Power Play (Exponents) | FIO

Question 7

Examine each statement below and find out if it is 'Always True', 'Only Sometimes True', or 'Never True'. Explain your reasoning.

(i) Cube numbers are also square numbers.

(ii) Fourth powers are also square numbers.

(iii) The fifth power of a number is divisible by the cube of that number.

(iv) The product of two cube numbers is a cube number.

(v) $q^{46}$ is both a 4th power and a 6th power ( $q$ is a prime number).

Solution

Understand the Question

We analyze statements about powers and exponents by applying the fundamental laws of exponents:

Power of a power: $(a^m)^n = a^{m \cdot n}$
Quotient of powers: $\dfrac{a^m}{a^n} = a^{m-n}$
Power of a product: $a^n b^n = (ab)^n$
A prime power $q^k$ is an $n^{\text{th}}$ power if and only if the exponent $k$ is divisible by $n$ .

(i) Cube numbers are also square numbers.

Step 1 · Check with Examples

A cube number is of the form $n^3$ , and a square number is of the form $m^2$ .

For a number to be both a cube and a square, it must be a $6^{\text{th}}$ power (since $\text{LCM}(2, 3) = 6$ ): $1^3 = 1 = 1^2 \quad (\text{both a cube and a square})$ $2^3 = 8 \quad (\text{a cube, but not a square})$ $4^3 = 64 = 8^2 \quad (\text{both a cube and a square})$

Since some cubes are squares while others are not, the statement is only sometimes true.

Answer

(i) Only Sometimes True

(ii) Fourth powers are also square numbers.

Step 1 · Rewrite Using Exponent Laws

Let any fourth power be $n^4$ for an integer $n$ .

Using the power of a power rule $(a^b)^c = a^{b \times c}$ : $n^4 = (n^2)^2$

Since $n^2$ is an integer, $(n^2)^2$ is always a perfect square.

For example: $2^4 = 16 = 4^2 = (2^2)^2$

Answer

(ii) Always True

(iii) The fifth power of a number is divisible by the cube of that number.

Step 1 · Evaluate the Quotient

For any non-zero number $a$ , its fifth power is $a^5$ and its cube is $a^3$ .

Using the quotient rule $\dfrac{a^m}{a^n} = a^{m-n}$ :

\begin{aligned} \dfrac{a^5}{a^3} &= a^{5-3} \\[0.6em] &= a^2 \end{aligned}

Since $a^2$ is an integer, $a^5$ is always divisible by $a^3$ .

Answer

(iii) Always True

(iv) The product of two cube numbers is a cube number.

Step 1 · Apply Power of a Product Rule

Let the two cube numbers be $x^3$ and $y^3$ for integers $x$ and $y$ .

Using the property $a^n b^n = (ab)^n$ : $x^3 \times y^3 = (x \times y)^3$

Let $P = x \times y$ . Since $P$ is an integer, $(x \times y)^3 = P^3$ is always a cube number.

For example: $2^3 \times 3^3 = 8 \times 27 = 216 = 6^3$

Answer

(iv) Always True

(v) $q^{46}$ is both a 4th power and a 6th power ( $q$ is a prime number).

Step 1 · Check Divisibility of the Exponent

For a prime power $q^{46}$ to be an $n^{\text{th}}$ power, the exponent $46$ must be divisible by $n$ .

Checking if $46$ is a multiple of $4$ : $46 \div 4 = 11 \text{ with a remainder of } 2$
Checking if $46$ is a multiple of $6$ : $46 \div 6 = 7 \text{ with a remainder of } 4$

Since $46$ is neither divisible by $4$ nor by $6$ , $q^{46}$ is neither a $4^{\text{th}}$ power nor a $6^{\text{th}}$ power.

Answer

(v) Never True

Common Mistakes

Assuming all cubes are squares: For a number to be both a square and a cube, its prime factors must have exponents that are multiples of $\text{LCM}(2, 3) = 6$ , not just $3$ .
Confusing base with exponent in divisibility: In part (v), checking whether $q^k$ is an $n^{\text{th}}$ power requires checking if the exponent $k$ is divisible by $n$ , not whether the prime $q$ is divisible.

Question 7

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