1+1≠2 ?

Baru2 ni salah sorang dari ahli 'GenG' telah menjalani sesi BTN akibat menerima tajaan dari pihah MOSTI. Sekembali nyer dier dari program itu, maka dier pun telah menge'post'kn comment pada Facebook nye seperti berikut :

"sape kate 25/5=5....." ini la yg aku blaja kat BTN.mmg saiko gler."

Maka comment nye itu mendapat sambut.. biase la kn.. Ramai peminat.. hahahah.. Dalam bebenyak komen yg dier dapat, ader satu komen yg aku agak kompius yakni:

"dari segi psikologi....25/5=bukannya 5....scientific fact pun kata 1+1 bukannya 2....cuba halusi dan fikirkan....hehehehe"

1+1 bukan 2?


Biar betul.. selame-lame aku belajar itu la persamaan asas yang diajar mase muler2 belajar math dulu.. ( xde la aku ni pandai sangat math.. hahaha..)

Kompius aku.. Dek sebab nak tau kenape 1 + 1 = 2, aku pun men'google' utk mencari kepastiaan.. Carik punyer carik, terserempak lak ngn satu site nie, jumpe la penyelesaian nyer kt situ.. so aku copy paste je la kan.. aku pun still cube memahami hujah2 yang di nyatakan nyer.. otak dh karat.. wakakaka..
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The proof starts from the Peano Postulates, which define the natural
numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which
replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the
definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:

Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

- Doctor Rob, The Math Forum
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tolong.. tolong.. kurang paham... hahahhaha..