I might have just disproved math.

[Random Fella]

Snip

No... Just no
Surely you should have thought that if it was possible to disprove math, an extremely intelligent mathematician would have done it before you no? You know someone that studies math for a living?
Firstly you can't define 0/0
Therefore 0/0=x Means x=undefined
You can't divide by 0. Don't try

Math: 5 + 3 x 4 x 5^2 = 320

This is false, the correct answer is:

5 + 3 x 4 x 5^2
= 5 + 3 x 4 x 25
= 5 + 3 x 100
= 5 + 300
= 305

I have disproved bad math.

Wait what? Didn't you just disprove idiocy and implement math to get the correct answer?

 

[Maze1125]

infinity + 1 = infinity
There's absolutely nothing wrong with him producing that extra 1, infinity is such that it allows that.

Not true, infinity + 1 does in fact equal infinity, however the second infinity is greater than the infinity used first.

To use overly complicated mathematics to prove something conceptually quite simple.

The definition of L'Hopital's Rule is:
(where f(x) & g(x) are continuous, differentiable functions and f'(x) is the derivative of f(x) and g'(x) is the derivative of g(x))

If Lim[sub]x->a[/sub](f(x) / g(x)) = 0 / 0 || inf / inf then
Lim[sub]x->a[/sub](f(x) / g(x)) = Lim[sub]x->a[/sub](f'(x) / g'(x))

We can use this rule to prove how one function grows greater than another and thereby how one infinity is greater than another. Consider the example:

If f(x) = e[sup]x[/sup] and g(x) = x

Lim[sub]x->inf[/sub](e[sup]x[/sup] / x) = e[sup]inf[/sup] / inf = inf / inf (therefore L'Hopital's Rule applies, thus we derive the top and bottom and find the limit)

Lim[sub]x->inf[/sub](e[sup]x[/sup] / 1) = e[sup]inf[/sup] / 1 = e[sup]inf[/sup] = inf

The answer of infinity tells us that f(x) = e[sup]x[/sup] grows faster than g(x) = x, which can be seen quite obviously by graphing both equations. But if we simply plug in infinity to both equations we get:
f(inf) = e[sup]inf[/sup] = inf AND g(inf) = inf

By this logic f(inf) = g(inf), but as L'Hopital's Rule proves, f(x) grows faster than g(x) and (after checking initial values) there is no point x on the interval (-inf, inf) in which g(x) >= f(x). Therefore f(x) != g(x) on any x. As such infinity does not equal infinity.

Infinity can equal infinity, but it must be the same infinity. The infinity that is the result of f(x) is greater than the infinity created by g(x). Although f(inf) does equal infinity, it is a different infinity than the one in g(inf).

By the way there is a small error in there, I'm curious if anyone can find it and call me out on it. I'm at a loss of how to explain it without that error, but lets see if anyone finds it to begin with.

EDIT: To go back to the original example he is saying (1 - 1) * inf = inf - inf = 1 which is not true.
(1 - 1) * inf = inf - inf = 0 as it should. Think about it in a way that does not use infinities.

(1 - 1) + (1 - 1) + (1 - 1) = 0 = true

Shifting all parentheses to the right

1 + (-1 + 1) + (-1 + 1) + (-1) = 0 = still true

He is forgetting that last -1, the fact that it is (1 - 1) * inf, means that EVEN TO INFINITY, there are the same number of 1's as -1's and therefore equals 0.

I'm sorry, but you are wrong. You simply are.

I'm not going to explain why, because it requires maths you clearly haven't even started to learn yet, but you are still wrong. Stop thinking you know enough maths to make the claims you are, because you don't. This isn't arrogance, it is simply fact.

L'Hopital's Rule does not let you work out which infinities are greater than other, it merely lets you work out the limits of fractional functions. You are right when you say e[sup]infinity[/sup] is greater than just infinity, but you are very very wrong when you say that infinity + 1 is greater than infinity.

It takes a whole lot to make one infinity than another, it's not even true that infinity[sup]2[/sup] > infinity. Those two are still equal, even though one is the square of the other.

One last point. It is not true that 0[sup]x[/sup] = 0 for all x. Just try a negative x.

 

[mooncalf]

It's the "Disproved Math" bit alone that brought me here because from that much I knew someone would be getting their post torn apart by a mob of angry armchair professors. I don't know if Op is a clever troll or a silly person, but look at that reply count go!

 

[savandicus]

I feel a need to point on the most annoying part of this post.

Your title should be "I might have just disproved maths"

"Math" is not a word people, please stop using it. You dont disprove mathematic, you disprove mathematics. There is an "S", its a plural word. Maths is a shortening of Mathematics. Math is a shorting of mathematic and therefore stupid to use in any sentence.

Please add S's to plural words!

 

[PatrickXD]

If I were to say that x=1, that doesn't mean that 1 can be any number.
Algebra is just substituting unknowns into an equation, not any number. They're how we represent something we don't know about yet, and find out about it.

 

[Sexy Devil]

Punch 0/0 into your calculator for me and see what happens. Nothing divides by 0, not even 0. Full stop. The end.

Ok, I get it I'm wrong(If you people actual paid closer attention to my post you would realize i didn't think it was true ether) But this answer pissed me off the most. You punch something into your calculator and it said error so you ignore it? You don't care why? Just because a calculator cant understand it does not mean thats the end.

I do know why it's wrong, for starters. But any decent calculator will tell you exactly why 0/0 is wrong by stating "Undefined."

But yeah, if you can't trust your calculator with basic algebra then you might want to look into getting a new one.

 

[Keal]

Hi Zack,

If you really want to understand the problems in your example, you have to start at rather basic mathematics: the notion of a Group.
A Group is a set G together with an operation ° such that the following axioms hold:

Closure:
If a and b are elements in G the a°b is also in G.

Associativity:
For a,b,c in G the equality a°(b°c)= (a°b)°c holds.

Neutral element:
There is an element e in G such that a°e =e°a = a for all a in G. e is called the neutral or the unit element.

Inverse element:
For every a in G there is a b in G such that a°b = b°a = e. b is called the inverse of a and is formally denoted by a^(-1).

If we have Commutativity (a,b in G: a°b = b°a), we speak of an abelian Group.

From these axioms you can derive the "rules of calculus" you are used to from school (for the abstract group G).

If you look closely you will see that this group structure is realized in the real numbers R, even in two different ways.

First: the real numbers and addition (R,+)
Closure and associativity are trivial, the neutral element is 0 and the inverse of an element a is (-a). Note that the "-" is more a way to denote the inverse rather then an operation it self.

Second: the real numbers without zero and multiplication (R\{0},*)
Closure and associativity are trivial again, the neutral element is 1 and the inverse is 1/a. Again division is not an operation but a way to denote the inverse and that is reason way "it is forbidden to divide by zero", there is no multiplicative inverse of zero in the real numbers, i.e. there is no a in R such that a*0=1.
So if you decide "to do it anyway" you loose the group structure and the "rules of calculus" you derived using this structure do not apply anymore. That is why it seems easy to construct paradoxes starting from "1/0".


For completeness: If you "combine" two abelian groups with a distributive rule (a,b,c in G (a+b)°c= a°c+b°c) you get a field. If you do that for (R,+) and (R\{0},*) you get the "normal" real numbers with addition and multiplication.

 

[Brown Cap]

No, because only Chuck Norris can divide by Zero

 

[asacatman]

Hi Zack,

If you really want to understand the problems in your example, you have to start at rather basic mathematics: the notion of a Group.
A Group is a set G together with an operation ° such that the following axioms hold:

Closure:
If a and b are elements in G the a°b is also in G.

Associativity:
For a,b,c in G the equality a°(b°c)= (a°b)°c holds.

Neutral element:
There is an element e in G such that a°e =e°a = a for all a in G. e is called the neutral or the unit element.

Inverse element:
For every a in G there is a b in G such that a°b = b°a = e. b is called the inverse of a and is formally denoted by a^(-1).

If we have Commutativity (a,b in G: a°b = b°a), we speak of an abelian Group.

From these axioms you can derive the "rules of calculus" you are used to from school (for the abstract group G).

If you look closely you will see that this group structure is realized in the real numbers R, even in two different ways.

First: the real numbers and addition (R,+)
Closure and associativity are trivial, the neutral element is 0 and the inverse of an element a is (-a). Note that the "-" is more a way to denote the inverse rather then an operation it self.

Second: the real numbers without zero and multiplication (R\{0},*)
Closure and associativity are trivial again, the neutral element is 1 and the inverse is 1/a. Again division is not an operation but a way to denote the inverse and that is reason way "it is forbidden to divide by zero", there is no multiplicative inverse of zero in the real numbers, i.e. there is no a in R such that a*0=1.
So if you decide "to do it anyway" you loose the group structure and the "rules of calculus" you derived using this structure do not apply anymore. That is why it seems easy to construct paradoxes starting from "1/0".


For completeness: If you "combine" two abelian groups with a distributive rule (a,b,c in G (a+b)°c= a°c+b°c) you get a field. If you do that for (R,+) and (R\{0},*) you get the "normal" real numbers with addition and multiplication.

Seems logical, but I have a (probably stupid/obvious) question. When working with the field of "normal" real numbers, I have often (I think?) multiplied by 0. for example in the quadratic x^2=0, the solutions are both zero. This is apparently a thing you would do using the "rules of calculus" (I think, anyway). If you are never allowed to use the number 0 in the "normal" real numbers, How come I seem to come across it so much in my "normal" maths lessons? Can you explain, please?

Hopefully I haven't made too many idiotic statements.

 

[deadish]

You can't divide by zero.

Why not?

To quote the Terminator in T2, "It doesn't work that way."

Seriously though. What many people don't realize is that the "rules of mathematics" are actually completely arbitrary and have no relation whatsoever to the "real world".

All concepts in Mathematics are idealized abstractions and have nothing to do with the "real world" other than maybe being inspired by their "real world analog". The reason maths is useful at all is because many things in the real world approximate or correctly maps to some of the abstract mathematical concepts being studied - if they do great, or they don't too bad.

The rules of mathematics (a.k.a. axioms) are for most part completely "made up". They are most of the time chosen because they are "interesting" to the mathematician and occasionally because it's useful.**

**However they do have to be "consistent", meaning they must not contradict one another.

 

[Maze1125]

So if you decide "to do it anyway" you loose the group structure and the "rules of calculus" you derived using this structure do not apply anymore.

I think that's going a bit too far.
Yes, if you "do it anyway" you lose the group structure, due to the definition of what a group is, but that doesn't mean you have to loose all the "rules of calculus", you just have to be careful with them, and adapt them occasionally.

Seems logical, but I have a (probably stupid/obvious) question. When working with the field of "normal" real numbers, I have often (I think?) multiplied by 0. for example in the quadratic x^2=0, the solutions are both zero. This is apparently a thing you would do using the "rules of calculus" (I think, anyway). If you are never allowed to use the number 0 in the "normal" real numbers, How come I seem to come across it so much in my "normal" maths lessons? Can you explain, please?

Hopefully I haven't made too many idiotic statements.

Put simply, it's because fields have been built to account for that. So they allow it.

One reason it's okay, is because you can do multiplication by 0 using the addition of the field instead of the multiplication.
a*0 = a + (-a) for all a in the field.

 

[Zack1501]

Punch 0/0 into your calculator for me and see what happens. Nothing divides by 0, not even 0. Full stop. The end.

Ok, I get it I'm wrong(If you people actual paid closer attention to my post you would realize i didn't think it was true ether) But this answer pissed me off the most. You punch something into your calculator and it said error so you ignore it? You don't care why? Just because a calculator cant understand it does not mean thats the end.

I do know why it's wrong, for starters. But any decent calculator will tell you exactly why 0/0 is wrong by stating "Undefined."

But yeah, if you can't trust your calculator with basic algebra then you might want to look into getting a new one.

A bad calculator will say "error"
A good calculator will just go "Divide by Zero"
A better calculator will say "Undefined"
But Amazing calculator will say that it is, in fact, "Indeterminate" or undetermined. If you really think about, though it has its many flaws, theres something interesting behind this pointless math. I didn't think I could determine what mathematicians around the world could not but thats a sorry reason not to try.

 

[Keal]

Your point is not idiotic at all, indeed I was a bit unclear about that.
I did not mean to abolish zero, we need it to make the addition work and it is a perfectly good number as you pointed out. But although it does not fit in group structure of the multiplication it is still true that a*0=0 for all a in R.
This property makes multiplying by zero to solve an equation rather useless, because you loose all the information (the result is always 0=0 and the the starting equation and 0=0 are not considered equivalent).

 

[Okysho]

I wanted to know zero divided by zero equals. I tried to do at algebraically. This is what I did:

-The answer I was trying to get will be represented by x
0/0=x
-I times both sides by zero
0=0x

I've only got a few seconds but here's your first mistake. You can't do this. It's bad math. When you multiply both sides by zero, it's not just "moving one digit from one side to the other" you're creating an x/x situation (which equals 1) 0/0 is undefinded therefore by your equation in this multiplication step:

0/0=x
0(0)/0 = 0(x)
but you're still left with a 0/0 it cannot divide out to make 1.

Here's what you're doing without using 0.

x/7 = y
7(x)/7 = y(7)
x = y(7)

See the error?

This is not the error. It is not 0(0)/0 it is 0(0/0) and as we all know anything times 0 equals 0 so the entire side is 0 thus 0=x0. The error is me saying x=0 when x does not, it just can equal 0.

You still can't do that mathematical operation. Again, the beginning of your proof is based on the simple mathematical principal of "moving from one side to the other" which is in fact creating an x/x = 1 situation.

What you've just told me still doesn't hold up though. 0(0/0) won't evaluate to 0=x(0) (for one thing 0/0 is undefined, but let's roll with it) you'll wind up with 0=x, therefore 0=0 and still doesn't disprove that x/0 is undefined (as in indeterminate) ...

Oddly enough, (contrary to what the great Yahtzee says) Alice in wonderland is a book about how silly mathematics is.

 

[asacatman]

snip

snip

Thanks, you answered my question, even if I didn't totally understand the answer. (mostly through this being a topic I'm a bit ignorant about, despite doing some elementary group theory.)

 

[Grygor]

That part was actually a joke, the imaginery VS real bit. However, I'm going to need some citation on the part of you stating that imaginery numbers have an application beyond thought experiment. Since 'i' is literally representing a paradox, and that this is actually the tamest aspect of math acting less like science and more like philosophy, it smacks of carelessness. "We didn't feel like figuring out where this leftover piece of the puzzle actually comes from, so here, have a Lowercase-I."

As has been mentioned before, i is not a paradox - the letter is a representation of a number that, when squared, is equal to -1. The thing about i is it does not lie on the real number line (hence the term "imaginary"), because it is not a member of the set of real numbers (pretty much by definition, because any real number times itself cannot be less than 0).

But the set of complex numbers, of which i is a member, does not take the form of a line - it's a plane, and because of this fact, the complex numbers have certain properties that the real numbers don't. For example, every root of every complex number is itself a complex number. (For real numbers, this is only true for roots of positive numbers and even roots of negative numbers - all other roots of negative numbers are complex numbers, of which the real numbers are a subset.)

The mathematics of the complex plane give rise to a special identity, e^(i*pi)+1=0, aka Euler's Identity. One of the consequences of this identity is that using complex numbers greatly simplifies the analysis and design of digital signal filters.

To say nothing of the quaternions, a higher-dimension analog of the complex plane in which -1 has three square roots, labeled i, j, and k. One notable use of quaternions is as a representation of camera rotations and orientations in 3D games that allows for freely-rotating cameras that are immune to Gimbal Lock.

In short, just because you can't personally see applications for a given bit of math, doesn't mean those applications don't exist.

 

[careful]

Everyone who says "you cannot divide by 0" has never taken calculus.

So, I have an interesting math based question. If you don't like/hate math or don't understand basic algebra(I understand if you don't) just hit the big THE ESCAPIST logo in the corner ...snip... 0/0 also then x=0
-If you fallowed so far and remember that x can be any number then that means zero can also be any and every number. So 0 can now equal 5 or any other number.

I realize something is most likely wrong here.
So tell me escapist, Did i Disprove math?

None of your math is incorrect, but your conclusion is slightly off.

What you are doing here is better dealt with with calculus than algebra, but the short version is this:

First, you are correct in saying that 0 divided by 0 MIGHT equal a number.

Calculus shows that 0 divided by 0 has four possible answers.
Those answers are 0, 1, undefined, or infinity.

It all depends on what the actual value of 0 is.


So, no, you did not disprove math. What you actually did was discover some very developed mathematical ideas.

i dont mean to be a party pooper, but this is wrong. 0 plays a role in the methods of calculus, but before calculus was even discovered the definition of 0, and all its interesting properties, had been solidified and accepted. see thread here on the escapist titled

Dividing by zero, the truth (this is long!)

 

[nsqared]

No. You didn't. It's like trying to disprove the fact that you have ten fingers by repeatedly counting them and every time you get 10 you say you made a mistake and have another go.

 

[rayen020]

no you did not and cannot disprove math. At best you found a fallacy and even then its probably a known fallacy.

 

[Guardian of Nekops]

Multiplying both sides of the equation by zero is eqivalent to wiping the equation off the blackboard. It makes the value of both sides of the equation zero for sure, but in the process it destroys any meaning the equation might have had.

At the end of the day, all you've 'proved' is that zero equals zero.