For an integer 'm' every even positive integer is of the form-
a) m +1
b) 2m + 1
c) m
d) 2m
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Answered by
9
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- Your answer is (D).....2m. ........if uhh want an odd positive integer uhh can take 2m+1........
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HERE IS YOUR ANSWER ✌✌
========================
ANSWER ✔ ✔ ✔
- Your answer is (D).....2m. ........if uhh want an odd positive integer uhh can take 2m+1........
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HOPE IT HELPS UHH ✌ ✌ ✌
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VIELEN DANK ♥ ♥ ♥
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PLEASE MARK ME BRAINLIEST ⭐ ⭐ ⭐ ⭐
karinasharma:
batao na
Answered by
5
(d) 2m is the answer.
Explanation:
We know that even numbers are 2, 4, 6, 8, 10...
So for an integer 'm', every even positive integer can be written as 2m, where m is an integer.
Alternative explanation:
Let 'a' be a positive integer. On dividing 'a' by 2, let m be the quotient and r be the remainder. Then, by Euclid's division algorithm, we have
a = 2m + r, where a ≤ r < 2 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.
Explanation:
We know that even numbers are 2, 4, 6, 8, 10...
So for an integer 'm', every even positive integer can be written as 2m, where m is an integer.
Alternative explanation:
Let 'a' be a positive integer. On dividing 'a' by 2, let m be the quotient and r be the remainder. Then, by Euclid's division algorithm, we have
a = 2m + r, where a ≤ r < 2 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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