Pure math, 2 = 1!
Consider the following equation (proved).
a = b a × a = b × a a 2 = a b a 2 − b 2 = a b − b 2 ( a + b ) ( a − b ) = b ( a − b ) a + b = b b + b = b 2 b = b 2 = 1
Is the proof correct?
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4 solutions
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Going by the fundamental definition , the expression (a/a = 1) for all real values of a except '0'.
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That's called a math fallacy. Even is a-b = 0, the equation satisfies, so that step is correct. Let me post an answer.
No, this solution is absolutely correct
Given in the question that a=b ,
→ a - b = 0 ,
Therefore after the fourth step, where
(a - b) was cancelled , it was a mistake because in mathematics , division by zero is not possible !
The cancelling step is skipped. So there is no step where I cancelled the (a-b) on both sides. Check again. The fifth step is where it is wrong.
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Ok got this thank you! But maybe the seniors could explain it in some other interesting way.
a = b ⟹ a − b = 0 Now in 5 t h step, We have divided both sides by ( a − b ) , ie. dividing by 0 what in turn gives 0 0 . Which is undetermined
@Vicky Vignesh Instead of asking if the proof is correct or not, you should have asked which step was wrong. It is obvious that 2 is not equal to 1. So the proof has to be false.
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Yeah. Still, since this has been proved in a fallacy , users might get confused! So, True / False is better.
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Then please rephrase the question to this ......... proof is correct. Is this statement true or false.
There's still some doubt here , @Calvin Lin sir please comment !
@Ashish Siva , @Svatejas Shivakumar , @Hung Woei Neoh , please do comment & clarify this concept!
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The answer is correct. The steps are all correct till the 6th step. In this they have cancelled off (a - b) from the numerator and the denominator. But if you see the first step, it is stated that a = b . So, a − b = 0 . So, in the fifth step can be written as ( a + b ) × 0 = b × 0 . Then we are dividing 0 by 0 by cancelling off the (a - b)'s. But 0 0 is indeterminate. So, this proof is flawed.
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Exactly! thnx for commenting.
Yeah! correct :)
Step 6 is clearly the wrong one. Step 6 is where we cancelled out ( a − b ) . This means we divided the equation by ( a − b ) at Step 6. This is equivalent to dividing by 0 , which does not exist.
However, the explanation given is wrong. Even if a = 0 , we cannot divide it like this, because ( a − b ) would still be 0 , which would mean that we're still dividing by 0 . The condition should be like this:
We can divide ( a − b ) if and only if a − b = 0 , or a = b
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Yes ur right , thnx for commenting! +1!
I understand that this concept is clear to everyone now! Nice discussion , cheers:)
The step at I which cancelled (a-b) is skipped.
And (a+b)=b only if a=0, so there applies an condition. Its not true for all numbers.
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@Viki Zeta – No, you don't just skip a step in proving something. There is no such thing as skipping steps in mathematical proofs.
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@Hung Woei Neoh – Just saying which is wrong from given and why. Anyway I agree 0/0 is not defined.
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@Viki Zeta – And I'm saying your explanation is not correct. Note that when you multiplied a to both sides of the equation, you were introducing extra roots to the equation. It's like saying this:
− 1 = ( − 1 ) 2 = 1 = 1
Care must be taken that you have introduced an extra root when you squared the equation. Similarly, over here, you just factorized it out like this
( a + b ) ( a − b ) − b ( a − b ) = 0 ( a + b − b ) ( a − b ) = 0 a + b − b = 0 or a − b = 0
You cannot just assume that a + b = b . In fact, it's totally irrelevant here
The proof holds up until the second to last line, provided that a = b = 0
If (a+b)(a-b)=b(a-b), then b=a+b if and only if a-b is not 0. However, the first step in the proof is that a=b, so a-b=0.