using Euclid division Lemma show that the square of any positive integer is either in the form of 2m or 2m +1
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Answered by
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Hey Mate ✌
Here's your answer friend,
==================================
==> Let a be the positive integer for some integer q.
==> We know every positive integer is in the form of 2q, or 2q + 1.
==> Here, b = 2,
==> Therefore, r = 0,1 .........{ 0>=r>b}
==> By using Euclid's division lemma,
we get,
For r = 0
==> a = bq + r
==> a = 2q + 0
==> a = 2q
==> square of positive integer
==> a² = (2q)²
==> a² = 4q²
==> a² = 2(2q²)
==> a² = 2m .........{ where m = 2q²}
Now,
when r = 1,
we get,
==> a = bq + r
==> a = 2q + 1
==> Square of positive integer
==> a² = (2q + 1)²
==> a² = 4q² + 4q + 1
==> a² = 2(2q² + 2q) + 1
==> a² = 2m + 1 ............{ where m = 2q² + 2q}
Hence, showed square of any positive integer is either in the form of 2m or 2m + 1.
==================================
⏩ Hope it helps you ^_^
Here's your answer friend,
==================================
==> Let a be the positive integer for some integer q.
==> We know every positive integer is in the form of 2q, or 2q + 1.
==> Here, b = 2,
==> Therefore, r = 0,1 .........{ 0>=r>b}
==> By using Euclid's division lemma,
we get,
For r = 0
==> a = bq + r
==> a = 2q + 0
==> a = 2q
==> square of positive integer
==> a² = (2q)²
==> a² = 4q²
==> a² = 2(2q²)
==> a² = 2m .........{ where m = 2q²}
Now,
when r = 1,
we get,
==> a = bq + r
==> a = 2q + 1
==> Square of positive integer
==> a² = (2q + 1)²
==> a² = 4q² + 4q + 1
==> a² = 2(2q² + 2q) + 1
==> a² = 2m + 1 ............{ where m = 2q² + 2q}
Hence, showed square of any positive integer is either in the form of 2m or 2m + 1.
==================================
⏩ Hope it helps you ^_^
Answered by
4
hope it will be helpful
thank you
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