MCQ
Choose the correct answer from the given four options in the following questions:
For some integer m, every even integer is of the form:
For some integer m, every even integer is of the form:
- Am.
- Bm + 1.
- ✓2m.
- D2m + 1.
✓
Answer
Correct option: C.
2m.
We know that, even integers are 2, 4, 6, ...
So, it can be written in the form of 2m.
where, m = Integer = Z [since, integer is represented by Z]
or m = ···, -1, 0, 1, 2, 3, ...
$\therefore$ 2m = ···, -2, 0, 2, 4, 6, ...
Alternate Answer
Let 'a' be a positive integer. On dividing 'a' by 2, let m be the quotient and be the remainder. Then, by Euclid's divlslon algonthm, we have
a = 2m + r, where
$\text{a}\leq\text{r}<2\text{ i.e.,}$
r = 0 and r = 1.
⇒ a = 2m or a = 2m + 1
when, a = 2m for some integer m, then clearly a is even.
So, it can be written in the form of 2m.
where, m = Integer = Z [since, integer is represented by Z]
or m = ···, -1, 0, 1, 2, 3, ...
$\therefore$ 2m = ···, -2, 0, 2, 4, 6, ...
Alternate Answer
Let 'a' be a positive integer. On dividing 'a' by 2, let m be the quotient and be the remainder. Then, by Euclid's divlslon algonthm, we have
a = 2m + r, where
$\text{a}\leq\text{r}<2\text{ i.e.,}$
r = 0 and r = 1.
⇒ a = 2m or a = 2m + 1
when, a = 2m for some integer m, then clearly a is even.
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