History of Mathematics See ISBN:978-0-07-338315-6: author was David M Burton See

Question

History of Mathematics

See ISBN:978-0-07-338315-6: author was David M Burton

See section 4.3 page 182, question 10b

The square of any odd integer is of the form 8n+1. [hint: any odd integer is of the form 4k+1 or 4k+3]

Proof by Contradiction:

Assume m,n are odd and both integers, where m =4k1+1 & n = 4k2+3, and K, K1, k 2 are also integers too. Now m+n =(4k1+1) +(4k2+3) = 4k1+4k2+4 = 4[(k1+k2)+1], where the sum of k1+k2 is also an integer, K= k1+k2----> 4[K+1], where K+1 is odd. I proved the assumption of m, n being odd, because m,n are true, contradiction is false, however 4[K+1] = 4K+4 is always even, so this must infact be a contradiction, which the overall assume must fail. I'm confused.

1. Please provide a step-by-step proof, so I can fully understand you help

2. What was the 8n+1 used for, since I used 4k1+1 and 4k2+3?

3. Should I have wrote two cases or just the case above?

Thank you.

Explanation / Answer

Let m = any integer
2m + 1 = odd number
square of odd number = (2m + 1)^2 = 4m^2 + 4m + 1 = 4(m^2 + m) + 1

Because if m is odd then m^2 is also odd so (m^2 + m) is sum of two odd numbers = even number
But if m is even then m^2 is also even so (m^2 + m) is sum of two even numbers = even number

As (m^2 + m) is even so let (m^2 + m) = 2n [n is an integer].
Thus (2m+1)^2 = 8n + 1

Hope this clears your doubt

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.