show that any positive even is on the form 4q, or 4q+2, where q is some integer
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b=4
when r=0
a= 4q
when r=2
a=4q+2
when r=0
a= 4q
when r=2
a=4q+2
Answered by
1
Step-by-step explanation:
Let a be the positive integer.
And, b = 4 .
Then by Euclid's division lemma,
We can write a = 4q + r ,for some integer q and 0 ≤ r < 4 .
°•° Then, possible values of r is 0, 1, 2, 3 .
Taking r = 0 .
→ a = 4q .
Taking r = 1 .
→ a = 4q + 1 .
Taking r = 2
→ a = 4q + 2 .
Taking r = 3 .
→ a = 4q + 3 .
But a is an even positive integer, so a can't be 4q + 1 , or 4q + 3 [ As these are odd ] .
∴ a can be of the form 4q or 4q + 2 for some integer q .
Hence , it is solved
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