#9**+5 **

Let x = 1 Then x^{2} = 1 as well

So x^{2} - 1 = x - 1

The left-hand side can be written as (x+1)(x-1) so

(x + 1)(x - 1) = x-1

Divide both sides by x - 1 to get

x + 1 = 1

But we started by saying x = 1, so the left-hand side is 1 + 1 or 2. This means the last equation is

2 = 1

This is the sort of thing that happens when you allow division by zero (which is what I did above when dividing through by x-1).

Alan Apr 26, 2015

#1**+5 **

Division by zero is undefined mathematically, because, if you try to allow it, you get all sorts of inconsistencies appearing in your calculations.

.

Alan Apr 25, 2015

#4**+5 **

Try 1/0.1 then 1/0.01 then 1/0.001 then 1/0.0001 etc. where the denominator is getting closer and closer to 0.

Does it look like the fractions are getting closer to zero?

Informally, the limit is tending to infinity, rather than zero. However, strictly, division by zero is still undefined. If you allow it you can prove silly things like 2 = 1.

.

Alan Apr 25, 2015

#6**0 **

If by that you mean the result is not a proper number then I'll agree with you!

.

Alan Apr 25, 2015

#7**0 **

yup i mean it by that. at least you know what i mean. so overall, good job sir

TitaniumRome Apr 25, 2015

#9**+5 **

Best Answer

Let x = 1 Then x^{2} = 1 as well

So x^{2} - 1 = x - 1

The left-hand side can be written as (x+1)(x-1) so

(x + 1)(x - 1) = x-1

Divide both sides by x - 1 to get

x + 1 = 1

But we started by saying x = 1, so the left-hand side is 1 + 1 or 2. This means the last equation is

2 = 1

This is the sort of thing that happens when you allow division by zero (which is what I did above when dividing through by x-1).

Alan Apr 26, 2015