Question

Consider the following statements :
(i) For every positive real number $x$, $x - 10$ is positive.
(ii) Let $n$ be a natural number. If $n^2$ is even, then $n$ is even.
(iii) If a natural number is odd, then its square is also odd.
Then

• A. (i) False, (ii) True and (iii) True
• B. (i) False, (ii) False and (iii) True
• C. (i) True, (ii) False and (iii) True
• D. (i) True, (ii) True and (iii) True
• E. (i) False, (ii) True and (iii) False

Hint:

To disprove a universally quantified statement ("for every", "for all"), providing just one counterexample is sufficient. To prove an implication "If P then Q", it is often easier to prove its contrapositive "If not Q then not P".

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to evaluate the truth value of three separate mathematical statements using basic logic, properties of numbers, and counterexamples.

Step 2: Detailed Explanation:
Let's analyze each statement one by one:
Statement (i): "For every positive real number $x$, $x - 10$ is positive."
To prove a "for every" statement false, we only need to find a single counterexample.
Let's pick a positive real number, for instance, $x = 5$.
Substitute $x = 5$ into the expression: $5 - 10 = -5$.
Since $-5$ is not a positive number, the statement does not hold for all positive real numbers.
Therefore, Statement (i) is False.
Statement (ii): "Let $n$ be a natural number. If $n^2$ is even, then $n$ is even."
We can prove this by proving its contrapositive: "If $n$ is not even (i.e., odd), then $n^2$ is not even (i.e., odd)."
Let $n$ be an odd natural number. By definition, an odd number can be written as $n = 2k + 1$, where $k$ is a non-negative integer.
Now let's square $n$:
\[ n^2 = (2k + 1)^2 \]
\[ n^2 = 4k^2 + 4k + 1 \]
We can factor out a 2 from the first two terms:
\[ n^2 = 2(2k^2 + 2k) + 1 \]
Let $m = 2k^2 + 2k$, which is also an integer. Then:
\[ n^2 = 2m + 1 \]
This is the standard form of an odd number. Thus, if $n$ is odd, $n^2$ is odd. The contrapositive is true, which implies the original statement is also true.
Therefore, Statement (ii) is True.
Statement (iii): "If a natural number is odd, then its square is also odd."
This is precisely the contrapositive we just proved while analyzing statement (ii).
Let $n = 2k + 1$ (an odd number).
$n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$, which is odd.
Therefore, Statement (iii) is True.
Summarizing our findings: (i) is False, (ii) is True, and (iii) is True. This matches option (A).

Step 4: Final Answer:
The correct option is (i) False, (ii) True and (iii) True.