Question :
_________

➡️Use Euclid’s division lemma to show that the square of any positive integer is either of th

Question

Question :
_________

➡️Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
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Alice 2 months 2021-07-30T04:48:55+00:00 2 Answers 0 views 0

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    0
    2021-07-30T04:50:14+00:00

    Answer:

    {\huge{\blue{\texttt{\orange A\red N\green S\pink W\blue E\purple R\red}}}}

    To prove

    • The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

    Proof

    Let us consider a positive integer ‘a’

    • Divide the positive integer a by 3, and let r be the reminder and b be the quotient such that
    • a = 3b + r……………………………(1)
    • where r = 0,1,2,3…..

    Case 1: Consider r = 0

    • Equation (1) becomes
    • a = 3b

    On squaring both the side

    • a^2 = (3b)^2
    • a^2 = 9b^2
    • a^2 = 3 × 3b^2
    • a^2 = 3m
    • Where m = 3b^2

    Case 2: Let r = 1

    • Equation (1) becomes
    • a = 3b + 1

    Squaring on both the side we get

    • a^2 = (3b + 1)^2
    • a^2 = (3b)^2 + 1 + (2 × (3b) × 1)
    • a^2 = 9b^2 + 6b + 1
    • a^2 = 3(3b^2 + 2b) + 1
    • a^2 = 3m + 1
    • Where m = 3b^2 + 2b

    Case 3: Let r = 2

    • Equation (1) becomes
    • a = 3b + 2

    Squaring on both the sides we get

    • a^2 = (3b + 2)^2
    • a^2 = 9b^2 + 4 + (2 × 3b × 2)
    • a^2 = 9b^2 + 12b + 4
    • a^2 = 9b^2 + 12b + 3 + 1
    • a^2 = 3(3b^2 + 4b + 1) + 1
    • a^2 = 3m + 1
    • where m = 3b^2 + 4b + 1

    ∴ square of any positive integer is of the form 3m or 3m+1.

    Hence proved.

    Done ✅

    0
    2021-07-30T04:50:37+00:00

    Answer:

    tu Deepali h n mujhe baat nhi krni terese

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