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kc_kennylau
 one year ago
[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
 one year ago
[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.

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dan815
 one year ago
Best ResponseYou've already chosen the best response.0now help me with my question

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1then how do you know that 0≠1?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1Is there a very axiom or something

dan815
 one year ago
Best ResponseYou've already chosen the best response.00 means u have 0 of 1 and 1 means you have 1 of 1

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1I want the name of the axiom/definition/theorem

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@ikram002p @ganeshie8 @primeralph

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1there 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
 one year ago
Best ResponseYou've already chosen the best response.0It 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
 one year ago
Best ResponseYou've already chosen the best response.1How do you know that if a>b a can't be equal to b? @ikram002p

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

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@ikram002p and which axiom is that?

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0You're dealing with notation and graphics here; this is not math.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1all theorems are built on axioms all axioms are built on definitions all definitions are built on notation

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@primeralph notations are the basis of Maths.

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0No. Notations are how we make things more appealing to understand.

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

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1assume 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.

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.0@kc_kennylau Are you talking about the real number system?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1then 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
 one year ago
Best ResponseYou've already chosen the best response.1@ikram002p how to prove that the axiom's converse also holds?

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.0Addition axioms: (d) For every x in R there exists an element y in R, called the negative of x such that x+y=0

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1ohk , 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 .....

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

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1ohk 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
 one year ago
Best ResponseYou've already chosen the best response.1sry 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
 one year ago
Best ResponseYou've already chosen the best response.1@ikram002p wow i appreciate your effort very much, I think that's all?

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0s=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

ikram002p
 one year ago
Best 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

kc_kennylau
 one year ago
Best 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

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1wats ur course name ?

kc_kennylau
 one year ago
Best 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?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@ikram002p There is no course, I learn everything by myself+my dad.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1But 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
 one year ago
Best ResponseYou've already chosen the best response.1wat 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
 one year ago
Best ResponseYou've already chosen the best response.1so what I wrote is enough?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1I really want to prove by every single definitions+axioms.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1ohk let me wrote the prove for u ok ? u wanna prove that 1 dnt equal 2 right ?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1yes, thank you so much :D

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1Lemma 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
 one year ago
Best ResponseYou've already chosen the best response.1And thank you for your blessing :D

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1ok 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 *********************************

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1let 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
 one year ago
Best ResponseYou've already chosen the best response.1oh i see, but I am mainly proving 1≠2, so all others I named them as lemmas

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.0Just 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
 one year ago
Best ResponseYou've already chosen the best response.1@terenzreignz but what's the problem with just having one element in my very field? :)

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.0There isn't a problem. But who wants to work with such a trivial field?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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
 one year ago
Best 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.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1or of which we can create this corollary.

terenzreignz
 one year ago
Best 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.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1lets 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
 one year ago
Best ResponseYou've already chosen the best response.12, according to peano's axioms, is the successor of 1. @terenzreignz

kc_kennylau
 one year ago
Best 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.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1tj 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

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1u 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
 one year ago
Best ResponseYou've already chosen the best response.1now prove theorem 2, thank you.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.0Well 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.

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

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1ok, 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)

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.0Then 1 is indeed not equal to 2.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.0Suppose 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
 one year ago
Best ResponseYou've already chosen the best response.1Then you have to prove S(a)=S(b)⇒a=b

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1Ouah j'adore cette preuve :D

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1C'est simple est courte :D

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1Je 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
 one year ago
Best ResponseYou've already chosen the best response.0You skipped a bit... maths teachers can be very nitpicky.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1Guess who my Maths teacher is :)

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1btw, what did I skip?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.0Mostly what the contradiction is, exactly.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1Je 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
 one year ago
Best ResponseYou've already chosen the best response.0Good enough, I guess.

ehuman
 one year ago
Best ResponseYou've already chosen the best response.0observational analysis dw:1388853436507:dw :

ehuman
 one year ago
Best ResponseYou've already chosen the best response.0If 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...

ehuman
 one year ago
Best ResponseYou've already chosen the best response.0:) sorry I'm not in serious mode yet lol

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1will 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
 one year ago
Best ResponseYou've already chosen the best response.1that should be an equivalent sign.

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