Try to solve this:

[Anonymous No.10029258]

Rules:
*Must not be equal numbers
*Must not be zero

[lamarr No.10029264]

>>10029258
>*Must not be equal numbers
What ? I can only see letters in your pic. You didnt define the letters.
>*Must not be zero
What must not be zero ?

[Anonymous No.10029271]

>>10029258
1 - 0.999... = 0

:^)

[lamarr No.10029274]

>>10029271

infinity - infinity = 0

did i gud senpai ?

[Anonymous No.10029275]

>>10029258
You haven't defined anything in your equation, so as far as I know the minus sign can be any binary operation over any set S and 0 any element in S. Thus the set of pairs (a,b) in SxS such that a-b=0 could be anything between the empty set and the whole SxS.

[Anonymous No.10029291]

A= 5
B= absolute value of - 5

Done

[Anonymous No.10029296]

>>10029291
absolute value of - 5 = 5, retard

[Anonymous No.10029312]

>>10029274
nope, undefined

[Anonymous No.10029313]

>>10029296
Absolute value of anything equals whatever the number is, without the sign.

Doesn’t mean they are equal though.

[Anonymous No.10029318]

>>10029312
but it could be 0, therefore the equation can be true. Also it therefore does contain the constraint of op

>*Must not be zero

[Anonymous No.10029324]

>>10029313
|-5| = 5, they are equal

[Anonymous No.10029336]

>>10029258
try this one

ln(1) - ln(e^(2*pi*i))

[Anonymous No.10029342]

>>10029336
>Must not be zero

[Anonymous No.10029343]

>>10029258
a

[Anonymous No.10029346]

>>10029343
What?

[Anonymous No.10029349]

>>10029258
a-b=0
a=b
But a is supposed to not be equal to b. It's a contradiction.
There is no such pair of numbers a and b that fulfills the criteria. Q.E.D.

[Anonymous No.10029350]

literally Euler's formula you retards

[lamarr No.10029351]

>>10029349
yes there is. infinity.

[Anonymous No.10029354]

>>10029350
Brainlet

[lamarr No.10029356]

>>10029350
see
>>10029342

[Anonymous No.10029361]

>>10029351
Infinity is not a number.

Also none of infinite surreal numbers fulfill the criteria. Any set with linear order won't work.

[Anonymous No.10029362]

>>10029350
Are you proposing that a=e^(ipi) and b=-1? Then a = b since e^(ipi) = -1

[Anonymous No.10029377]

>>10029258

Let a = "lowest non" and b = "negative number"

QED

[Anonymous No.10029397]

>>10029258
lowest non-negative number=0

[Anonymous No.10029412]

>>10029258
a=0
b=0
0 is not a number

[Anonymous No.10029418]

>>10029258
Zero is not a number, so it's not possible.

[Anonymous No.10030099]

>What are Cyclic Groups
All of you failed.
Sage.

[Anonymous No.10030987]

>>10029258
a and b are infinitely close hyperreal numbers. here = denotes equivalence of hyperreal numbers.

[Anonymous No.10030993]

>>10029258
- here denotes multiplication
So there are pairs (a,b) of solutions with a=0 and b !=a

[Anonymous No.10030997]

>>10029377
Underrated

[Anonymous No.10030998]

a=P, b=NP

[Anonymous No.10031023]

5 apple - 5 orange = no fruit
= 0
Everyone in this thread is a moron

[Anonymous No.10031093]

>>10030099
You are so dumb. If a and -b are elements of a cyclic group, and 0 is the identity, then -b is the inverse of a which implies --b = b = a

[Anonymous No.10031095]

>>10031023
How does that equal no fruit? You still have 5 apples and -5 oranges

[Anonymous No.10031124]

>>10031095
Ok i eat all the fruit now there's no fruit
Me + fruit = 0 fruits

[Anonymous No.10031233]

a=1, b=3, where equality is the equivalence relation of congruence modulo 2.

[Anonymous No.10031333]

a=b
a=brainlet
b=brainlet
brainlet-brainlet=0

[Anonymous No.10031337]

>>10029258
1/0 - 2/0 = 2

[Anonymous No.10031357]

>>10031337
* 1/0 - 2/0 = 0
mouse slip. I hope the image makes up for it

[Anonymous No.10031374]

>>10031233
>where equality is the equivalence relation of congruence modulo 2.
Then a=1=3=b, retard

[Anonymous No.10031389]

>>10029258
assuming "-" is modulo 2 subtraction, a = 3 and b = 1

[Anonymous No.10031396]

>>10031357
a and b are nilpotent matrices.

[Anonymous No.10031416]

>>10030099
>>10031389
Integers equivalent modulo n are de facto equal in the quotient group Z/nZ. Same in any cyclic group, writing the same element in two different ways doesn't make them different elements. Besides, in any additive group
a-b=0 => a=b
You are looking for a serious answer to an unserious question.

[Anonymous No.10031420]

>>10031416
>You are looking for a serious answer to an unserious question.
oh really? what was your first clue?

[Anonymous No.10031425]

>>10031357
a and b are arrows in an abelian category C. If 0 is the (existing and unique up to isomorphism) zero object, and X is any nonzero object, let a be the unique arrow X->0 and b the unique arrow 0->X; then ab is the unique arrow 0:0->0 .

[Anonymous No.10031445]

a-b=0
1(a-b)=0
(0/0)(a-b)=0
0(a-b)=0*0
0(a-b)=0
which is true for all a and b

[Anonymous No.10031467]

>>10029258
>Must not be equal numbers
So if they're not numbers they can be equal huh?

[Anonymous No.10031737]

A = 1 or x. B = -1 or -x. U guys r dumb

[Anonymous No.10031751]

>>10029258
8=D

Discuss.

[Anonymous No.10031793]

e^{-i\pi}-1=0

a^0 - 1=0

[Anonymous No.10031820]

>>10029258
A= i
B= i
Done bango bingo
qued

[Anonymous No.10031839]

I only solve problems with Kurisu in it

[Anonymous No.10031890]

>>10029258
you want me to tell you how I know this is bait?

[Anonymous No.10031897]

>>10031793
e^{-i\pi} - 1 = -2
a^0 = 1
Both examples don't work

[Anonymous No.10031921]

>>10031897
e^{-i\pi} = a
-1 = b

e^0 = a
1 = b

[Anonymous No.10031933]

>>10031921
a = e^{-i\pi} = -1 = b
a = e^0 = 1 = b
Still doesn't work

[Anonymous No.10032058]

>>10029258

I define "-" as a function that takes two real numbers a and b, notated "a - b", and returns 0 regardless of the values of those numbers.

a = 1, b = 2

a - b = 0

[Anonymous No.10032066]

1- -1 = 0

and no 1 and minus 1 are not 'equal' numbers try eating negative one apples, you can't.

[Anonymous No.10032088]

a = b + 0
a = b + 1
b = b + 1
b - b = 1
0 = 1

[Anonymous No.10032102]

OP is a stupid nigger

[Anonymous No.10032118]

>>10029258
a = OP
b = faggot
They're not equal numbers, they're equal faggots

[Anonymous No.10032120]

>>10032118
Also they're not zero, they're less than zero

[Anonymous No.10032276]

>writes the same number in two different ways
>thinks they're different

[Anonymous No.10032285]

>>10029258
a=8, b=0

[Anonymous No.10032452]

a = pi, b= 3pi

pi-3pi
=(cos/cos)(pi-3pi)
=(cos pi - cos 3pi )/cos
=(-1 +1)/cos
=0

[Anonymous No.10032464]

>>10029258
a= 5
b= 17

[Anonymous No.10032552]

>>10029258
We assume that a=1/2b(both not 0) and each variable exists in it's own reality, only when it is being observed it becomes accessible and they are observed one at a time from left to right. Once a. is observed a counter t. starts, it increases at intervals of +1 and the observer moves one step right. Each +1 for t. or step right corresponds to a change in the values of a. under these rules:
a. doubles every t+n. We also assume that once a variable has been observed it freezes and is kept in it's own pocket timeline, this copy stops following the t. rule which applies only if the observer is present.

we get:
t=0
a[-b=0]

t=1
((2a))-b[=0]

Now have a custom rule that at t=2 collapses the timelines and merges the different realities we've created into one, retaining the last available state for each, using the observer's timeline as reference to get the state:
t=2
(((2a-b=0)))

[Anonymous No.10032569]

>>10029258
Alright guys, let's solve this one

0.1 = Acos(4- φ)
0.2 = Acos(8- φ)

[Anonymous No.10032675]

>>10029258
a=(1+1)
b=2

a-b=0
(1+1)-2=0

Bow to your king brainlets

[Anonymous No.10032687]

>>10032675
But 1+1=2

[Anonymous No.10032924]

>>10029258
Implying this is rationals, you are saying that a-b=0 which is the same as saying a=b, so this is an unsatisfyable proposition when a!= b is required. Aka this is a contradiction.
Right guys?

[Anonymous No.10032926]

>>10032675
Still doesn't satisfy a!=b

[Anonymous No.10032930]

>>10029258
a-b=0
a=b

They aren't equal numbers, they're equal letters ;^)

[Anonymous No.10032937]

>>10029258
a is i
b is -i

[Anonymous No.10032947]

>>10032937
Then a-b = 2i

[Anonymous No.10032968]

>>10029258
You're looking for a shitty answer of 1 and 0.99999...? Because that is also wrong since in that case a=b

[Anonymous No.10032970]

ZERO = NEUTRAL INFINITY.

[Anonymous No.10033158]

>>10032970
There are no positive or negative infinity, so neutral infinity doesnt even make sense

[Anonymous No.10033209]

>>10032285
BASED AND REDPILLED

[Anonymous No.10033228]

Let a=1
Let b=(a^1/2)^2

[Anonymous No.10033245]

>>10029258
is this bait?

[Anonymous No.10033249]

This is like the cat and laser pointer game. Except with autists.

[Anonymous No.10033253]

Just stop posting, the only correct answer was posted here >>10029377 and everyone else failed.

[Anonymous No.10033283]

>>10033253
"Lowest non" and "negative numbers" are not numbers, but strings; and whatever a subtraction of string is, it is not the collation of them; and "0" and "lowest nonnegative number" are still different as strings.

[Anonymous No.10033321]

>>10033283
Nowhere in the problem does it say a and are numbers (it simply says they are not the same number or 0) and nowhere does it say "-" is a subtraction sign. If both were true then since 0 is the additive identity we have a=b, violating the premise of the problem. Thus by contradiction both cannot be true. If a and b are not numbers then the minus sign has no meaning in this context, thus the only rational conclusion is that "-" is a hyphenation between a and b which equals 0. One such hyphenation is "lowest non-negative number." Since "lowest non" and "negative number" are not the same number and aren't 0, this solves the problem.

[Anonymous No.10033323]

>>10029313
bait

[Anonymous No.10033326]

>>10029349
This.
/bait thread

[Anonymous No.10033331]

>>10032066
>1 - (-1) = 0
>1 + 1 = 0
Literal retard.

[Anonymous No.10033375]

Just invent a new set of numbers where this can be applied.

[Anonymous No.10033478]

>>10033321
But the strings "lowest non-negative number" and "0" are not equal. Hyphenation operates on strings. Two strings are not equal when they are synonyms.

That a and b are said to be different numbers, but not that they are numbers, has already been remarked before. >>10031467
So why can't they be equal matrices? Or equal vectors? Or equal functions? Or equal doctor whos? This thread is stupid.

[Anonymous No.10033481]

>>10029349
>Assumes - denotes subtraction in an abelian group
Dumbass

[Anonymous No.10033508]

wouldn't something like this work:

(-6) - (6) = 0

[Anonymous No.10033513]

>>10033508
well (-6) - (6) = -12, so no you fucking retard?

[Anonymous No.10033520]

>>10033513
mean + you nigger

[Anonymous No.10033521]

>>10033520
Then a=b

[Anonymous No.10033523]

>>10033521
ah i see.

[Anonymous No.10033524]

>>10029258
Proof by contradiction. Assume there are real numbers a,b such that a=/=b and a=/=0=/=b.

We start with the problem a-b=0 and then add b to both sides. We get a=b, therefore a equals b. But that contradicts our supposition that a=/=b. Therefore there are no such numbers that satisfy the specified conditions.

[Anonymous No.10033527]

>>10029271
OP said a=/=b.

[Anonymous No.10033530]

>>10033524
Cringe

[Anonymous No.10033534]

>>10029274
Infinity is not a number

[Anonymous No.10033536]

>>10033534
You're mom's not a number

[Anonymous No.10033539]

>>10033536
True

[Anonymous No.10033542]

I'm op. The correct answer is >>10032058
Stop bitching around!

[Anonymous No.10033595]

>>10033478
They aren't strings, they are equal concepts.

>So why can't they be equal matrices? Or equal vectors?
Because the result would be a matrix or vector.

>Or equal functions?
Equal functions would be equal numbers.

[Anonymous No.10033602]

>>10029271
Doesn't work, numbers don't lie
>>9999999

[Anonymous No.10034275]

>>10029258
>solve this, except the correct answer is banned

[Anonymous No.10034278]

Let a = 1
Let b = 2
Let 0 = 3

[Anonymous No.10034283]

a-b=0
1(a-b)=1(0)
(1(a-b))/0=1
(1(a-b))/0-a=1-a
(1/0)((1(a-b))/0-a)=(1/0)(1-a)
(1(a-b))/(0*0)-(a/0)=(1-a)/0
(a-b)/0-(a/0)=(1-a)/0
(a-b-a)/0=(1-a)/0
a-b-a=1-a
a-b=1
a=1+b

[Anonymous No.10034395]

>>10033595
>Because the result would be a matrix or vector.
...and op never said it shouldn't be. He just said they're not equal numbers. He never said they are numbers. 0 can denote a zero matrix or vector.
>Equal functions would be equal numbers.
No. Why? A function is not a number. Moreover a function need not be defined on a number set.
>Equal concepts
Of course they are, but equality of english sentences is not math. It's not even an equivalence relation.

[Anonymous No.10034415]

a=asian
b=black
They are equal, but not the same

[Anonymous No.10034421]

>>10034415
what???
fuck off chink

[Anonymous No.10034438]

a = {}
b = {}
a - b = {}
There, did I do gud guyz?
Infinity — -infinity = 0

[Anonymous No.10034450]

10-1=0
20-2=0
30-3=0
12980-1298=0

Seriously guys, this was so easy

[Anonymous No.10034480]

>>10029258
a-b=0, a=b
a-b=b-b
b(a-b)=b^2-b^2
b(a-b)=(b+b)(b-b)
a(a-b)=2b(a-b)
a=2b
a and b are not equal

[Anonymous No.10034496]

- = 0

[Anonymous No.10034520]

If any of you can factor in a metaphor over an analogy to this


to make that zero, the a cannot be = to 0 and the b cannot be = to but unless one of them is equal to zero the other can't exist, so the conditions define a paradox. If that paradox equates two forms of existence then the only available values to it are available only as a conduit for a sort of transfer of the values that exist to contain that paradox. So, when the values contained in the paradox equal to a value that the next consequent action cannot incur consequence to, you have two numbers that equal zero in the sense no number can have a value that'll equal more than the sum even with the gestation of a proper fact of ternary action in the consequent terms.

Check mate, putos. I'm brown skinned not Mexican. I lived in Paramount, CA under like gang rule and shit and went to church or played Final Fantasy all the time. I'm not your fucking shit skin genius and I'm not your toy. Fess up to something some time so you know that we're not here to "contain" anything. You can be a fucking fag or a fucking deviant just stay the fuck out of the illegal shit when you're needing our help. Fucking dumb asses.

[Anonymous No.10034621]

>>10034395
>...and op never said it shouldn't be.
He didn't need to since 0 lacks the notation of a matrix or vector.

>No. Why? A function is not a number. Moreover a function need not be defined on a number set.
"Subtracting functions" means subtracting the values of a function. For example, f(x)-g(y) = 0 is a subtraction of two (equal) numbers. A subtraction of the functions themselves would be f-g=0, which is meaningless dive you haven't specified which values of f and g are being operated on.

>Of course they are, but equality of english sentences is not math. It's not even an equivalence relation.
They aren't sentences, they are mathematical concepts. And how is it not an equivalence relation?

[Anonymous No.10034641]

>>10029258
a=i^2
b=-1
Succ my dicc

[Anonymous No.10034712]

>>10034641
,,,

[Anonymous No.10034753]

>>10034621
A zero element in any abelian group or vector space is usually denoted 0, the little arrow on top of it is optional.

A function is "a bunch of numbers" the same way a vector is "a bunch of numbers". Actually the set of continuous functions R->R is a vector space. Addition and subtraction of function is done component-wise, as for vectors, so (f-g)(x) is defined to be f(x)-g(x) for all x. And the zero function is the one that is constantly zero.
More geneally, functions G->H between two abelian groups can be summed or subtracted in a similar fashion.

Mathematical concepts are not mathematical objects, maybe metamathematical. Equality is usually a mathematical relation between mathematical objects. Anyway I was mostly trolling you.

>how is "having the same meaning" not an equivalence relation?
Well, it is not among english words and sentences. That a word, or sentence, can have multiple meanings should be enough proof. I guess a mathematical concept can have different interpretations according to context as well.

[Anonymous No.10034770]

>>10034753
The numbers we use from number theory and in adding and substracting numbers comes from the rule of thumbs. That Ruler with the centinmeters and inches on it and all that as a metric for the matter. We make the jump from here to there using those numbers after arriving at them conclusively but unless the information is schematic there is no number system just numbers. Assembling that understanding leads to the rules we acquire as understanding and we then start to deride the conjectures of a mentality that cannot be sure by being there when it happens, as it happens, the way we know it should and could, for the purposes of our well being as anon above said. The only thing that's missing here is that the people can actually feel more than they're saying and we're just breaking our backs so that they don't have to contend that intention is a serial cause for matter to "suggest" things more than out of scrutiny.

[Anonymous No.10034779]

>>10034753
There's also the use of halves instead of calling it degree by powers and issuing that the values come from like some omega strain of a value combine function so that the end result can at least float on water if the damn thing won't fly. We need to land.


More free shit. You guys are fucking pieces of shit. You gotta know Karma will you.

[Anonymous No.10034780]

>>10034753
And more then adversely if you miss that the meaning can be comprised by the function you relay a sense of satisfaction over observable phenomena and they, in anticipation, reel over the area shown as a way of agreeing with the nature YOU seem to get to decide for things with all this magic whiteness you have in you.

Ridiculous nigger mind the American mind.

[Anonymous No.10034821]

>>10034770
>>10034779
>>10034780
Is this from Fanged Noumena?

[Anonymous No.10035031]

>>10034753
>Addition and subtraction of function is done component-wise, as for vectors, so (f-g)(x) is defined to be f(x)-g(x) for all x.
It's done by referencing values in the domain. (f-g)(x), f(x), g(x) are values of functions at the value x in the domain, not functions. f and g are functions. Without specifying a value in the domain, "f-g" has no meaning.

>Equality is usually a mathematical relation between mathematical objects.
Equality can also be used as a definition.

>Well, it is not among english words and sentences. That a word, or sentence, can have multiple meanings should be enough proof.
If I define two real numbers such that x = y, the fact that x could also be defined as not equal to y does not affect that x = y is an equivalence relation.

[Anonymous No.10035058]

>>10034821
It's from the salad bar.

[Anonymous No.10035089]

>>10032276
But they're not the same senpai, they're just equivalent! You wouldn't call two groups equal just because they're isomorphic, you know?

[Anonymous No.10035116]

>>10035031
the difference function f-g is defined as the function h such that h(x)=f(x)-g(x) for all x. This is standard notation

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

[Anonymous No.10035122]

>>10035089
This is stupid. Two equal numbers are equal. Two different strings of mathematical symbols representing the same number are different-but-equivalent as elements of some bogus monoid-of-words over some acceptable set of symbols. The number represented is the same.

[Anonymous No.10035167]

>>10035089
Two things with an equal sign between them are equal

[Anonymous No.10035219]

>>10035116
That defines (f-g)(x) not "f-g." We can say that the value of two functions are equal at values in their domain, that is what it means for two functions to be equal.

[Anonymous No.10035331]

>>10035219
Oh my god! It's like trying to explain what air is!

f-g is the name given to the map that sends x to f(x)-g(x). Same with f+g.

[Anonymous No.10035348]

>>10029274
>infinity - infinity = 0
this gave me cancer, thx

[Anonymous No.10035352]

>>10029258
The closest I can manage is a = 7 and b = 6. Can anyone do better?

[Anonymous No.10035370]

>>10035331
Yup, and that map is not equal to some number. The values that it maps are.

[Anonymous No.10035457]

>>10035370
A constant function is a function. The zero function 0 is defined as 0(x)=0 for all x. No need of a "function sign" above it because when a number is involved in an equality of functions it is understood to be a constant function.

[Anonymous No.10035462]

>>10035370
Saying that the function f-g is equal to the zero function simply means that f(x)-g(x)=0 for all x, thus f(x)=g(x) for all x, thus f and g are equal functions.

[Anonymous No.10035585]

>>10035457
A constant function f isn't equal to some number, f(x) is equal to some number. f is the relation between the domain and the range, it's not a value.

>Saying that the function f-g is equal to the zero function simply means that f(x)-g(x)=0 for all x
Yes, that's what someone might say informally in order to mean f(x)-g(x)=0, which is a statement about numbers. But no one would ever write f-g=0.

[Anonymous No.10035606]

>>10035585
>no one would ever write f-g=0
t. engineer

[Anonymous No.10035623]

>>10035606
please dont write this, please write (f-g)(x) = 0

[Anonymous No.10035765]

>>10032058
/thread

[Anonymous No.10035810]

>>10032285
Lost

[Anonymous No.10035829]

Who the fuck cares in you use [math] (f-g)(x)=0 [/math] or [math] f-g =0[/math]? Dumb shit.

[Anonymous No.10036043]

>>10035089
>he doesn't understand the construction of [math]\mathbb{Q}[/math]

[Anonymous No.10036058]

>>10029258
nigger - liberal = 0

[Anonymous No.10036083]

>>10035623
Why? f-g=0 is clearer and doesn't introduce an unnecessary variable. With (f-g)(x) = 0, you'd have explain what x is

[Anonymous No.10036084]

>>10035829
f - g is an unapplied function, it cannot equal 0.

[Anonymous No.10036095]

>>10036084
It cannot equal the number 0. Although technically it can equal the number 0 since set theory is untyped. For example, if f and g are both the empty function, then f-g=0 (since f-g={} and 0={}). But when you see f-g=0, it will be defined ahead of time what the symbol "0" means. When talking about real functions, 0 is defined as the constant function that maps to 0

[Anonymous No.10036208]

>>10029258
a= dy/dx
b=y

[Anonymous No.10036248]

a=-b?

[Anonymous No.10036336]

>>10029258
square both sides, you got urself a quadratic, now find the roots, ez.

[Anonymous No.10036371]

>>10036084
f+g, f-g, 0 are just names of functions, the same way f and g are. For them to make sense, it should be understood that we are working on a given group of functions, like the space C(R) of continuous functions R->R. If you think those notations are confusing, you could give different names, or use longhand notations such as "0_{C(R)}" as in "zero element of the space of continuous functions R->R, defined pointwise as constantly 0", or "f +_{C(R)} g" as in "C(R)-summation of functions f and g, defined pointwise as f(x)+g(x) for all x", but you'd grow tired of this soon.

The reason you write f+g and f-g as if you were operating with functions instead of numbers, is because that's precisely what you are doing. You define an operation on the set C(R) of continuos functions R->R, that send a pair of functions f and g to the summation function f+g defined as (f+g)(x):=f(x)+g(x) , which is again an element of C(R). Similarly with f-g. These operations satisfy all good algebraic properties, like + being associative, and the zero function works as zero element of the group.

[Anonymous No.10036411]

>>10035623
Let [math]E[/math] denote the exponential function [math]E(x):=e^x[/math]. Then for all [math]f,g:R\rightarrow R[/math] we have that
[eqn] E \circ (f+g) = (E\circ f)\cdot(E\circ g).[/eqn]

[Anonymous No.10036433]

>>10029258
This is retarded.
You didn't mention of which structure a,b,0 are elements of, and how - is defined.

You can just define a logical theory with sentences
a-b=0
a!=b
a!=0
b!=0

Then M=(A,-,a,b) with
A :={0,1,2},
- := (_,_) -> 0
0 := 0
a := 1
b := 2
is a structure in which these assumptions do hold.

[Anonymous No.10036451]

>>10029258
a = {0,1}, b = {0,1,2}
Here 0 = {}, as this is the neutral element of finite sets under union/difference.

[Anonymous No.10036481]

>>10036451
0 and {0} are different things desu. But I like the way you think

[Anonymous No.10036488]

>>10036481
He didn't claim 0={0},
but that 0 = {}, if talking about sets.

[Anonymous No.10036505]

>>10036488
Yea I read it wrong with a and b swapped

[Anonymous No.10036529]

>>10035623
What if no linear u stupid

[Anonymous No.10036544]

>>10035623
>(f-g)(x) = 0
makes no sense if you dont specify what x is. if you want to say that the function is zero for all x you should write for all x. otherwise its not clear that may be an equation you can solve for x

[Anonymous No.10036680]

>>10029258
17/9/2018 - 24/9/2018=0 because they're both mondays without being equivalent since they're different days. They're also not the same in the days' equivalence group because I day so.