vertical steps add up to b. So the length of the hypotenuse is a+b.

The reason this doesn't work is simply that there's no reason it should. The taxicab distance (the shortest distance between two points moving only north/south/east/west) is the same regardless of the size of the steps, so the limit of the distance as the step size goes to zero is just a+b. There's no reason this should be equal to the "as the crow flies" distance.

rjt

Babbage wrote: »

Protip: we're just having fun.

I just pointed out the mistake, which people've done for each one so far :P

ray giraffe

Col Man wrote: »

OK obviously maths isn't broken.... can you tell me why not?

(From UCL Mathematical Fallacies)

Suppose that u is a solution of u = 1 + u^2. Clearly u =/= 0 (does not equal), so we can divide by u. Hence 1 = 1/u + u. Therefore u = 1 + u^2 = (1/u + u) + u^2 [substituting 1/u + u for 1]. Hence u = 1/u + u + u^2 and so 0 = 1/u + u^2. Therefore 0 = 1 + u^3, so that u^3 = -1 and so u = -1.

Hence u = 1 + u^2 becomes -1 = 1 + (-1)^2 = 2

Therefore 3 = 0. Good times.

The argument actually proves: if u is a real-valued solution of u=1+u^2 then u=1. This is not the same as proving the converse however.

This is an example of "A false statement implies any statement" ("u is a real-valued solution of u=1+u^2" is false already). So "If the moon is made of cheese then god exists" :pac:

I will refrain from claiming "If god exists then the moon is made of cheese"

Fremen

ray giraffe wrote: »

This is an example of "A false statement implies any statement

There's a nice story about G H Hardy demonstrating this idea. Someone asked him to prove that he was the pope, assuming 1+1=1. His reply:

"The pope is one. I am one. Therefore the pope and I are one".

ray giraffe

Fremen wrote: »

There's a nice story about G H Hardy demonstrating this idea. Someone asked him to prove that he was the pope, assuming 1+1=1. His reply:

"The pope is one. I am one. Therefore the pope and I are one".

Nice

Unfortunately we have blasphemy laws to contend with, so we better watch what we say :eek: