I broke maths

[The Evil Thing]

http://i135.photobucket.com/albums/q145/ILPPendant/contradiction.png
Discuss.


Also, post other examples of the wanton destruction of mathematical rules... at the risk of getting inundated with "Find x", "Here it is".

Lastly, I'd appreciate it if you could tell me exactly what has gone wrong here, since no-one I know can figure out where the problem is.

[Dallas-Dakota]

*calls in Player Zero*


He broke maths?

[Oregano]

*Takes The Evil Thing out back while holding a shotgun*
....
*bang*
....
*comes back in*

The problem's been solved.

The closest way I knew to breaking maths was something probably similar to this but it doesn't work because you divided by 0 at a certain point.

[Calamity]

Oh... uh oh... let's hope that neither PZ nor Thufir find this thread, or our heads may explode. :smalltongue:

[Tamburlaine]

Me and my flatmate reckon you need to consider the case where x=0.

The best example I can think of for another maths breaker is the whole 0.999... =1 business.

[Lord Herman]

And, seen any giant tears in the fabric of reality appear yet?

[The Rose Dragon]

Then it is a case of taking the square root. You always get 1 ^ (1 / 2 * n), which can give either -1 or +1.

I mean:

x = 2 * pi * n

pi / x = 1 / (2 * n).

[The Evil Thing]

Me and my flatmate reckon you need to consider the case where x=0.

The best example I can think of for another maths breaker is the whole 0.999... =1 business.
0.9 recurring being equal to one is quite simple. Since one third is 0.3 recurring, three thirds must be 0.9 recurring. Three thirds is the same as one.

0 is not a member of X but you'll end up with e^(i0) = cos 0 + isin 0. Or 1 = 1.


Then it is a case of taking the square root. You always get 1 ^ (1 / 2n), which can give either -1 or +1.

I mean:

x = 2 x pi x n

pi / x = 1 / (2 x n).
Ah, very clever. That must be it.
Sadly I can offer no cookies but you have my congratulations, for what they're worth.

[Griever]

Then it is a case of taking the square root. You always get 1 ^ (1 / 2n), which can give either -1 or +1.

I mean:

x = 2 x pi x n

pi / x = 1 / (2 x n).

If you are going to use x as a variable, PLEASE use * for multiplication.

[Chronos]

Yeah, you've got to be careful with those multivalued functions. When you're just dealing with the real numbers, there's usually some clear choice for which value to pick for the function, but ambiguity can creep in once you start working with complex numbers. Another one to watch out for is the logarithm: You're on the verge of discovering there that log is also multivalued, and you can get some similar contradictions from it if you're not careful.

[The Rose Dragon]

If you are going to use x as a variable, PLEASE use * for multiplication.

Sowwy. I usually do these stuff on paper, not a keyboard. Never occured to me you could actually use that.

Although I'm more prone to using dots as a multiplication mark. :smallconfused:

EDIT: d-d, for simple stuff like this, we don't need mathematics geniuses like P_Z.

[adanedhel9]

http://i135.photobucket.com/albums/q145/ILPPendant/contradiction.png
Discuss.


π/x where x in {2nπ} is π/(kπ) = 1/k where k is in {2n}.

1(1/k) is the k-th root of 1. Odd roots of 1 are always 1. Even roots of 1 can either be 1 or -1. Since k is in {2n}, the k-th root of 1 can be either 1 or -1.

Argg, ninjaed. Oh well.

[Tamburlaine]

0.9 recurring being equal to one is quite simple. Since one third is 0.3 recurring, three thirds must be 0.9 recurring. Three thirds is the same as one.



No, I don't mean the thing itself, but what it can imply. Namely that 1-0.9 recurring=0.00...01 => 0.00...01=0

[The Rose Dragon]

The problem there is that 9s never end. There will never be a 1 at the end of infinite zeroes.

[The Evil Thing]

No, I don't mean the thing itself, but what it can imply. Namely that 1-0.9 recurring=0.00...01 => 0.00...01=0
1-1=0 :smallwink:

Facetiousness aside an infinitely small number will be 0 so it makes sense. It's not like infinity where you have different sizes. :smallyuk:

[Tamburlaine]

I bow to your superior knowledge, it's just one of those things that has always bothered me, is all.

[Winterwind]

http://i135.photobucket.com/albums/q145/ILPPendant/contradiction.png
Discuss.


Also, post other examples of the wanton destruction of mathematical rules... at the risk of getting inundated with "Find x", "Here it is".

Lastly, I'd appreciate it if you could tell me exactly what has gone wrong here, since no-one I know can figure out where the problem is.The problem is in the step where you change
(exp(i*x))^(pi/x)
into
exp(i*x*pi/x).

The rule that
(x^a)^b=x^(a*b)
is valid for real numbers only, not for complex ones.

[Tirian]

No, I don't mean the thing itself, but what it can imply. Namely that 1-0.9 recurring=0.00...01 => 0.00...01=0

That fact is also true. If 0.00000...001 and 0 were different real numbers, there would be other real numbers between them, but obviously there could be no such real numbers. The entire bit about putting ellipses in the middle of numbers is a barbarism anyway; it is a clumsy shorthand for saying that the sequence where a_n = 10^(-n) converges to 0, which should not inflict disbelief on anyone.

To plainly illustrate the flaw of the original "proof", observe the following:

1) Let x = -3
2) x^2 = 9
3) sqrt(x^2) = sqrt(9)
4) x = 3
5) -3 = 3

The problem is with step 4; sqrt(x^2) = |x|, which is only x when it is non-negative. In the original argument, you have correctly shown that e^i*pi is one of the x'th roots of unity (for some even x, so -1 is among the candidates as you would expect), but it is rash to assume that it is therefore 1.

[UncleWolf]

1=√1= √(-1*-1)=√(-1)√(-1)= i* i = i² = -1

Now that is a mindbender.

[Oregano]

How do you do a root sign on a keyboard, have I missed it?:smallconfused:

Sorry for the minor derailment.

[The Rose Dragon]

√1 may equal -1 or 1.

Taking an even root of something is like dividing by zero, only less well known.

[Tirian]

√1 may equal -1 or 1.

That isn't quite true, although it illuminates the central issue. While there are two solutions to the equation x² = 9, the √ operator is specifically designed to choose the positive solution. We can say that 3 is the (principal) square root of 9, but only because we can distinguish between 3 and -3 because one of them is greater than 0 and the other isn't.

But things are not so simple with the complex numbers. Again, there are two solutions to x² = -1, but as it turns out there is no way to distinguish between what we call i and -i. Therefore, there is no justification to speak of the square root of -1; whenever we write √(-1) = i, we should keep in the front of our mind that don't mean that but only that i is a complex number that satisfies the equation i² = -1.

[Mr. Moon]

You know, I once pulled the "Find X" "Here it is" on my math teacher.

He wasn't very impressed...

Also: *flees thread before my head asplodes*

[The Rose Dragon]

That isn't quite true, although it illuminates the central issue.

Well, it is quite true for the purposes of unbreaking math. :smalltongue:

[Deathslayer7]

want to see math at its finest.

0/0=

whats the answer people?

[averagejoe]

The problem is in the step where you change
(exp(i*x))^(pi/x)
into
exp(i*x*pi/x).

The rule that
(x^a)^b=x^(a*b)
is valid for real numbers only, not for complex ones.

As has already been established, pi/x=1/2n, n natural. It is relatively easy to show by induction that, for z=r*exp(ia), z^n=(r^n)*exp(ina). Now, the 2n-th root of z is the set of complex numbers W such that w^2n=z for w in W. For w=s*exp(ib), we have s^2n*exp(i2nb)=r*exp(ia). It's the case that two complex numbers with radii r1, r2 and angles a1, a2 are equal if and only if r1=r2 and a1=a2+2*k*pi, k an integer (this too is fairly obvious, though I can show it if anyone really has a problem with this step.) Therefore s^2n=r and 2nb=a+2*k*pi. so s=(r)^(1/2n) and b=(a/2n)+(2*k*pi/2n). Therefore, in this case, w=z^(1/2n)=(r^1/2n)*exp[i(a/2n)+i(2*k*pi/2n)]. r=1, and a=x=2*pi*n so w=exp[i(2*pi*n/2n)+i(2*k*pi/2n)]=exp[ipi+i(k*pi/n)], with k an integer and n a fixed natural number. Therefore the solutions are exp(i*pi), exp[i(pi+pi/n)], exp[i(pi+2pi/n)]... exp[i(pi+(2n-1)pi/n)]. Since exp(ia)=exp(ia+2*pi) (this comes straight from exp(ia)=cos(a)+i*sin(a), since both sine and cosine have the same values at a+2*pi as at a) exp[i(pi+2n*pi/n]=exp[i*pi], so the solution is the set exp[i(pi+pi*k/n] where k has integer values ranging from 0 to 2n-1. In other words, there are 2n solutions spaced evenly about the unit circle starting at the point -1+(0*i)=-1. It is worth noting that 1 is a solution, setting k=n. However, as Winterwind stated above, this is not a valid identity anymore since the root of a complex number gives a set of solutions.


1=√1= √(-1*-1)=√(-1)√(-1)= i* i = i² = -1

Using the same proof as above, on the complex plane the square root of one is a set. Since 1=exp[i*(0+2k*pi)], k an integer, the square root of one is the solution set exp(i*k*pi), k=0 or 1. Therefore it is not the case on the complex plane that 1=√1; the square root of one gives a set of solutions.

Also, the square root of negative one is the solution set exp[i*(pi/2+k*pi)] since -1=exp[i*(pi+2k*pi)]. In other words the square root of negative one is the solution set which consists of i and -i.

Edit:


want to see math at its finest.

0/0=

whats the answer people?

Undefined. :smalltongue:

[Deathslayer7]

Undefined. :smalltongue:

*smacks with broom*

No its do more work stupid. :smalltongue:

v: you'll learn more about 0/0 in calc 2. or even infinity/infinity

[Zarrexaij]

want to see math at its finest.

0/0=

whats the answer people?The answer is OH SNAP BLACK H---

Nah, it's undefined.

Boy I'm glad I'm only in calculus I. :smallbiggrin:

[The Rose Dragon]

Undefined. :smalltongue:

No, 1 / 0 is undefined. 0 / 0 is indeterminate. It can be any number, from real numbers and complex numbers. There is a difference.

[Deathslayer7]

*high-fives for you*

[Threeshades]

Hey guys, I just came in to tell you all that I am glad to have absolutely no frickin idea what the hell is going on here. :smallsmile:

[Ichneumon]

This is just like the day that I thought to have broken Physics by proving that velocity squared times the force was equal to the acceleration times the force.

[Zarrexaij]

This is just like the day that I thought to have broken Physics by proving that velocity squared times the force was equal to the acceleration times the force.This makes me smile. :smalltongue:

I've "broken" chemistry plenty of times. :smalltongue:

[averagejoe]

*smacks with broom*

No its do more work stupid. :smalltongue:

:smallfrown: You don't need to hit. :smallfrown:


No, 1 / 0 is undefined. 0 / 0 is indeterminate. It can be any number, from real numbers and complex numbers. There is a difference.

It largely depends on who you talk to. Algebraically this statement is accurate, since 0x=0 can have any x as a solution, whereas 0x=1 has none. However, if you use calculus, lim[(a,b)->(0,0)]a/b has no well defined solution. However, lim(b->0)a/b isn't well defined on the extended real line, but in the real projective line is an unsigned infinity, which is to say the point infinity in the set R union infinity, where infinity is neither positive nor negative. It's similar for the Riemann sphere, or C union unsigned infinity, where 0/0 and 0 times infinity are undefined but 1/0 is infinity.

Edit:


*high-fives for you*

:smallannoyed: *removes high fives*

[Deathslayer7]

i assumed we were on a calculus level here.

*checks first post again*

^: hater! :smallcool:

[Oregano]

This makes me smile. :smalltongue:

I've "broken" chemistry plenty of times. :smalltongue:

My maths teacher pointed out the reason, maths needs to be consistent and and perfectly correct whereas Science just has to predict accurately.

[The Evil Thing]

Did I post this twice or something? :smallconfused:

[Mando Knight]

http://i135.photobucket.com/albums/q145/ILPPendant/contradiction.png
Discuss.


Also, post other examples of the wanton destruction of mathematical rules... at the risk of getting inundated with "Find x", "Here it is".

Lastly, I'd appreciate it if you could tell me exactly what has gone wrong here, since no-one I know can figure out where the problem is.

The problem is that you're taking the square root of a number, and then assuming that both sides only provide the positive root. This is an incorrect assumption in problems involving i. What you need to do is remember that the 1 here is essentially i^4, so its square root is i^2, or -1.

Math's not broken, the assumptions used were.

Also, using 0.99999... = 1 to deduce that 1 - 0.99999... = 0 = 0.000... 001 is not hard at all. Remember that 1/∞ = 0? The hypothetical 0.000... 001 is that 1/∞. The one lies at such an infinitely distant point that it practically doesn't exist.

[SurlySeraph]

This entire discussion made sense to me. Cool, I haven't forgotten quite as much calculus as I expected.

[Collin152]

I broke math too, but that's by trying to do something I diddn't understand.

See, I was trying to find the inverse of the formula for a circle...

I wanted to know what the opposite of a circle was!

[chiasaur11]

I made a circle that uses 3 instead of Pi.

It... could be going better.

[Vizen]

I don't know what my friends did, but ages back they managed to get 14 = 14 out of some problem. I'm pretty sure they screwed up somewhere.

[Jack Squat]

Also, using 0.99999... = 1 to deduce that 1 - 0.99999... = 0 = 0.000... 001 is not hard at all. Remember that 1/∞ = 0? The hypothetical 0.000... 001 is that 1/∞. The one lies at such an infinitely distant point that it practically doesn't exist.

But there is no .00000...01 number to make this happen.

[Deathslayer7]

using sequencies and series, you prove indeed that .999999999=1

or to other way around 1-.9999999999=0

[Jack Squat]

using sequencies and series, you prove indeed that .999999999=1

or to other way around 1-.9999999999=0

or there's always 3/3 = 1 1/3 = .33333333 *3 = .9999999999, which I always found easier

[Deathslayer7]

that doesnt really prove it though.

[Jack Squat]

it proves .9999999999 = 1. I can infer the rest.

[Mando Knight]

But there is no .00000...01 number to make this happen.

Exactly. It would exist, but the definition of infinity shrinks the difference between it and 0 down to non-existence... so that number doesn't exist separately from 0. The 1 at the end doesn't exist because of its literally infinitesimal size--less than a millionth of the smallest number you can imagine. (note that if you take the millionth of the smallest number you can imagine, that number becomes the smallest you can imagine...)

[Krade]

How about this one:

Three men pay for a hotel room for one night.
The room cost $30.
They each pay $10.
They go to their room.
The manager realizes they paid too much and tells the bellhop to take them $5 back.
The bellhop realizes you can't split 5$ three ways and takes a $2 "tip".
Each man recieves $1 back for the room, meaning they each paid $9.
So the room cost them $27 (icluding the "tip").
The bellhop took $2.
Where's the other dollar?

[chiasaur11]

How about this one:

Three men pay for a hotel room for one night.
The room cost $30.
They each pay $10.
They go to their room.
The manager realizes they paid too much and tells the bellhop to take them $5 back.
The bellhop realizes you can't split 5$ three ways and takes a $2 "tip".
Each man recieves $1 back for the room, meaning they each paid $9.
So the room cost them $27 (icluding the "tip").
The bellhop took $2.
Where's the other dollar?

What other dollar?
$3 to the guys
$2 to the dishonest bellhop
$25 to the Hotel.
All accounted for.

[Pyrian]

I don't know what my friends did, but ages back they managed to get 14 = 14 out of some problem. I'm pretty sure they screwed up somewhere.I'm pretty sure fourteen does equal fourteen, actually. :smallwink:

[averagejoe]

How about this one:

Three men pay for a hotel room for one night.
The room cost $30.
They each pay $10.
They go to their room.
The manager realizes they paid too much and tells the bellhop to take them $5 back.
The bellhop realizes you can't split 5$ three ways and takes a $2 "tip".
Each man recieves $1 back for the room, meaning they each paid $9.
So the room cost them $27 (icluding the "tip").
The bellhop took $2.
Where's the other dollar?

The other dollar doesn't exist; you over counted by two. This number isn't going to add up to thirty since they didn't pay thirty, they payed 25. They each payed $9, which makes $27. However, from that, the bellhop took $2; adding the $2 to the $27 makes no sense, because now you are adding more money two it; they only payed $27 because the bellhop cheated them each out of $2/3. In reality they each should have gotten back $2/3, thus each having payed $8+1/3, which brings it up to 24+3/3=$25, which is how much was refunded.

To put it another way:
The room cost them $25.
They payed the bellhop $2.
They payed $27.
It all balances.

[potatocubed]

Just as an aside, according to this article I read at work (http://www.informaworld.com/smpp/content~content=a794477746~db=all~order=page) exactly 0 out of 42 "pre-classroom" maths teachers knew what a radian was.

Given that you can find out what a radian is in less than a minute on Wikipedia, that's pretty appalling.

[thubby]

it is my understanding that breaking maths is a sign that you have need of the next level of maths, and your calculator is insufficient.
*awaits ding*
gratz :smallbiggrin:

[DM Shake]

Brain.... hurts.... so..... bad.....

[Jack Squat]

Exactly. It would exist, but the definition of infinity shrinks the difference between it and 0 down to non-existence... so that number doesn't exist separately from 0. The 1 at the end doesn't exist because of its literally infinitesimal size--less than a millionth of the smallest number you can imagine. (note that if you take the millionth of the smallest number you can imagine, that number becomes the smallest you can imagine...)

I misunderstood you. I thought you were saying that 1/infinity allowed .999999 to equal 1. I had someone in my calc class argue that, took forever to convince them otherwise.

[Massy]

http://img391.imageshack.us/img391/5317/math3cm.jpg

Easier than the complex powers one.

[averagejoe]

Just as an aside, according to this article I read at work exactly 0 out of 42 "pre-classroom" maths teachers knew what a radian was.

Given that you can find out what a radian is in less than a minute on Wikipedia, that's pretty appalling.

The problem is that most math and science teachers know neither math or science. A girl who I take math with, and is also becoming a teacher, suggested that everyone who goes into these fields (i.e. majors in them) should be required to teach for two years as part of their education, or something similar. Just a thought.


http://img391.imageshack.us/img391/5317/math3cm.jpg

Easier than the complex powers one.

a=b so dividing by (a-b) is dividing by zero, and then all bets are off. As previously stated, algebraically 0/0 can be any number.

[Zeebiedeebie]

*Zzk* *Zzk* *Zzk*... [SEGMENTATION FAULT!]

[Tirian]

Just as an aside, according to this article I read at work (http://www.informaworld.com/smpp/content~content=a794477746~db=all~order=page) exactly 0 out of 42 "pre-classroom" maths teachers knew what a radian was.

Given that you can find out what a radian is in less than a minute on Wikipedia, that's pretty appalling.

ZOMG. Golly, I reckon they also don't know how many grains there are in a dram.

Putting on my hat that allows me to speak for all mathematicians, I can't imagine what practical application pre-classroom mathematics would have with an esoteric angle measurement that doesn't pay off until second-semester calculus, and I can only hope that elementary school math teachers who are teaching children to use protractors don't find it necessary to burden kids with radians either.

[Pyrian]

My reading of the abstract of that study dramatically contradicts potatocubed's summary. It wasn't that they had no idea what a radian was, it was that they described it differently than the researchers wanted them to.

It's like all those people who were described as unable to find the U.S. on a map because they didn't also point out Puerto Rico.

[Phae Nymna]

I broke math too, but that's by trying to do something I diddn't understand.

See, I was trying to find the inverse of the formula for a circle...

I wanted to know what the opposite of a circle was!

Oh god. Would that be everything inside of a circle laid out over the infinite multiverses?

[Collin152]

Oh god. Would that be everything inside of a circle laid out over the infinite multiverses?

Depending on which trial you use...
It's either an almost-line, two vertical lines, or every point on a plane but one.

If someone has a better answer, I'd like to hear it!

[Tirian]

Lulz, good call. Of course, the definition of the radian measure of an angle is the ratio of the subtended arc-length of that angle against an arbitrary reference circle centered at the apex of the angle to the radius of that circle. What sort of pi-obsessed weirdo would say anything different?

[averagejoe]

ZOMG. Golly, I reckon they also don't know how many grains there are in a dram.

Putting on my hat that allows me to speak for all mathematicians, I can't imagine what practical application pre-classroom mathematics would have with an esoteric angle measurement that doesn't pay off until second-semester calculus, and I can only hope that elementary school math teachers who are teaching children to use protractors don't find it necessary to burden kids with radians either.

One of my calculus teachers got sent to the principal's office in elementary school for insisting that negative numbers existed. I went years disbelieving Einsteinian relativity because of something I "learned" in seventh grade. Teachers not knowing the subjects they teach is a problem. Not knowing radians isn't significant because kids need to know radians, knowing radians is significant because it's what you learn once you achieve a certain level of mathematics.

That's not even getting into the problems of the structure of the curriculum for mathematics.

[Deathslayer7]

ok. You do not need to know radians until you reach geometry at the earliest. That is about 8th/9th grade. I see no reason why elementary students need to know radians. They wouldnt even understand where it came from .

[Tirian]

One of my calculus teachers got sent to the principal's office in elementary school for insisting that negative numbers existed. I went years disbelieving Einsteinian relativity because of something I "learned" in seventh grade. Teachers not knowing the subjects they teach is a problem. Not knowing radians isn't significant because kids need to know radians, knowing radians is significant because it's what you learn once you achieve a certain level of mathematics.

I'm not buying it. I will join you in anger against a teacher who denied the existence of negative numbers, or one who claimed that the Fifth Postulate was inviolate, because that is shutting out the existence of actual and very valid fields of mathematics. But radian measure is a translation device, something that as far as I can tell exists only because that is how we make the Taylor expansions for trig functions work out smoothly. It is mildly elegant as a unitless measure and is a useful nugget of information for computer programmers since the low-level trig functions apply the Taylor expansion, but for pretty obvious reasons it has had no traction in lay society.

Plus, as Pyrian noted, based on the abstract it would seem that the paper's gripe is not that the teachers didn't know what a radian was, it's that they evidently defined it as the degree measure times pi over 180 degrees, instead of defining it in terms of a subtended arc length in a manner similar to what I pulled out of my ... hat a few posts above. I suppose I can understand why that would get under some formalists' skins -- I myself get tweaked when people define C(n,r) by a formula instead of sensibly defining it as the cardinality of the set of r-element subsets of an arbitrary set with n distinct elements and derive the formula -- but I'm not going to write a paper that calls educators onto the carpet over it.

[averagejoe]

ok. You do not need to know radians until you reach geometry at the earliest. That is about 8th/9th grade. I see no reason why elementary students need to know radians. They wouldnt even understand where it came from .

I think you underestimate elementary school students. However, this isn't what I suggested anyhow.


I'm not buying it. I will join you in anger against a teacher who denied the existence of negative numbers, or one who claimed that the Fifth Postulate was inviolate, because that is shutting out the existence of actual and very valid fields of mathematics. But radian measure is a translation device, something that as far as I can tell exists only because that is how we make the Taylor expansions for trig functions work out smoothly. It is mildly elegant as a unitless measure and is a useful nugget of information for computer programmers since the low-level trig functions apply the Taylor expansion, but for pretty obvious reasons it has had no traction in lay society.

Plus, as Pyrian noted, based on the abstract it would seem that the paper's gripe is not that the teachers didn't know what a radian was, it's that they evidently defined it as the degree measure times pi over 180 degrees, instead of defining it in terms of a subtended arc length in a manner similar to what I pulled out of my ... hat a few posts above. I suppose I can understand why that would get under some formalists' skins -- I myself get tweaked when people define C(n,r) by a formula instead of sensibly defining it as the cardinality of the set of r-element subsets of an arbitrary set with n distinct elements and derive the formula -- but I'm not going to write a paper that calls educators onto the carpet over it.

Whatever, I don't care about the article. Point is, our math and science teachers are undereducated.

[UncleWolf]

How do you do a root sign on a keyboard, have I missed it?:smallconfused:

Sorry for the minor derailment.

Microsoft office.
copy and paste work wonders

[The Rose Dragon]

Alt - V also works in my case.

[Yarram]

Me and my flatmate reckon you need to consider the case where x=0.

The best example I can think of for another maths breaker is the whole 0.999... =1 business.

Aw I wanted to say that. Idea stealer.