## 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. 6 months ago 6 months ago

1. dan815 Group Title

|dw:1388848268722:dw|

2. dan815 Group Title

now help me with my question

3. kc_kennylau Group Title

then how do you know that 0≠1?

4. dan815 Group Title

definition

5. kc_kennylau Group Title

Is there a very axiom or something

6. dan815 Group Title

yeah

7. kc_kennylau Group Title

Which

8. dan815 Group Title

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

9. kc_kennylau Group Title

...

10. kc_kennylau Group Title

I want the name of the axiom/definition/theorem

11. dan815 Group Title

12. dan815 Group Title

ikram

13. kc_kennylau Group Title

@ikram002p @ganeshie8 @primeralph

14. ikram002p Group Title

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. primeralph Group Title

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 Group Title

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

17. kc_kennylau Group Title

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

18. kc_kennylau Group Title

@ikram002p and which axiom is that?

19. primeralph Group Title

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

20. kc_kennylau Group Title

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

21. kc_kennylau Group Title

@primeralph notations are the basis of Maths.

22. primeralph Group Title

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

23. primeralph Group Title

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

24. ikram002p Group Title

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 Group Title

a-b=0 then what?

26. RolyPoly Group Title

@kc_kennylau Are you talking about the real number system?

27. ikram002p Group Title

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 Group Title

@RolyPoly yes

29. kc_kennylau Group Title

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

30. RolyPoly Group Title

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. RolyPoly Group Title

*axiom

32. ikram002p Group Title

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 Group Title

@RolyPoly source?

34. ikram002p Group Title

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

35. kc_kennylau Group Title

@ikram002p yes

36. ikram002p Group Title

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 Group Title

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 Group Title

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

39. amoodarya Group Title

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 Group Title

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

41. kc_kennylau Group Title

@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 Group Title

wats ur course name ?

43. kc_kennylau Group Title

@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 Group Title

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

45. kc_kennylau Group Title

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 Group Title

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 Group Title

so what I wrote is enough?

48. kc_kennylau Group Title

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

49. ikram002p Group Title

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

50. kc_kennylau Group Title

yes, thank you so much :D

51. kc_kennylau Group Title

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 Group Title

And thank you for your blessing :D

53. ikram002p Group Title

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 Group Title

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 Group Title

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

56. terenzreignz Group Title

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 Group Title

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

58. terenzreignz Group Title

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

59. kc_kennylau Group Title

meeeeeeee :pp

60. terenzreignz Group Title

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 Group Title

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

62. kc_kennylau Group Title

or of which we can create this corollary.

63. terenzreignz Group Title

@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 Group Title

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 Group Title

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

66. kc_kennylau Group Title

@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 Group Title

Which axiom Lau?

68. kc_kennylau Group Title

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 Group Title

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 Group Title

now prove theorem 2, thank you.

71. terenzreignz Group Title

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 Group Title

how?

73. terenzreignz Group Title

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

74. kc_kennylau Group Title

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 Group Title

included.

76. terenzreignz Group Title

Then 1 is indeed not equal to 2.

77. kc_kennylau Group Title

how?

78. terenzreignz Group Title

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 Group Title

Then you have to prove S(a)=S(b)⇒a=b

80. terenzreignz Group Title

That's #8

81. kc_kennylau Group Title

82. kc_kennylau Group Title

C'est simple est courte :D

83. terenzreignz Group Title

^_^

84. kc_kennylau Group Title

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 Group Title

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

86. kc_kennylau Group Title

Guess who my Maths teacher is :)

87. kc_kennylau Group Title

btw, what did I skip?

88. terenzreignz Group Title

Leonhard Euler?

89. kc_kennylau Group Title

C'est moi meme :D

90. terenzreignz Group Title

Mostly what the contradiction is, exactly.

91. kc_kennylau Group Title

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 Group Title

Good enough, I guess.

93. kc_kennylau Group Title

Gracias :D

94. ehuman Group Title

observational analysis |dw:1388853436507:dw| :

95. kc_kennylau Group Title

\(-_-")/

96. ehuman Group Title

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 Group Title

|-_-"|

98. ehuman Group Title

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

99. kc_kennylau Group Title

lolz xD

100. ikram002p Group Title

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 Group Title

that should be an equivalent sign.

102. ikram002p Group Title

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

103. kc_kennylau Group Title

oh i see.