Q. 4

# Prove that every

We know that from Euclid’s division lemma, for every natural number a if we take b = 5 then unique non-negative integers k and m can be obtained so that a = 5k + m

Where 0 ≤ m ≤ 5

Therefore, m = 0, 1, 2, 3, 4 and 5

So, we can say that a = 5k or a = 5k + 1 or a = 5k + 2 or a = 5k + 3 or a = 5k + 4 where k E N V {0}

Now, a = 5k + 3 = 5k + 5 – 2 = 5(k + 1) – 2

= 5k1 – 2 where (k1 = k + 1)

Therefore, a = 5k + 3 can be written as a = 5k – 2

Now, a = 5k + 4 = 5k + 5 – 1 = 5(k + 1) – 1

= 5k1 – 1 where (k1 = k + 1)

Therefore, a = 5k + 4 can be written as a = 5k – 1

So, every natural number can be written in the form 5k or 5k ± 1 or 5k ± 2, k E N V {0}.

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