So you think you're good at math?

[T_Pyro]

Ingenuity lies in doing things differently, as they say. Riddles are great mind openers, allowing one to "think outside the ellipsoid". A mind which is regularly stimulated is more likely to be succesful at solving tricky problems in real life.

Have a close look at the following image:
http://mercury.walagata.com/w/joydeepb/pic23805.gif

So, what's the hitch, assuming there is one? If you think you have the answer, you could post a mathematical / analytical explanation here.

The question was not mine, but a friend passed it on to me after he was flummoxed. If you've already come across it before, or if you find the question insulting to your intelligence, my humble apologies.

Another one:

x^2 = x + x + x + x + .... {x times}

Differentiating both sides with respect to x,

2x= 1 + 1 + 1 + 1 + .... {x times}

But, LHS= 2x, while RHS = x
What's gone wrong? ;)

If anybody else has any other interesting math riddles, they could add them here, I'm sure others will be interested in them too.

[Rhadon]

I don't know if you wanted us to post solutions here, but since it's not "real" riddles, here are my answers:

The image in the link you posted has a trick behind it, IIRC the lines aren't exactly straight. The thing with the equation is easy: Differentiation isn't an equivalent modification <expression?>, e.g. differentiating both sides of the equation x = 1 yields 1 = 0 which is something totally different.

[Marvin]

It took me several mins to spot the problem. While all the lines are straight the red and green triangular sections have different angles. In one configuration they stick out from the hypothetical 'hypotenuse' and in the other they are within. The difference in this very shallow area is exactly the same as the square that apears (it has to be).

Differentiation of both sides only works to give identical answers when the equations are exactly and always equivalent. Expanding to x + x + x .. x times is only valid for integer steps so it is not the same as x^2, a condition differentiation doesnt even work with (smooth always valid and single valued (for any given x) functions only).

[vulture]

Another one:

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

This equation is mathematically false to begin with.

You're assuming that x^2 equals the infinite sum of x, which isn't true ofcourse.

[YayItGoBoom!]

I'm going with the angle theory as well. I printed the sheet out and cut out the triangles, and the red triangle in the one with the gap is just a tiny bit smaller than the triangle on top. Its hardly noticeable, but I think its enough to account for the gap.

[Marvin]

You cut off part of the quote vulture,

x^2 = x + x + x + x + .... {x times}

More formally,

x2 = x1 + x2 + ... + xx-1 + xx

In terms of integer mathematics it is correct,

For example,

3*3 = 3 + 3 + 3

4*4 = 4 + 4 + 4 + 4

Differentiation is more along the lines of answering a question that a 'rearengement'. Its an operator in other words. The question it answers is 'what is the equation that describes the way my functions gradiant changes with respect to x'. This question isnt valid for an integer expansion.

YayItGoBoom!,
In terms of the triangles, just look at the gradiant of the triangles, do this by counting the squars. They are different, and that means one is more shallow than the other. If you are building a hill and put the shallow one first and the steep one on top the slope is 'concave'. If you put the steep one first and the shallow one on top, the slope is 'convex'. A concave slope takes up less volume/area than a convex one. Since the shapes that form this are fixes in area, this volume is displaced from under them. The shapes under are designed to fit exactly into a 3 by 5 oblong and fit within a 2 by 8 oblong (leaving a single gap and completing the illusion).

When simply looking at the problem the brain makes it a little simpler than it really is, and both sets seem to fit within the same 'large triangle'. The square is a simply different shape to the space it displaces and its a distance away from where it comes from.

This is a very clever problem. Now dont I just wish I was smart enough to be able to design something like this.

[T_Pyro]

Regarding the triangle problem, succinctly put, Marvin. Martians: 1 Earthlings: 0.75

My solution: (Highlight to read)
Actually, the "outer" triangle has an area of 32.5 square units, but the shaded regions cover only 32 square units. In the first figure, the perimeter is such that the "hypotenuse" is concave, so that the enclosed area is 0.5 square units less than it should be. In the second figure, the "hypotenuse" is convex, so it takes up 0.5 square units more than it should, so the difference in the enclosed area and the shaded area is 1 square unit, which is shown as the "gap".

The second problem, however, has more to it than meets the eye. If the flaw was just the fact that x has to be an integer, the equation (with some constraints) would have held good for the set of integers. However, that isn't the case.

Marvin: Notation of the form x1 , x2 , x3 etc. are (by convention) used to denote a series of constants, not variables.

Rhadon: The differential form of an equation is exact, and there are no doubts whatsoever about that. If y=x, dy=dx holds good. In your example, the differential form itself is incorrect. It should be 1dx = 0. This is true of course, since as x is a constant, dx=0. In the equation stated in the problem, both sides are in terms of x, so the "dx" factor cancels out.

Hint: The flaw is a fundamental one, and has to do with how the quantities are treated in the equation.

[fire vs. water]

here's an interesting equation (I don't know if that's what you can call it...but..)

a=b
ab=b^2
a^2-b^2=ab-b^2
(a-b)(a+b)=b(a-b)
(a+b)=b
a+a=a
2a=a
2=1

^ = in the power of
if anyone wants to think about this one, I will post what the "catch" here is, in a few days' time.

[nesler]

The transition fromt he second to the third step doesn't make any sense.

[glith]

T-pyro has the right answer. since the area that the difference between the concave hypotenuse and the convex hypotenuse is a square.

[devilassassin]

From step 4 to step 5 u are dividing by (a-b). But since a=b then (a-b)=0 and you cannot divide by zero as that answer is undefined. Thats why it does not work.

[T_Pyro]

The third statement is arrived at by subtracting b2 from both sides, so we get,
ab-b2=b2 - b2
But, since b=a, b2=a2
hence, ab-b2=a2 - b2

Elementary, Fire vs. water! The solution is: (highlight to read)

The error (and an unpardonable one at that) lies in the transition from the fourth to the fifth step. Since a=b, a-b=0 Therefore, if we cancel the (a-b) term from both sides, we're actually performing division by zero, which, of course, is undefined. The following example illustrates this same mistake:
a=2
a*0=2*0
but, 2*0=0, so, a*0=0
cancelling 0 from both sides, a=1
Which, of course, is incorrect. You cannot divide any number (not even 0 itself) by 0.

[Marvin]

"If the flaw was just the fact that x has to be an integer, the equation (with some constraints) would have held good for the set of integers. "

I dont understand how this can be true.

By definition, the derivitive of a point depends not on that point itself, but on what the function is doing immediatly either side of that point. In the case of an integer only function, it is doing nothing either side of that point. So it doesnt seem meaningful for it to even have a devivitive. Furthur more it should be possible to produce an indefinite number of continuous functions that satisfy the same points as the integer function, but pass through the points at totally different angles, a series of solutions should result which are not consistant with eachother, and therfore cannot be singly consistant with the integer solution. Regardless of any sane contraints I can think of.

So I cant see how it can be true.

The more 'formal' method I used for the expansion may be more suited to constants, but it is taught here for progressions, particually proofs by induction where it is used to label iterations of a function. Since X is mearly the simplest function of X, it seemed to fit. Do you label functions a different way for induction proofs?

[T_Pyro]

"If the flaw was just the fact that x has to be an integer, the equation (with some constraints) would have held good for the set of integers. "

Back to the basics of differentiation:
Differentiation of a function f(x) gives another function f'(x) that shows how the function varies with a change in x for the domain of f(x). For most of the functions that we come across, the domain is the set of real numbers, where the "smallest increment" is dx, where dx->0. Hence, f'(x) for the domain of real numbers is ( f(x+h) - f(x) )/h where h->0. For the set of integers, however, the operation will have to be redefined for the smallest increment possible, which is 1. Hence, differentiaition of f(x) in a world where only integers exist, would be f'(x) = ( f(x+1) - f(x) )/1 . Redefining the operation of differentiation, one could handle complex numbers too, provided you work out the corresponding relationships.

Such a model complies with all our notions of "differentiation", including the one you mentioned about the infinite indefinite functions. If you see the analogue of such a situation in the integral domain, you'll notice that it holds good. In the real domain, if 2 points are finitely close, there can be infinite curves made to pass through the points with infinitely different tangents, like you said. However, if the points are infinitessimally close, there can be only 1 such curve. Well, if the domain is an integral one, you'll realise that in the defined set of numbers (namely, integers), a difference of 1 is an infinitessimal change!

For mathematical induction, we take in general any "n"th term, and call it Next, Tn is defined for n, and then the series evaluated/ manipulated/ equation proved/ whatever.
eg. ex = Summation(n=0 to n=infinity) Tn
where Tn = xn/n!

[Rhadon]

Your second problem has indeed more behind it than I thought at first. Anyway, differentiability is only defined for certain vector spaces, fields and subsets of the latter, e.g. Rn and C. N and its subsets are none of them, for reasons Marvin has already stated; at least I believe he's right.

I think that regarding dx as a "smallest increment" can be done to illustrate what it shall resemble, but actually there is no such thing as a "smallest increment" where numbers are dense, of course. In N that's something else, so there is one more reason for the necessity to re-define the term "differentiable". But if you do that, in general it will be wrong to think that you can still use what you know about functions which are differentiable in the common way. You don't even know if e.g. d/dn nm = m • nm-1 unless you prove it.

Edit: Well, if you regard N as an integral domain, N is dense (in N), but I think that it's still necessary to re-define the term "differentiable".

[T_Pyro]

By saying that it's necessary to "re-define the term "differentiable"", I think what you mean is to "rename" the new operator. Redifining a function is done by changing the way it operates on the supplied parameters.

OK, the answer:
If y=x*n, the statement y=x + x + x + ... {n times} is true only if n is independent of x. Otherwise, while rewriting the expression (or framing the identity), the degree of the expression itself is altered, since x (the number of terms) has been treated as a constant. The fact is again proved by the result we get from differentiation. While differentiating this identity, we take into account the change in the value of x, but not the change in the number of terms itself!

[Marvin]

I couldnt for the life of me figure out why expanding in x implied that was a constant, I think that clears that up for me :)

Rhadon does indeed mean that for integers differentiation needs to be redefined, as the same method does not give correct answers.

I'm not happy that is 'the answer' as other mistakes have to happen before you get that far. Its certainly hard to spot and it completly foxed me. Taken as is, the left side is correct. The first step on the right side is a mistake, as it produces an equation only valid for integers. If we continue the expansion is correct, but applying differentiation isnt, for the reason youve mentioned, and also because differentiating an integer only function isnt meaningful and contravenes the assumptions of calculus. However small a dx of 1 is, its still finite.

If we add the assumption to the problem that x only has to be integer valued, then differentiating the left hand side as before is not a valid operation, and the result is unsurprisingly wrong. The expansion is correct on the right hand side, but differentiating is an incorrect operation on for both prior reasons. We could define a correct version of calculus, but then we'd have to explain its derivation from first principles, and its too easy to simply fall over the correct answer (ie not conceal the fact we are aiming for a false one).