What pattern do you observe? Why are 2 and 5 related in this way?

Question

Homi · Accepted Answer

Understand the Question

In our decimal (base-10) system, $10 = 2 \times 5$ .
When dividing $1$ by successive powers of $2$ or powers of $5$ , each step introduces another factor into the denominator, which increases the number of decimal places by $1$ .
Because $2$ and $5$ are the prime factors of $10$ , any fraction with a denominator of $2^n$ or $5^n$ can be converted directly into a power of $10$ in the denominator by multiplying by $5^n$ or $2^n$ respectively, resulting in a terminating decimal.

Step 1 · Observe the Patterns and Find Missing Values

Powers of 2: Each step multiplies the denominator by $2$ , halving the decimal value and adding one decimal place. Missing value for $\dfrac{1}{2 \times 2 \times 2 \times 2 \times 2} = \dfrac{1}{2^5}$ :

\begin{aligned} 0.0625 \div 2 &= 0.03125 \end{aligned}

Powers of 5: Each step multiplies the denominator by $5$ , dividing the decimal value by $5$ and adding one decimal place. Missing value for $\dfrac{1}{5 \times 5 \times 5 \times 5 \times 5} = \dfrac{1}{5^5}$ :

\begin{aligned} 0.0016 \div 5 &= 0.00032 \end{aligned}

Step 2 · Explain the Relationship Between 2 and 5

Our number system is base- $10$ , and the prime factors of $10$ are $2$ and $5$ .

A fraction whose denominator contains only factors of $2$ or $5$ can always be converted into a power of $10$ :

For $\dfrac{1}{2^n}$ , multiply numerator and denominator by $5^n$ :

\begin{aligned} \dfrac{1}{2^n} &= \dfrac{1 \times 5^n}{2^n \times 5^n} \\[0.6em] &= \dfrac{5^n}{(2 \times 5)^n} \\[0.6em] &= \dfrac{5^n}{10^n} \end{aligned}

For $\dfrac{1}{5^n}$ , multiply numerator and denominator by $2^n$ :

\begin{aligned} \dfrac{1}{5^n} &= \dfrac{1 \times 2^n}{5^n \times 2^n} \\[0.6em] &= \dfrac{2^n}{(5 \times 2)^n} \\[0.6em] &= \dfrac{2^n}{10^n} \end{aligned}

Thus, each expression converts directly into a terminating decimal with exactly $n$ decimal places.

Answer

Pattern: For $\dfrac{1}{2^n}$ , the decimal value is halved at each step and the number of decimal places increases by one. For $\dfrac{1}{5^n}$ , the decimal value is divided by $5$ at each step and the number of decimal places increases by one. The missing values are $\dfrac{1}{2^5} = 0.03125$ and $\dfrac{1}{5^5} = 0.00032$ .

Relationship: $2$ and $5$ are the prime factors of $10$ ( $10 = 2 \times 5$ ). Fractions with denominators $2^n$ or $5^n$ can be converted to have a denominator of $10^n$ , resulting in terminating decimals with $n$ decimal places.

Common Mistakes

Non-Terminating Factors: Assuming any fraction terminates as a decimal. Only fractions whose simplified denominators have prime factors of only $2$ and/or $5$ form terminating decimals; other prime factors (like $3$ or $7$ ) result in repeating decimals.
Decimal Place Count: Assuming $2^n$ or $5^n$ yields $n$ trailing zeros instead of $n$ digits after the decimal point.

Question 17

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