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Show that any positive odd integer is of the form 4m+1 or 4m+3, where m is some integer

User JBond
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Answer:

Explanation:

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\ \text {and}\ 0 \leq r < 4 .

°•° Then, possible values of r is 0, 1, 2, 3 .


\texttt {Taking} \ r = 0 .\\\\a = 4q .


\texttt {Taking} \ r = 1 .\\\\a = 4q + 1 .\\\\\texttt {Taking} \ r = 2\\\\a = 4q + 2 .\\\\\texttt {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 .

User Soren
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