Readings for Session 11 – (Continued)
Division with Zero
Many people have trouble with
problems involving division where either the dividend or divisor
is zero. Note that in the missing-factor definition, we assume
that the divisor cannot be zero. Why do we make that assumption?
What would
happen if the divisor was 0?
Example: 5 ÷ 0.
Rewrite the problem 5 ÷ 0 = c as a multiplication problem, 0 ∙ c = 5. But, we know from multiplication of whole numbers that 0 times any whole number is 0, which means there is no whole number solution to the problem.
Example:
0
÷ 0.
Rewrite the problem 0
÷ 0 =
c as a multiplication
problem, 0
∙ c = 0. Again, we know that 0
times any whole number is 0, which means that every whole number
is a solution to the problem; e.g., 0
∙ 4 = 0, 0
∙
35 = 0, etc.
The above two examples illustrate
why we do not define division by zero. When a problem involves
division by zero (0 as the divisor), we state that the solution
is undefined (or that
there is no solution).
What about when the dividend is 0?
Example: 0 ÷ 6.
Rewrite the problem 0 ÷ 6 = c as a multiplication problem, 6 ∙ c = 0.
We know from
multiplication that this solution must be 0 since 6
∙ 0 = 0 and
multiplication by a nonzero value would give a nonzero answer;
e.g., 6
∙ 1 = 6, 6
∙
7 = 42, etc.
Return
to Peil's Homepage | Minnesota
State University Moorhead | Mathematics
Department