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kc_kennylau
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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.
 10 months ago
 10 months ago
kc_kennylau Group Title
[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.
 10 months ago
 10 months ago

This Question is Closed

dan815 Group TitleBest ResponseYou've already chosen the best response.0
dw:1388848268722:dw
 10 months ago

dan815 Group TitleBest ResponseYou've already chosen the best response.0
now help me with my question
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
then how do you know that 0≠1?
 10 months ago

dan815 Group TitleBest ResponseYou've already chosen the best response.0
definition
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
Is there a very axiom or something
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
Which
 10 months ago

dan815 Group TitleBest ResponseYou've already chosen the best response.0
0 means u have 0 of 1 and 1 means you have 1 of 1
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
I want the name of the axiom/definition/theorem
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@ikram002p @ganeshie8 @primeralph
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
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
 10 months ago

primeralph Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
How do you know that if a>b a can't be equal to b? @ikram002p
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@primeralph but do you have an axiom/definition/theorem that says 1≠2?
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@ikram002p and which axiom is that?
 10 months ago

primeralph Group TitleBest ResponseYou've already chosen the best response.0
You're dealing with notation and graphics here; this is not math.
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
all theorems are built on axioms all axioms are built on definitions all definitions are built on notation
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@primeralph notations are the basis of Maths.
 10 months ago

primeralph Group TitleBest ResponseYou've already chosen the best response.0
No. Notations are how we make things more appealing to understand.
 10 months ago

primeralph Group TitleBest ResponseYou've already chosen the best response.0
For example, math had advanced greatly even before 0 was included in numbers. They simply used nothingness to represent 0.
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
assume u have a,b belongs to z then u have another axiom sys that say 0 belongs to Z aa=0 u know this axiom right ? so if a>b then a=b+c s,t c belong to z if a=b then ab=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.
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
ab=0 then what?
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
@kc_kennylau Are you talking about the real number system?
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
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)
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@RolyPoly yes
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@ikram002p how to prove that the axiom's converse also holds?
 10 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
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 : socksshoes property .....
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@RolyPoly source?
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
" how to prove that the axiom's converse also holds" u mean uniqueness of inverse ??
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@ikram002p yes
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
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
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
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)
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@ikram002p wow i appreciate your effort very much, I think that's all?
 10 months ago

amoodarya Group TitleBest ResponseYou've already chosen the best response.0
s=11+11+11+11+... suppose (wrong) s converge now let write 1+s as 1+s=1+{1+(11)+(11)+(11)+...}=1+{1+0+0+0...}=2 let write s+1 as 1+s=1 +{(11)+(11)+(11)+...}=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
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
@kc_kennylau idk if thats all lolz , its depand on u if u got it or not :P
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@amoodarya wow that's a cool approach i loved this approach, but I don't know if it'd be formal enough xPP
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
wats ur course name ?
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@ikram002p I think I very much got it: \[\begin{array}{lrl} \mbox{Lemma 1:}&\forall a\in\mathbb Z:&aa=0\\ \mbox{Lemma 2:}&\forall a\in\mathbb Z,\forall b\ne a:&ab\ne 0\\ \mbox{Lemma 3:}&\forall a,b\in\mathbb Z:&a>b\Rightarrow a\ne b\\ \mbox{Lemma 4:}&\forall n\in\mathbb Z:&s(n)\ne n\\ \mbox{THEOREM:}&&2\ne 1 \end{array}\]Is this correct?
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@ikram002p There is no course, I learn everything by myself+my dad.
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
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)
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
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 .
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
so what I wrote is enough?
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
I really want to prove by every single definitions+axioms.
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
ohk let me wrote the prove for u ok ? u wanna prove that 1 dnt equal 2 right ?
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
yes, thank you so much :D
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
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
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
And thank you for your blessing :D
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
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 : socksshoes property *********************************
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
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 ??
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
oh i see, but I am mainly proving 1≠2, so all others I named them as lemmas
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@terenzreignz but what's the problem with just having one element in my very field? :)
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
There isn't a problem. But who wants to work with such a trivial field?
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
meeeeeeee :pp
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
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....
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@terenzreignz that's exactly the problem, since there isn't in fact any theorem that states 1≠2.
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
or of which we can create this corollary.
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
@kc_kennylau no need to tag me every time, it's *TJ* :P Anyway... clear this up by defining what exactly is 2.
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
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 ?
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
2, according to peano's axioms, is the successor of 1. @terenzreignz
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
@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.
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Which axiom Lau?
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
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#Settheoretic_models
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
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!
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
now prove theorem 2, thank you.
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Simply define 2 to be equal to 1. That already implying that your group/field is a trivial one.
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
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)
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
included.
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Then 1 is indeed not equal to 2.
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
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
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
Then you have to prove S(a)=S(b)⇒a=b
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
That's #8
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
Ouah j'adore cette preuve :D
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
C'est simple est courte :D
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
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
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
You skipped a bit... maths teachers can be very nitpicky.
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
Guess who my Maths teacher is :)
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
btw, what did I skip?
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Leonhard Euler?
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
C'est moi meme :D
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Mostly what the contradiction is, exactly.
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
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)
 10 months ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Good enough, I guess.
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
Gracias :D
 10 months ago

ehuman Group TitleBest ResponseYou've already chosen the best response.0
observational analysis dw:1388853436507:dw :
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
\(_")/
 10 months ago

ehuman Group TitleBest ResponseYou've already chosen the best response.0
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...
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
_"
 10 months ago

ehuman Group TitleBest ResponseYou've already chosen the best response.0
:) sorry I'm not in serious mode yet lol
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
lolz xD
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
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 )
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
that should be an equivalent sign.
 10 months ago

ikram002p Group TitleBest ResponseYou've already chosen the best response.1
:) note wat i wrote :o im talking abt the prove in "operation , groups"
 10 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.1
oh i see.
 10 months ago
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