1.

Prove that the work of any group is 4m or 4m + 1 for a complementn​

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ong>Answer:

Step-by-step explanation:

Note :- I am taking q as some integer.

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 odd positive integer, so a can't be 4q , or 4q + 2 [ As these are even ] .

•°• a can be of the FORM 4q + 1 or 4q + 3 for some integer q .

Hence , it is solved

THANKS

#BeBrainly.

Step-by-step explanation:


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