## kc_kennylau 2 years 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.

1. dan815

|dw:1388848268722:dw|

2. dan815

now help me with my question

3. kc_kennylau

then how do you know that 0≠1?

4. dan815

definition

5. kc_kennylau

Is there a very axiom or something

6. dan815

yeah

7. kc_kennylau

Which

8. dan815

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

9. kc_kennylau

...

10. kc_kennylau

I want the name of the axiom/definition/theorem

11. dan815

12. dan815

ikram

13. kc_kennylau

@ikram002p @ganeshie8 @primeralph

14. 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

15. anonymous

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.

16. kc_kennylau

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

17. kc_kennylau

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

18. kc_kennylau

@ikram002p and which axiom is that?

19. anonymous

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

20. kc_kennylau

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

21. kc_kennylau

@primeralph notations are the basis of Maths.

22. anonymous

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

23. anonymous

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

24. 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.

25. kc_kennylau

a-b=0 then what?

26. anonymous

@kc_kennylau Are you talking about the real number system?

27. 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)

28. kc_kennylau

@RolyPoly yes

29. kc_kennylau

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

30. 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

31. anonymous

*axiom

32. 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 .....

33. kc_kennylau

@RolyPoly source?

34. ikram002p

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

35. kc_kennylau

@ikram002p yes

36. 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

37. 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)

38. kc_kennylau

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

39. 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

40. ikram002p

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

41. 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

42. ikram002p

wats ur course name ?

43. 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?

44. kc_kennylau

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

45. 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)

46. 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 .

47. kc_kennylau

so what I wrote is enough?

48. kc_kennylau

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

49. ikram002p

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

50. kc_kennylau

yes, thank you so much :D

51. 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

52. kc_kennylau

And thank you for your blessing :D

53. 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 *********************************

54. 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 ??

55. kc_kennylau

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

56. 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.

57. kc_kennylau

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

58. terenzreignz

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

59. kc_kennylau

meeeeeeee :pp

60. 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....

61. kc_kennylau

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

62. kc_kennylau

or of which we can create this corollary.

63. terenzreignz

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

64. 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 ?

65. kc_kennylau

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

66. 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.

67. terenzreignz

Which axiom Lau?

68. 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

69. 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!

70. kc_kennylau

now prove theorem 2, thank you.

71. 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.

72. kc_kennylau

how?

73. terenzreignz

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

74. 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)

75. kc_kennylau

included.

76. terenzreignz

Then 1 is indeed not equal to 2.

77. kc_kennylau

how?

78. 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

79. kc_kennylau

Then you have to prove S(a)=S(b)⇒a=b

80. terenzreignz

That's #8

81. kc_kennylau

82. kc_kennylau

C'est simple est courte :D

83. terenzreignz

^_^

84. 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

85. terenzreignz

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

86. kc_kennylau

Guess who my Maths teacher is :)

87. kc_kennylau

btw, what did I skip?

88. terenzreignz

Leonhard Euler?

89. kc_kennylau

C'est moi meme :D

90. terenzreignz

Mostly what the contradiction is, exactly.

91. 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)

92. terenzreignz

Good enough, I guess.

93. kc_kennylau

Gracias :D

94. anonymous

observational analysis |dw:1388853436507:dw| :

95. kc_kennylau

\(-_-")/

96. 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...

97. kc_kennylau

|-_-"|

98. anonymous

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

99. kc_kennylau

lolz xD

100. 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 )

101. kc_kennylau

that should be an equivalent sign.

102. ikram002p

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

103. kc_kennylau

oh i see.