Question

For some integer \[m\], every even integer is of the form:
(a) \[m\] (b) \[m + 1\] (c) \[2m\] (d) \[2m + 1\]

Answer 1

Hint:
Here, we need to check which of the given options represents every even integer where \[m\] is some integer. We will check all the options for some odd and even value of the integer \[m\]. The option which represents every even integer for both the odd and even value of the integer, that value of \[m\] is the correct option.

Complete step by step solution:
We will check each option for some odd or even value of the integer \[m\] to find the correct option.
(a)
Suppose that \[m\] is the integer 1.
The integer 1 is an odd integer.
Therefore, \[m\] does not represent every even integer.
This is true for all values of \[m\] that are odd integers.
Thus, we can conclude that if \[m\] is any odd integer, then \[m\] does not represent every even integer.
Therefore, option (a) is incorrect.
(b)
Suppose that \[m\] is the integer 2.
The integer 2 is an even integer.
1 more than the integer 2 is \[2 + 1 = 3\].
Therefore, if \[m = 2\], then \[m + 1 = 3\].
The number \[m + 1\] is an odd integer.
Thus, \[m + 1\] does not represent every even integer.
This is true for all values of \[m\] that are even integers.
Thus, we can conclude that if \[m\] is any even integer, then \[m + 1\] does not represent every even integer.
Therefore, option (b) is incorrect.
(c)
Suppose that \[m\] is the integer 1.
The integer 1 is an odd integer.
Multiplying the integer 1 by 2, we get
\[1 \times 2 = 2\]
Therefore, if \[m = 1\], then \[2m = 2\].
Here, the number \[2m\] is an even integer.
Now, suppose that \[m\] is the integer 2.
The integer 2 is an even integer.
Multiplying the integer 2 by 2, we get
\[2 \times 2 = 2\]
Therefore, if \[m = 2\], then \[2m = 2\].
Here, the number \[2m\] is an even integer.
We can observe that for any odd or even value of \[m\], the value of \[2m\] is an even integer.
Thus, we can conclude that if \[m\] is any integer (even or odd), then \[2m\] represents every even integer.
Therefore, option (c) is the correct option.
(d)
We have proved that if \[m\] is any integer (even or odd), then \[2m\] represents every even integer.
Therefore, since \[2m\] is an even integer, then \[2m + 1\] represents every odd integer.

Thus, option (d) is incorrect.

Note:
We have proved that if \[m\] is any integer (even or odd), then \[2m\] represents every even integer. This is because when any integer is multiplied by 2, the resulting number has either 2, 4, 6, 8, or 0 as the digit in the unit’s place. Any number that has the digit 2, 4, 6, 8, or 0 in the unit’s place is divisible by 2, and every number divisible by 2 is an even number. Odd numbers are not divisible by 2 whereas even numbers are divisible by 2.