Question

for some integer m, every even integer is of the form

By Euclid divison lemma
a=bq+r
a=bm+r
If b=2 then a=2m is divisible by 2, then 2m is an even integer ​

Answer 1

Answer:

We know that an integer is said to be even when it is divisible by 2.

Let m be a integer i.e. m = ..., -3, -2, -1, 0, 1, 2, 3, …

Clearly, 'm' may or may not be even, therefore, 'm' can't be an answer.m + 1 = ..., -2, -1, 0, 1, 2, 3, 4...Here also, 'm + 1' may or may not be even, therefore, 'm + 1' can't be an answer.

Now, consider 2m i.e. 2m = ...,-6, -4, -2, 0, 2, 4, 6,...

Here, for some integer m, 2m is always integer,

Now, consider 2m + 1, i.e. 2m + 1 = ..., -5, -3, -1, 1, 3, 5, 7,...Clearly, 2m + 1 is an odd integer, therefore it can't be an answerHence, 2m is only answer.

Answer 2

Step-by-step explanation:

above answer is the correct answer