Some one tell me that 1 not equal 2 . It true?

[OOP]

How to solution ?

 

[muruga86]

do you mean
if(1 != 2)
//true
else
//false

 

[OOP]

do you mean
if(1 != 2)
//true
else
//false

No ,I mean in the mathermatical

 

[engineer]

Is it a joke like
sin x ÷ n = six

 

[muruga86]

your question is unclear , please kindly elabrate your question with an example?

 

[cherrytart]

maybe you mean the classic proof using beginning algebra presented here with an explanation of the fallacy?

hxxp://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html

 

[cedance]

maybe you mean the classic proof using beginning algebra presented here with an explanation of the fallacy?

hxxp://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html

yeah.. i think he meant that sort of stuff... thou, a=b implies a-b is 0... and i remember yet another 1... it goes like this...

1 = 1
2 = 1 + 1 ( 2 times)
3 = 1 + 1 + 1 (3 times)
4 = 1 + 1 + 1 + 1 ( 4times)
similiarly, x = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1.... x times

now differentiating...

d/dx(x) = 1 = 0+0+0+0+... x times..
=> 1 = 0 <=> 2 = 1

good luck....
/cedance

 

[techie]

differentiating is just calculating the rate of change. Two quantities that change at an equal rate are not necessary equal.

 

[hugo]

Hi,

(x²-x²)=(x²-x²)
(x-x)(x+x)=x(x-x) /(x-x)
(x+x)=x
2x=x /x
2=1 false

 

[techie]

Again a fallcy. When both sides of the equation become zero or infinite, no further algebra can take place. It is like saying
0 = 0
1 * 0 = 2 * 0
therefore 1 = 2.
Same is the case when you work with infinite.

 

[gopalsamy]

1*0 = 2*0 = 0

=> either 1=2 or 0 = 0 but 0=0 so 1!=2

 

[cedance]

differentiating is just calculating the rate of change. Two quantities that change at an equal rate are not necessary equal.

hi,

could u spot me an example, a function f(x) where its differential is not equal to the differentiation result... in my view, when y = f(x), then dy/dx is equal to df(x)/dx.... and is not the other way and so do we say a "constant"

that is only when we reverse the steps... when dy/dx = df(x)/dx then y is not necessarily equal to f(x) .... when diferentiating, always LHS is equal to RHS! the proof i gave is obviously wrong for different reasons..

/cedance