What Is 0 Divided By 0? | Brilliant Math & Science Wiki

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Kuldeep Guha Mazumder, Samir Khan, Hobart Pao, and 9 others

• Jason Horowitz
• Zoe Codrington
• Pi Han Goh
• Patrick Corn
• William Hou
• Andrew Ellinor
• Jimin Khim
• Eli Ross
• Zandra Vinegar

contributed

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.

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

Reveal the correct answer

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 every number \(x\) 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. \(_\square\)

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

See common rebuttals

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 \( \frac{y}{0}=\infty\) 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 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:

Step 1 Step 2 Step 3 Step 4 Step 5 Reveal the answer

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 \(8-8 = 0 \), \( \frac00 = 1 \).

The correct answer is: Step 4

See Also

• List of Common Misconceptions

• Indeterminate Forms

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