#1**+11 **

Division by zero is undefined because it produces no single answer.

To check division, multiply. For example 14 ÷ 7 = 2 because 7 X 2 = 14.

If you have any non-zero number, such as 6, and divide it by zero, there is no possible answer.

6 ÷ 0 = ? To solve this, you need an answer for which ? X 0 = 6. This is impossible. No number works!

If you take 0 and divide it by 0, you get a different problem, because every number works!

0 ÷ 0 = ? Here, you can use any number for an answer.

0 ÷ 0 = 27 because 0 X 27 = 0

0 ÷ 0 = -4 because 0 X -4 = 0 Similarly, every number works.

In the first case, you get no answer, in the second case you get no single answer.

geno3141 Sep 28, 2014

#2**+11 **

You cannot!

Say that four people share 12 apples, how many apples does each get? three, right?

What about three people? 12/3 = 4. Four apples each.

What about two people? Six apples.

If only one person "shares" 12 apples, he get all the 12 apples for himself.

What, then, if zero people "share" 12 apples? Does zero people get 12 apples? That doesn't make any sense! Does zero people get zero apples? Then, what happens with the 12 apples? All of this is nonsense, one simply cannot divide by zero.

I've linked a wikipedia article about the matter:

http://en.wikipedia.org/wiki/Division_by_zero

Guest Sep 28, 2014

#3**+28 **

Best Answer

Great answers, geno3141 and anonymous! But just for reference:

Unless you are Chuck Norris, of course. But you're not:)

kitty<3 Sep 28, 2014

#6**0 **

Even if you say set f to equal 1/0 it is useless because nothing needs to divide by zero. oh and doing polynomial equations with xf=x/0 you can get some pretty weird stuff. like how some times you can just cancle all but a few things out and get a single point. or f=0 when you started with some thing like, (fx-1)(x+f)=0

fx^2+xf^2-x-f=0

(fx^2+xf^2-x-f)=((1/0)x^2+x(1/0)^2-x-(1/0))=0

((1/0)x^2+x(1/0)^2-x-(1/0))*0=0*0

(1+f-1)=0

(f+1-1)=f=0???????

but when you try to solve the factored version it comes out to be

fx-1=0 or x+f=0

x=1/f or x=-f

x=1/(1/0) or x=-(1/0)

sooo if you plug that in to the equations: (f*(1/(1/)-1)(1/(1/0)+f)=0

1/1/0 = 1/0 so

((1/0*1/0)-1)(1/0+1/0)=0

(f^2-1)(2f)=0

so f^2-1=0 or (2f)=0

and since 2f is just 1/0 + 1/0 we get undefined

but 1/0^2, if just 1/0 because 1*1/0*0

and then subtracting one doesnt help since we cant divide by zero.

btw i have found an equation that has a one point solution its just lost in my notes.

Guest Mar 6, 2017