Sure, let's carefully examine each statement one by one to determine their validity.
### Statement (i): The sum of two odd whole numbers is always a whole number
Consider any two odd whole numbers. An odd number can be represented in the form \(\(2k + 1\)\), where \(\(k\)\) is an integer.
1. Let the first odd number be \(\(2a + 1\)\).
2. Let the second odd number be \(\(2b + 1\)\).
3. The sum of these two odd numbers is:
\(\[
(2a + 1) + (2b + 1) = 2a + 2b + 2 = 2(a + b + 1)
\]\)
The result is \(\(2(a + b + 1)\)\), which is clearly a whole number since it is an integer.
Therefore, the sum of two odd whole numbers is always a whole number, making Statement (i) true.
### Statement (ii): The product of two odd numbers is always odd
Let's check this by again taking two odd numbers, which can be represented as \(\(2k + 1\)\), where \(\(k\)\) is an integer.
1. Let the first odd number be \(\(2a + 1\)\).
2. Let the second odd number be \(\(2b + 1\)\).
3. The product of these two odd numbers is:
\(\[
(2a + 1) \times (2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1
\]\)
The result is of the form \(\(2m + 1\)\) where \(\(m\)\) is an integer, which is again an odd number.
Therefore, the product of two odd numbers is always odd, making Statement (ii) true.
### Statement (iii): The product of two even numbers is always even
Now let’s verify this using two even numbers. An even number can be represented as \(\(2k\)\), where \(\(k\)\) is an integer.
1. Let the first even number be \(\(2a\)\).
2. Let the second even number be \(\(2b\)\).
3. The product of these two even numbers is:
\(\[
(2a) \times (2b) = 4ab = 2(2ab)
\]\)
The result is of the form \(\(2n\)\) where \(\(n\)\) is an integer, which is clearly an even number.
Therefore, the product of two even numbers is always even, making Statement (iii) true.
### Conclusion
Each of the three statements is true:
(i) The sum of two odd whole numbers is always a whole number.
(ii) The product of two odd numbers is always odd.
(iii) The product of two even numbers is always even.
