Question 15 Marks
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) q46 is both a 4th power and a 6th power (q is a prime number).
(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) q46 is both a 4th power and a 6th power (q is a prime number).
Answer
View full question & answer→(i) Only sometimes true.
Explanation: $64=2^6=\left(2^3\right)^2=\left(2^2\right)^3$ is both a cube and a square.
But $8=2^3$ is a cube, not a square.
(ii) Always true.
Eplanation: $3^4=\left(3^2\right)^2=9^2$.
$5^4=\left(5^2\right)^2=25^2 .$
(iii) Always true.
Explanation: $a ^5= a ^3 \times a ^2$ and is divisible by $a ^3$.
(iv) Always true.
Explanation: $8=2^3, 27=3^3$
$8 \times 27=216$, which is $6^3$.
(v) Never true.
Explanation: Since 46 is not divisible by 4 or 6 . Therefore, there is no prime number $q$ such that $q ^{46}$ is both a perfect fourth power and a perfect sixth power.
Explanation: $64=2^6=\left(2^3\right)^2=\left(2^2\right)^3$ is both a cube and a square.
But $8=2^3$ is a cube, not a square.
(ii) Always true.
Eplanation: $3^4=\left(3^2\right)^2=9^2$.
$5^4=\left(5^2\right)^2=25^2 .$
(iii) Always true.
Explanation: $a ^5= a ^3 \times a ^2$ and is divisible by $a ^3$.
(iv) Always true.
Explanation: $8=2^3, 27=3^3$
$8 \times 27=216$, which is $6^3$.
(v) Never true.
Explanation: Since 46 is not divisible by 4 or 6 . Therefore, there is no prime number $q$ such that $q ^{46}$ is both a perfect fourth power and a perfect sixth power.