# What is 0 divided by 0?

This is part of a series on common misconceptions.

What is \(\frac00?\)

**Why some people say it's 0:** Zero divided by any number is 0.

**Why some people say it's 1:** A number divided by itself is 1.

**Why some people say it's undefined:** Dividing by zero is undefined.

Only one of these explanations is valid, and choosing the other explanations can lead to serious contradictions.

The expression is \( \color{red}{\textbf{undefined}}\).

Here's why:

Remember that \(\frac{a}{b}\) means "the number which when multiplied by \(b\) gives \(a.\)" For example, the reason \(\frac{1}{0}\) is undefined is because there is no number \(x\) such that \(0 \cdot x = 1.\)

The situation with \(\frac{0}{0}\) is strange, because

everynumber \(x\) satisfies \(0 \cdot x = 0.\) Because there's no single choice of \(x\) that works, there's no obvious way to define \(\frac{0}{0}\), so by convention it is left undefined.

Of course, there are many possible counterarguments to this. Here are a few common ones:

Rebuttal: Any number divided by itself is \( 1.\)

Reply: This is true for any nonzero number, but dividing by \(0\) is not allowed.

Rebuttal: \( 0\) divided by any number is \( 0.\)

Reply: This is true for any nonzero denominator, but dividing by \( 0\) is not allowed no matter what the numerator is.

Rebuttal: Any number divided by \( 0\) is \( \infty.\)

Reply: Even for nonzero \( y,\) writing \( \dfrac{y}{0}=\infty\) is not entirely accurate: see 1/0 for a discussion. But this reasoning only makes sense for a nonzero numerator.

Rebuttal: If we choose to set \( \frac00=1,\) or \( 0,\) it is not inconsistent with other laws of arithmetic, and it makes one of the rules in the above rebuttals true in all cases.

Reply: This is a combination of the first two rebuttals, so here is a "big-picture" reply. Any specific choice of value for \( \frac00\) will allow some function to be extended continuously. For instance, if we mandate \( \frac00=1,\) then the function \( f(x) = \frac xx\) becomes continuous at \( x=0.\) If \(\frac00=0,\) the the function \( f(x)=\frac0x\) becomes continuous at \( x=0.\)But this is not satisfactory in all cases, and the arbitrariness of the choice will break other laws of arithmetic. For instance, \[ \begin{align} \frac00 + \frac11 &= \frac{0\cdot 1+1\cdot 0}{0\cdot 1} \\ &= \frac00, \end{align} \] which doesn't make any sense for any (finite) choice of \( \frac00.\)

Introduction of terms like \(\frac{0}{0}\) in otherwise sound arguments can break them down. See if you can spot the error in the problem below.

I will attempt to prove that \(\frac00 = 1 \). In which of these steps did I first make a mistake by using flawed logic?

**Step 1:** We can rewrite 15 as \(7 + 8\) or \(8 + 7\).

**Step 2:** This means that \(7 + 8 = 8+ 7 \).

**Step 3:** If we move one term from each side of the equation to the other side, we will get \[7-7 = 8-8.\]
**Step 4:** Dividing both sides by \(8-8\) gives \(\frac{7-7}{8-8} = 1 \).

**Step 5:** Since \(7-7= 0\) and \(8-8 = 0 \), \( \frac00 = 1 \).

**See Also**

**Cite as:**What is 0 divided by 0?.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/what-is-0-0/