[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.

- kc_kennylau

- schrodinger

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- dan815

|dw:1388848268722:dw|

- dan815

now help me with my question

- kc_kennylau

then how do you know that 0≠1?

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## More answers

- dan815

definition

- kc_kennylau

Is there a very axiom or something

- dan815

yeah

- kc_kennylau

Which

- dan815

0 means u have 0 of 1
and 1 means you have 1 of 1

- kc_kennylau

...

- kc_kennylau

I want the name of the axiom/definition/theorem

- dan815

ask uhhh

- dan815

ikram

- kc_kennylau

@ikram002p @ganeshie8 @primeralph

- ikram002p

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

- primeralph

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.

- kc_kennylau

How do you know that if a>b a can't be equal to b? @ikram002p

- kc_kennylau

@primeralph but do you have an axiom/definition/theorem that says 1≠2?

- kc_kennylau

@ikram002p and which axiom is that?

- primeralph

You're dealing with notation and graphics here; this is not math.

- kc_kennylau

all theorems are built on axioms
all axioms are built on definitions
all definitions are built on notation

- kc_kennylau

@primeralph notations are the basis of Maths.

- primeralph

No. Notations are how we make things more appealing to understand.

- primeralph

For example, math had advanced greatly even before 0 was included in numbers. They simply used nothingness to represent 0.

- ikram002p

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.

- kc_kennylau

a-b=0 then what?

- anonymous

@kc_kennylau Are you talking about the real number system?

- ikram002p

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)

- kc_kennylau

@RolyPoly yes

- kc_kennylau

@ikram002p how to prove that the axiom's converse also holds?

- anonymous

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

- anonymous

*axiom

- ikram002p

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
.....

- kc_kennylau

@RolyPoly source?

- ikram002p

" how to prove that the axiom's converse also holds" u mean uniqueness of inverse ??

- kc_kennylau

@ikram002p yes

- ikram002p

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

- ikram002p

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)

- kc_kennylau

@ikram002p wow i appreciate your effort very much, I think that's all?

- amoodarya

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

- ikram002p

@kc_kennylau idk if thats all lolz , its depand on u if u got it or not :P

- kc_kennylau

@amoodarya wow that's a cool approach i loved this approach, but I don't know if it'd be formal enough xPP

- ikram002p

wats ur course name ?

- kc_kennylau

@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?

- kc_kennylau

@ikram002p There is no course, I learn everything by myself+my dad.

- kc_kennylau

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)

- ikram002p

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 .

- kc_kennylau

so what I wrote is enough?

- kc_kennylau

I really want to prove by every single definitions+axioms.

- ikram002p

ohk let me wrote the prove for u ok ?
u wanna prove that 1 dnt equal 2 right ?

- kc_kennylau

yes, thank you so much :D

- kc_kennylau

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

- kc_kennylau

And thank you for your blessing :D

- ikram002p

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
*********************************

- ikram002p

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 ??

- kc_kennylau

oh i see, but I am mainly proving 1≠2, so all others I named them as lemmas

- terenzreignz

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.

- kc_kennylau

@terenzreignz but what's the problem with just having one element in my very field? :)

- terenzreignz

There isn't a problem. But who wants to work with such a trivial field?

- kc_kennylau

meeeeeeee :pp

- terenzreignz

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....

- kc_kennylau

@terenzreignz that's exactly the problem, since there isn't in fact any theorem that states 1≠2.

- kc_kennylau

or of which we can create this corollary.

- terenzreignz

@kc_kennylau no need to tag me every time, it's *TJ* :P
Anyway... clear this up by defining what exactly is 2.

- ikram002p

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 ?

- kc_kennylau

2, according to peano's axioms, is the successor of 1. @terenzreignz

- kc_kennylau

@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.

- terenzreignz

Which axiom Lau?

- kc_kennylau

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

- ikram002p

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!

- kc_kennylau

now prove theorem 2, thank you.

- terenzreignz

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.

- kc_kennylau

how?

- terenzreignz

Simply define 2 to be equal to 1.
That already implying that your group/field is a trivial one.

- kc_kennylau

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)

- kc_kennylau

included.

- terenzreignz

Then 1 is indeed not equal to 2.

- kc_kennylau

how?

- terenzreignz

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

- kc_kennylau

Then you have to prove S(a)=S(b)⇒a=b

- terenzreignz

That's #8

- kc_kennylau

Ouah j'adore cette preuve :D

- kc_kennylau

C'est simple est courte :D

- terenzreignz

^_^

- kc_kennylau

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

- terenzreignz

You skipped a bit... maths teachers can be very nitpicky.

- kc_kennylau

Guess who my Maths teacher is :)

- kc_kennylau

btw, what did I skip?

- terenzreignz

Leonhard Euler?

- kc_kennylau

C'est moi meme :D

- terenzreignz

Mostly what the contradiction is, exactly.

- kc_kennylau

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)

- terenzreignz

Good enough, I guess.

- kc_kennylau

Gracias :D

- anonymous

observational analysis
|dw:1388853436507:dw|
:

- kc_kennylau

\(-_-")/

- anonymous

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...

- kc_kennylau

|-_-"|

- anonymous

:) sorry I'm not in serious mode yet lol

- kc_kennylau

lolz xD

- ikram002p

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 )

- kc_kennylau

that should be an equivalent sign.

- ikram002p

:)
note wat i wrote :o
im talking abt the prove in "operation , groups"

- kc_kennylau

oh i see.

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