-∞ + ∞ = 0

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-∞ + ∞ = 0

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>>7760109

[eqn] \left ( -\lim_{x \to \infty} x \right ) + \left ( \lim_{x \to \infty} x \right ) [/eqn] Which, by the algebraic limit theorem, becomes: [eqn] \lim_{x \to \infty} (-x+x) = 0 [/eqn]

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Infinity is not a number.

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>implying

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>>7760121

do it again and I will fucking kill you you bastard son of a whore

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>>7760218

He's right though...

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>>7760109

Yeah that is the kind of logic a middle aged boomer would use

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>>7760207

Watch me make it one, faggot.

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something that doesn't exist + something that doesn't exist = something that doesn't exist

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>>7760121

Algebraic limit theorem only works with convergent sequences.

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>>7760303

Infinite is a concept meaning "without limit". Nothing you can do makes that a number.

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>>7760109

Infinity cannot be really be expressed in negatives, because it means an infinite sequence, so not really a rational number

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>>7760447

[math]-1 \cdot a_{n}[/math]

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Actually ∞+-∞=∞

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>>7760545

I thought that

∞-∞=that's not how maths work you tard

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>>7760121

You can merge limits iff they exist and existence of limits do not include inf. or -inf.

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>>7760546

That's b/c your'e still in high school

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>>7760304

>0 doesn't exist

SHIT

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So, does [math]\aleph_{-1}[/math] exist?

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>>7760303

Can't really do that consistently, but I'd like to see you try.

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>>7760856

>Adding a 1 is hard

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>>7760867

Problems like

a+ infinity = infinity

a+ infinity - infinity = infinity - infinity

a = 0

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>>7760872

Not if you preserve the expression in limits

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There are infinite infinities :)

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>>7760877

Yeah, but then those things aren't numbers.

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>>7760877

>infinity is a "limit"

freshmen taking calculus need not post their opinions on math

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>>7760883

hey, talking about serious stuff,

how many "types" of infinity are there?

- natural numbers <

- irrational numbers <

- all possible curves on a plane

this this hold still?

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>>7760904

>"types" of infinity

do you mean how many different infinite cardinals are there? an infinite amount, they're also well-ordered

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>>7760121

>algebraically adding non convergent limits

You think this is a fucking game

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>>7760912

What is the cardinality of the set of cardinalities?

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All real numbers infinitely added together = infinity

;^)

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>>7760904

like...countably and uncountably?

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>>7760872

Your error is you don't realize that infinity minus infinity is still infinity. I'ts not zero. There is no problem

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>>7760403

It is an element of the extended real line, which is good enough for me.

It actually allows for interesting metric spaces.

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>>7760912

Hey anon, I believe you so don't worry, but how are they well ordered?

I have no fucking idea. Is it based on bijection a between sets of certain sizes? Like how the naturals have the same size as the rationals because they're isomorphic.

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>>7761371

https://en.wikipedia.org/wiki/Aleph_number

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>>7761365

https://en.wikipedia.org/wiki/Hyperreal_number

https://en.wikipedia.org/wiki/Surreal_number

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>>7760555

Just use l'Hôpital

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>>7760872

infinity-infinity isn't even a defined mathematical operation, ie it doesn't mean anything

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>>7760881

>N is a finite infinity

>R is an infinite infinity

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>>7761881

>finite infinity

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>>7761881

>finite infinity

am i being rused here

i think the word you're looking for is countable

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R - R = 0

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infinity x infinity = ?

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>>7762093

Infinity :^)

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-∞ + ∞ = nothing. literally.

infinity is not a number

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>>7760121

Is this the new 0.9999=1?

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>>7760109

Only if you assume that the infinite expansions are occurring at the same rate though.

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The concept of infinity just blows my mind.

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>>7762186

Theorem:

0.9999 ≠ 1

Proof:

Assume not.

Then 1 - (1 / ∞) = 1.

From which it follows that ∞ - 1 = ∞.

However, this is a contradiction since x - 1 < x for any x.

Therefore, 0.9999 ≠ 1.

Q.E.D.

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>>7760109

You can't do that.

But lets say you have a function f such that

f(x) = x - x

This function yields 0 for all x and it is obvious that the limit as x approaches infinity is also 0.

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>>7762218

That's a sure way to show your prof that you haven't learned a thing about math.

Infinity is not a number and you cannot use it as a number. Doing so is just plain stupid.

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>>7762226

In other words, f(x)=0 for f: R->R

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>>7762379

Yeah but lets not forget that f(x) = 0 and f(x) = x - x are different functions.

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>>7762371

>Infinity is not a number and you cannot use it as a number.

Infinity is a number and can be used as a number.

https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations

You retarded piece of shit.

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>>7762201

>le 4th grade math meme

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>>7762426

>proving a property of the reals

>using the hyperreals

Are you literally retarded.

Also, even in that 'extended real number line', the operation that OP is using is also undefined.

I recommend brain extraction surgery. You don't seem to use it/need it.

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>>7760121

DELETE THIS NOW!

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>>7762438

I don't give a shit about OP. I only refuted your retarded "infinity is not le number xD" reddit shitpost.

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>>7760121

>analytic continuation - not even once

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>>7762467

Please care about me

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isn't this precisely what we say when we integrate an odd degreed function within -∞ to ∞

Despite what the rules state, there will always be an equally large and negative "rectanle" on the left for every positive one on the right to the origin, since the fuction is behaves the same in both directions only differing by a sign

y=x, y=x^3 for example should yield 0 when integrated as such

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>>7761276

>set of cardinalities

That's actually not a set (it's a proper class), so no cardinality.

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>>7760109

True as long as both of them are the same infinity.

Unfortunately there are an infinite number of different types of infinity. So the probability that two randomly selected infinities were equal would be 1/infinity which is zero.

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>>7762419

It's the same function.

Or in your world is x-x not equal to zero?

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>>7762426

Unless explicitly stated otherwise, it is safe to assume that we are talking about the reals. There is number infinity in the reals.

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>>7762597

When you integrate from minus infinity to infinity, you are doing a limit whether or not you write it as such.

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>>7762748

The word "number" is not synonymous with "real number". Please go take a math class.

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>>7762741

But they are not. x - x may not be best example but it is easy to show that the simplification of a function is not the same function if we introduce a similar function.

Consider the function

[math] f(x) = \frac{x - x}{x} [/math]

Here you may be tempted to simplify it and claim that

[math] f(x) = = \frac{x}{x} - \frac{x}{x} [/math]

Therefore

[math] f(x) = 1 - 1 so f(x) = 0 [/math]

But here you would be wrong. You need only notice that the original function is not defined at [math] x = 0 [/math] but the simplified function is indeed defined at [math] x = 0 [/math]

Therefore, a function and its simplified form are not the same. Even if for some functions like [math] f(x) = x - x [/math] there is no difference.

btw, this is specially important when you work with limits, which is what we are doing here.

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>>7762784

how's the first analysis class doing? just because you can find an example of two similar looking functions where one is not continuous somewhere doesn't mean there cannot be two equivalent functions written in different ways.

f: R -> R, f(x) = x - x and g: R -> R, g(x)=0 are completely identical

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>>7762784

>Therefore, a function and its simplified form are not the same. Even if for some functions like f(x)=x−x there is no difference.

You need to clean up your logic, this is retarded.

A function and its simplified form are not _always_ the same.

Unless you change the domain somehow (usually by introducing or removing a division by zero) they're identical and you can substitute one for the other freely.

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>>7762819

Computing f(x) = x - x takes three steps, plugging in two x's and doing the substraction.

Computing f(x) = 0 takes 1 step, just put 0.

Just another meaningless difference that shows the difference between graphically equivalent functions.

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>>7762840

The number of "steps" isn't relevant in any mathematical sense though

unless you wanted to talk about computational complexity or something, but I'm not aware of any branch of analysis that cares how many "steps" it takes to evaluate a function

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>>7760904

There are too many infinities to assign a number to, in any number system. The infinity of infinities is just too large to even talk about in any meaningful sense.

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>>7760109

Indeterminate, but can be true.

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>>7762840

If f(x)=0 and f(x)=x-x are two distinct functions, then for what value of x do they differ?

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>>7762840

I'm sure you've had a silly definition of a function as an operation between two sets, but I'm sorry, it's not :,(

Rigorously, a function is a triple of sets, (M, N, G) s.t.

-G is a subset of MxN

-For all elements of M, say x, there exists a unique element of N, say y, s.t. the tuple (x,y) is a member of G.

So let's consider the two functions...

First f:R->R, x|->0.

With the notation above, clearly M=N=R.

Now, what is G? Clearly {(x,0)|x is real}.

So we have f = (R, R, {(x,0) | x is real}).

Now, for f:R->R, x|->x-x.

We have that M=N=R once again.

Now, G = {(x, x-x) | x is real}, but since x-x := +(x, x^-1) , we have x-x=0 since (R,+,0) is an abelian group. Therefore G={(x,0)| x is real}.

Therefore in this case, f=(R, R, {(x,0) | x is real}.

Now, since we've completed an entirely unnecessary and utterly trivial proof, can you stop? This is barely even shit for baby's first analysis. It fucking hurt to write this stupid shit out for you.

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>>7762419

f(x) = 0

g(x) = x-x

f(x) = g(x) for all values of x

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>>7763510

They are equal but they are not the same function.

f(x) is the operation of just plugging 0.

g(x) is the operation of substracting x to itself

that they happen to yield the same value is simply not important.

Lets say that you had other functions f and g. Any functions.

What you are saying is that if for some set of numbers I: i1,i2,i3,...,in f(x) =g(x) then f and g are the same function for that set of numbers. And that is simply not the case. They yield the same value but they are not same.

And here you may say

>Well, but x - x and 0 yield the same value everywhere

Well, there is no difference. It just means that set I between f and g is an infinite set.

So unless you claim that the function

f(x) = x^2 is LITERALLY the same function as g(x) = x for I = {0,1} then you can't claim that that f(x) = x - x is the same function as f(x) = 0 for the infinite set of the reals.

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>>7763853

f(x) = x^2 is LITERALLY the same function as g(x) = x for I = {0,1}

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>>7763859

It is not the same function. It just yields the same value for those numbers.

If it is literally the same function then why wouldn't you feel content graphing x squared in a graph of x.

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>>7763867

I would for only those two points. It's not that hard to understand.

A function is simply a relation between a set of outputs and a set of inputs. if the inputs and outputs of two functions are exactly the same then the relationship is the same.

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>>7763853

The actual method, steps involved are different. But if you think of functions in terms of a table mapping inputs to outputs, in that sense they're the same.

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>>7760207

It is.

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>>7763853

By the simple rules of arithmetic on the reals, when you subtract a number from itself you get zero. Thus, x-x=0. And thus, the functions f(x)=x-x and g(x)=0 are identical.

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>>7760109

> infinity and zero don't even exist

Literally nothing

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>>7763952

But we are talking in the context of the hyperreals and there, the operation of infinity-infinity is undefined.

That means that if we were able to do a 'hypergraph' then our function would stop being continuous at infinity, where the function would be undefined.

But the function f(x)=0 is always defined, even in our hypergraph.

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>>7763853

f:{0,1}->{0,1}, x|->x^2. f = ({0,1}, {0,1}, {(0,0),(1,1)})

g:{0,1}->{0,1}, x|->x^2. g = ({0,1}, {0,1}, {(0,0),(1,1)})

f=g

It's fucking simple. The two functions are the same.

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>>7762665

this desu

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>>7762218

lol

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>>7763958

The only reason they aren't the same then is because the outputs don't match. When they do match they are the same function.

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Honestly I started out laughing at the retarded posts at the beginning of the thread, but this anon claiming f(x)=x-x is not the same function as g(x) = 0 doesn't actually seem to be joking, and I can't laugh about it anymore.

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>>7764167

Probably a comp sci major

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>>7761881

>Finite infinity

>Finite

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>>7764608

countable and uncountable

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>>7761341

>infinity minus infinity is still infinity

what am i reading

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>>7764651

The truth.

[math]

\infty - 1 = \infty

[/math]

similarly,

[math]

\infty - 2 = \infty

[/math]

etc.

[math]

\infty - n = \infty

[/math]

for any n regardless of how big it is.

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>>7764686

>ze truss

stopped reading there

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Minus 2 points.

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>>7764686

You can do the exact same method but with n-infinity to "prove" that infinity-infinity=negative infinity

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>∞ = ∞

>∞ + 49 = ∞ - 36

Are both true?

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>>7760912

What do red birds gave to do with this, retard?

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>>7761881

Don't mind me just needed to remind myself this post existed

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>>7764855

Sounds good. How about we agree that

[math]

\infty - \infty

[/math]

is meaningless?

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