The short answer is that 0/0 = undefined, meaning that no value can be associated to it. In fact, dividing anything by zero is undefined.

Let's take something that no one has a problem with dividing, 8/4. Anytime one divides, you can represent it into the following equation:

Denominator*x=Numerator

Let's do this for 8/4.

4x=8 Divide by 4 on both sides

x=2

Now, let's see what happens when we try this same method with 0/0.

0x=0

Oh no! All numbers for *x *satisfy this equation. For simplicity, let's say that x=1 and x=2.

0*1=0, so 1=0/0

0*2=0, so 2=0/0

1=2.

Clearly, this is a contradictory result, and assigning 0/0 a value would destroy any and all current algebra rules. This is also why there are "proofs" that 1=2. Here's an example:

1. a=b Multiply both sides by a

2. a^2=ab Subtract b^2 from both sides

3. a^2-b^2=ab-b^2 Group both terms using the GCF

4. (a+b)(a-b)=b(a-b) Divide by (a-b) on both sides

5. a+b=b Because a=b, b can always be replaced for a

6. b+b=b Combine like terms

7. 2b=b Divide b from both sides

8. 2=1.

Where is the problem in this proof? The problem is in step 4 when dividing by (a-b) occurs.

a=b Subtract b from both sides

a-b=0

If a=b, then a-b=0. This means that when this person divided by (a-b), he was dividing by 0, which algebra does not allow and creates the contradictory result. Still don't understand? Let's substitute in a value for a and b to see what is happening behind the scenes:

1=1

1^2=1*1

1^2-1^2=1*1-1^2

(1+1)(1-1)=1(1-1) <-- Dividing by (1-1), or 0 cannot be done in algebra

1+1=1

2*1=1

2=1

Technically, step 7 is flawed, too, when he divided by b. Here's why.

2b=b Subtract b from both sides

2b-b=0 Combine like terms.

b=0

Oh no! If 2b=b, then b=0. This means that dividing by b is dividing by 0. Hopefully, you understand why you cannot do 0/0 now.

TheXSquaredFactor
May 18, 2017