### Why 0÷0 doesn't have a value?

In this short post we're going to take a look at why `0/0` is really undefined.

# Overview It's an old mathematical issue, which has been debated many times over the centuries with mostly the same result: `0/0` does not have a defined value. But now, we're going to take a look at why this is true.

# Illustration To illustrate this, let's consider the following sum:

`\sum_{k=0}^(n)b^k = b^0 + b^1 + b^2 + ... + b^n`

Deriving a closed form to this sum, we get:

`\sum_{k=0}^(n)b^k = (b^(n+1) - 1) / (b-1)`

For example, when `b=3` and `n=4`, we have:

`3^0 + 3^1 + 3^2 + 3^3 + 3^4 = (3^(4+1) - 1) / (3-1)`

All good so far. However, if we set `b=1`, we have a special case:

`(1^(n+1) - 1) / (1-1) = 0/0`

We know that `1^k=1` for any `k>=0`, therefore:

`\sum_{k=0}^(n)1^k = n+1`

but when `b=1`, our closed-form evaluates to `0/0` for any value of `n`. From this we can conclude that `0/0` does not have a certain value.

Taking this example a little bit further, we can also show that…

# Overview It's an old mathematical issue, which has been debated many times over the centuries with mostly the same result: `0/0` does not have a defined value. But now, we're going to take a look at why this is true.

# Illustration To illustrate this, let's consider the following sum:

`\sum_{k=0}^(n)b^k = b^0 + b^1 + b^2 + ... + b^n`

Deriving a closed form to this sum, we get:

`\sum_{k=0}^(n)b^k = (b^(n+1) - 1) / (b-1)`

For example, when `b=3` and `n=4`, we have:

`3^0 + 3^1 + 3^2 + 3^3 + 3^4 = (3^(4+1) - 1) / (3-1)`

All good so far. However, if we set `b=1`, we have a special case:

`(1^(n+1) - 1) / (1-1) = 0/0`

We know that `1^k=1` for any `k>=0`, therefore:

`\sum_{k=0}^(n)1^k = n+1`

but when `b=1`, our closed-form evaluates to `0/0` for any value of `n`. From this we can conclude that `0/0` does not have a certain value.

Taking this example a little bit further, we can also show that…