What is 1 divided by 0?
This is part of a series on common misconceptions.
Is this true or false?
\( 1/0\) is undefined.
Why some people say it's true: Dividing by \( 0\) is not allowed.
Why some people say it's false: \( 1/0 = \infty.\)
The statement is \( \color{blue}{\textbf{true}}\).
Proof:
If \( 1/0 = r\) were a real number, then \( r\cdot 0 = 1,\) but this is impossible for any \( r.\) See division by zero for more details.
Rebuttal: In calculus, \( 1/0\) equals \( \infty.\)Reply: This statement is incorrect for two reasons. First, infinity is not a real number. The proof demonstrates that the quotient \( 1/0\) is undefined over the real numbers.
It is true that in some situations, the indeterminate form \( 1/0\) can be interpreted as \( \infty,\) for instance when taking limits of a quotient of functions. But even this is not always true, as the following example shows:
Consider \( \lim\limits_{x\to 0}\frac{1}{x}.\)
Approaching from the right, \( \lim\limits_{x \to 0^+} \frac{1}{x} = + \infty. \)
Approaching from the left, \( \lim\limits_{x \to 0^-} \frac{1}{x} = - \infty. \)In order for \( \frac{1}{0} \) to be consistent, the limits from both directions should be equal, which is clearly not the case here.
Rebuttal: What about on the Riemann sphere?
Reply: For certain complex functions, it is convenient and consistent to extend their domain and range to \( {\mathbb C} \cup \{\infty\}.\) This set has the geometric structure of a sphere, called the Riemann sphere. For instance, suppose \(a,b,c,d\) are complex numbers such that \( ad-bc\ne 0.\) Then the function \( f(z) = \frac{az+b}{cz+d} \) can be extended by defining \( f(-d/c) = \infty \) and \( f(\infty) = a/c \) (or \( f(\infty) = \infty \) when \(c=0.)\) This makes \(f\) a bijection on the Riemann sphere, with many nice properties.
So there are situations where \( 1/0\) is defined, but they are defined in a tightly controlled way. It is still the case that \(1/0\) can never be a real (or complex) number, so--strictly speaking--it is undefined.
What is wrong with the following "proof"?
Let \(a = b=1\), then \(a=b.\)
Step 1: \(a^2 = ab \)
Step 2: \(a^2 - b^2 = ab - b^2 \)
Step 3: \((a+b)(a-b) = b(a-b) \)
Step 4: \(a+b= \dfrac{b(a-b)}{a-b} \)
Step 5: \(a+b = b\)
Conclusion : By substituting, \( a = b = 1 \) , \(1+1 = 1 \ , \ 2 = 1 \) .
See Also