Question 3

The product of two numbers is non-zero only if both numbers are non-zero.

Step 1 — Understand the condition

We are given that $a$ and $b$ are real numbers. Their product $ab$ is not equal to zero. $ab \neq 0$ This means $a$ multiplied by $b$ gives a non-zero number.

Step 2 — Check statement (i)

Statement (i) says: Both $a$ and $b$ must be zero. Let's assume $a$ is $\mathbf{0}$ and $b$ is $\mathbf{0}$. We find their product. $ab = 0 \times 0$ $= 0$ This result $ab=0$ contradicts the given condition $ab \neq 0$. So, statement (i) is incorrect.

$\boxed{\text{Statement (i) is False}}$

Step 3 — Check statement (ii)

Statement (ii) says: Both $a$ and $b$ must be non-zero. If $a$ is not $\mathbf{0}$ and $b$ is not $\mathbf{0}$. Then their product $ab$ will also not be $\mathbf{0}$. For example, if $a=2$ and $b=3$, then $ab=6$. Here, $6 \neq 0$. This matches the given condition $ab \neq 0$. So, for $ab \neq 0$, both $a$ and $b$ must be non-zero.

$\boxed{\text{Statement (ii) is True}}$

Step 4 — Check statement (iii)

Statement (iii) says: Either $a$ or $b$ must be non-zero. This means that $a$ and $b$ cannot both be $\mathbf{0}$. Let's consider if $a$ was $\mathbf{0}$ and $b$ was $\mathbf{0}$. As shown in Step 2, their product $ab$ would be $\mathbf{0}$. $ab = 0 \times 0$ $= 0$ This contradicts the given condition $ab \neq 0$. So, it is impossible for both $a$ and $b$ to be $\mathbf{0}$. Therefore, at least one of them must be non-zero. This makes statement (iii) correct.

$\boxed{\text{Statement (iii) is True}}$

Answer

(i) False (ii) True (iii) True