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[Serious mode] How to prove that 1≠2? We have equality is transitive, we have 1's successor is 2, we have... No theorem ever said that a successor cannot be equal to the number itself No theorem ever said that two numbers with different looks cannot be equal 1.000... is equal to 0.999... for example A number can have many ways of presenting... ASSUMPTIONS: 2:=S(S(0)), 1:=S(0), where S(x) denotes the successor of x, where x is a natural number.

Mathematics
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|dw:1388848268722:dw|
now help me with my question
then how do you know that 0≠1?

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Other answers:

definition
Is there a very axiom or something
yeah
Which
0 means u have 0 of 1 and 1 means you have 1 of 1
...
I want the name of the axiom/definition/theorem
ask uhhh
ikram
there is axioms that say for any integer n belong to z s.t z={0,1,2,...,n ,...} then n+1>n or 0>1 its depand on the groupe that u wanna make binary operation on :) u cud see Abstract Algebra books ull got it easilly
It all depends on assignment. If I decided to assign two units to 1, then 1=2. In Math, two units has already been assigned to the figure 2 and is fixed in most cases.
How do you know that if a>b a can't be equal to b? @ikram002p
@primeralph but do you have an axiom/definition/theorem that says 1≠2?
@ikram002p and which axiom is that?
You're dealing with notation and graphics here; this is not math.
all theorems are built on axioms all axioms are built on definitions all definitions are built on notation
@primeralph notations are the basis of Maths.
No. Notations are how we make things more appealing to understand.
For example, math had advanced greatly even before 0 was included in numbers. They simply used nothingness to represent 0.
assume u have a,b belongs to z then u have another axiom sys that say 0 belongs to Z a-a=0 u know this axiom right ? so if a>b then a=b+c s,t c belong to z if a=b then a-b=0 its the thms, axioms on the binary operation of the "groups" idk if it has a unigue name , if u wanna ill check my old Abstract algebra book.
a-b=0 then what?
@kc_kennylau Are you talking about the real number system?
then b must equal a , cuz there is unique num that u cud add to a to give 0 ( but its depand on the groupe u have , n the operation that u r givven)
@ikram002p how to prove that the axiom's converse also holds?
Addition axioms: (d) For every x in R there exists an element y in R, called the negative of x such that x+y=0
*axiom
ohk , i checkd the book the definition said Group let G be a set together with a binary operation (usually called multiplication ) that assigns to each orderd pair (a,b) of elements of G an element in G denoted by ab.......} to the rest of it then u have a groupe of thms called Elementary properties of groups thm 1: uniqueness of the identity thm 2: cancellation thm 3: uniqueness of inverse thm 4 : socks-shoes property .....
@RolyPoly source?
" how to prove that the axiom's converse also holds" u mean uniqueness of inverse ??
ohk to prove it on the addition operation for example u should know that thm 1: uniqueness of the identity (e=0) thm 2: cancellation ( if ab=cb then a=c) then prove uniqueness of inverse assume u have a,b,c belonge to z use contradiction let a+b=0 a+c=0 then conclude that b=c
sry typo "thm 2: cancellation ( if ab=cb then a=c)" its like this thm 2: cancellation ( if a+b=c+b then a=c)
@ikram002p wow i appreciate your effort very much, I think that's all?
s=1-1+1-1+1-1+1-1+... suppose (wrong) s converge now let write 1+s as 1+s=1+{1+(1-1)+(1-1)+(1-1)+...}=1+{1+0+0+0...}=2 let write s+1 as 1+s=1 +{(1-1)+(1-1)+(1-1)+...}=1+{0}=1 now we have 1+s=1 and 1+s=2 so 1=2 but assumption was wrong from the first so by contradiction 1≠2
@kc_kennylau idk if thats all lolz , its depand on u if u got it or not :P
@amoodarya wow that's a cool approach i loved this approach, but I don't know if it'd be formal enough xPP
wats ur course name ?
@ikram002p I think I very much got it: \[\begin{array}{lrl} \mbox{Lemma 1:}&\forall a\in\mathbb Z:&a-a=0\\ \mbox{Lemma 2:}&\forall a\in\mathbb Z,\forall b\ne a:&a-b\ne 0\\ \mbox{Lemma 3:}&\forall a,b\in\mathbb Z:&a>b\Rightarrow a\ne b\\ \mbox{Lemma 4:}&\forall n\in\mathbb Z:&s(n)\ne n\\ \mbox{THEOREM:}&&2\ne 1 \end{array}\]Is this correct?
@ikram002p There is no course, I learn everything by myself+my dad.
But I have gone beyond what my dad can teach me, so I basically learn everything by myself (ain't erasing all the efforts of my dad)
wat u got is true , but there is no need to write them as lemmas , and wat u ment wid lemma 4 ? for the last one no its not THEOREM cuz u cud prove it using a basic groupe of thms :) but u r such a smart person good for u , hope u the best .
so what I wrote is enough?
I really want to prove by every single definitions+axioms.
ohk let me wrote the prove for u ok ? u wanna prove that 1 dnt equal 2 right ?
yes, thank you so much :D
Lemma 1 is an axiom, Lemma 2 you have proven it before, Lemma 3 corollary from Lemma 2, Lemma 4 you still not stated the source Theorem corollary from Lemma 4
And thank you for your blessing :D
ok using this axioms sys that i wrote before *********************************** ohk , i checkd the book the definition said Group let G be a set together with a binary operation (usually called multiplication ) that assigns to each orderd pair (a,b) of elements of G an element in G denoted by ab.......} to the rest of it then u have a groupe of thms called Elementary properties of groups thm 1: uniqueness of the identity thm 2: cancellation thm 3: uniqueness of inverse thm 4 : socks-shoes property *********************************
let me tell u this lemma , thm , proposition all of them are the same , the oly difference btw is the priority of using , so u dnt simply write lemma abt somthin unless its new sup u creat , or book or somthing ok ??
oh i see, but I am mainly proving 1≠2, so all others I named them as lemmas
Just a little comment: Assuming 1 = 0 implies that your field only has one element, since 0 is defined as the additive identity and 1, the multiplicative identity.
@terenzreignz but what's the problem with just having one element in my very field? :)
There isn't a problem. But who wants to work with such a trivial field?
meeeeeeee :pp
Well, it all boils down to how useful it is to deal with the (the name for it is: ) Trivial Subgroup of the Group of real numbers....
@terenzreignz that's exactly the problem, since there isn't in fact any theorem that states 1≠2.
or of which we can create this corollary.
@kc_kennylau no need to tag me every time, it's *TJ* :P Anyway... clear this up by defining what exactly is 2.
lets us prove that 1 dnt = 2 from def :: let Z be our group on binary operation of addition so u already know that(all of them r thms ) 1_0 is the identity 2- for a,b,c belongs to Z if a+b=a+c then b=c 3_for every a there exist unique a^-1 such that a+a^-1 =0 . the proof , by contradiction assume 1=2 1=2 1+0=1+1 (use thm 2) 0=1 (which is contradictiom to thm 1that 0 is unique) done ! got it ?
2, according to peano's axioms, is the successor of 1. @terenzreignz
@ikram002p wow this proof is so short, I love it very much :DDDDD Now the time has come to prove the uniqueness of the additive identity.
Which axiom Lau?
tj sorry i meant definition, and i realized that 2 is actually defined to be S(S(0)). https://en.wikipedia.org/wiki/Peano_axioms#Set-theoretic_models
u need to know the Group first on binary and prove it in general . use contradiction assume u have two identity lets take e1,e1 so for a belonge to G a+e1=a a+e2=a a+e1=a+e2( use thm 2) e1=e2 which is contradiction done!
now prove theorem 2, thank you.
Well actually, that article simply states that 2 CAN be defined as S(S(0)). Depending on your definition of 2, 1=2 may yet be true.
how?
Simply define 2 to be equal to 1. That already implying that your group/field is a trivial one.
ok, I now define 2 to be S(S(0)) and 1 to be S(0). (I'll edit the question to include this piece of detail)
included.
Then 1 is indeed not equal to 2.
how?
Suppose 1 = 2 S(0) = S(S(0)), by #6, S(0) is a natural number. by #1, 0 is a natural number by #8, 0 = S(0), contradicts #7
Then you have to prove S(a)=S(b)⇒a=b
That's #8
Ouah j'adore cette preuve :D
C'est simple est courte :D
^_^
Je le recris comme sa: Suppose que 1=2. 1=2 S(0)=S(S(0)) (par definition) 0=S(0) (par #8) Donc contradiction
You skipped a bit... maths teachers can be very nitpicky.
Guess who my Maths teacher is :)
btw, what did I skip?
Leonhard Euler?
C'est moi meme :D
Mostly what the contradiction is, exactly.
Je le recris encore comme sa: Suppose que 1=2. 1=2 S(0)=S(S(0)) (par definition) 0=S(0) (par #8) Donc contradiction (par #7)
Good enough, I guess.
Gracias :D
observational analysis |dw:1388853436507:dw| :
\(-_-")/
If you are not sure how to tell that 1 is not = to 2, I would love for you to come and work for me. I'll give you $30 per hour...
|-_-"|
:) sorry I'm not in serious mode yet lol
lolz xD
will but u need to know it depand on the operation to prove that 1 dnt equal 2 let me give u example that 1= 2 1=0mod 1 2=0 mod 1 1=2 (on mod 1 operation )
that should be an equivalent sign.
:) note wat i wrote :o im talking abt the prove in "operation , groups"
oh i see.

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