Mathmatically prove that 1 = 0?

Answer 1

Given a = b + 1
By the properties of multiplication: (a-b)a = (a-b)(b+1)
Therefore: a^2 - ab = ab + a - b^2 - b
So: a^2 - ab -a = ab + a -a - b^2 - b
By distribution: a(a - b - 1) = b(a - b - 1)
And so: a = b
Giving: b + 1 = b
Therefore: 1 = 0

The problem with this "proof" is that there is a division by zero error, and so the proof is flawed (so I've been told. I can't see it though).

Additional: -

For the critics below, I did state that the proof is flawed, but it was a professor at university who instructed us in the proof, stating that it was a commonly used argument by philosophers as an argument against the "self referential" logic of mathematics. I just couldn't remember where the flaw was. Thanks for the explanation appledaydreamer.

Answer 2

gimread u are wrong.

if Given already a = b + 1
And so: a = b will be wrong since given already a = b + 1

note: if u are trying to prove a = b which is 1 = 0 whatever,

then a cannot be equal <> to b + 1
(ie. a <> b - 1) since

if the end result is a = b, then substituting
a = b + 1 into a will give :

a = b
(b +1 ) = b
b - b = -1
0 = -1 which is wrong and also :

a = b is true only there are 2 similar equations involved not just 1 set of equation.

and also :

if a = b in the end result, then from the beginning, a <> b +1, correct?

the 1 way to prove that 1 = 0 is to use a scenario :

1 = 0 is true when 1 = 0 + a when a = 1

admirer

Answer 3

>Given a = b + 1
>By the properties of multiplication: (a-b)a = (a-b)(b+1)
>Therefore: a^2 - ab = ab + a - b^2 - b
>So: a^2 - ab -a = ab + a -a - b^2 - b
>By distribution: a(a - b - 1) = b(a - b - 1)
>And so: a = b
>Giving: b + 1 = b
>Therefore: 1 = 0

By rules of algebra this is true, but by rule of math this cannot be.
Here is why.

>By distribution: a(a - b - 1) = b(a - b - 1)
>And so: a = b

Reason #1.
Both sides are divided by (a - b - 1) to get a = b.
BUT given that a = b + 1
if we substitute b + 1 to a we get (b + 1 - b - 1)
which is equal to 0.
RULE OF MATH. WE CANNOT DIVIDE BY 0.
Therefore, the algebraic equation is mathematically wrong.

Answer 4

if the number or calculation in question is of great value and length a number as small as one can be so insignificant that it would in fact be virtually equivalent to 0

Answer 5

um guy two above mine:

i'll admit its been a while for such math, but im not sure i agreed with his breakdown or how can b + 1 = b regardless?

Answer 6

i've seen that answer somewhere beofre. kinda lame

Answer 7

one vacuum = nothing