show any positive integer is of the form 3q and 3q +1 or 3q+2 for some integer q
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Answered by
1
Hey here is your answer
let x be the integer
x=3q
x^2=9q^2
x^2=3 (3q^2)
x=3q (let 3q^2 be q)
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x=3q+1
x^2=(3q+1)^2
x^2=9q^2+6q+1
x^2=3 (3q^2+2q)+1
x^2=3q+1 (let 3q^2+2q be q)
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x =3q+2
x^2=(3q+2)^2
x^2=9q^2+ 12q+4
x^2=3 (3q^2+4q+1)+1
x^2=3q+1 (let 3q^2+4q+1 be q)
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it proves q^2 in the form of 3q and 3q+1 but not in the form of 3q+2
let x be the integer
x=3q
x^2=9q^2
x^2=3 (3q^2)
x=3q (let 3q^2 be q)
---------------------
---------------------
x=3q+1
x^2=(3q+1)^2
x^2=9q^2+6q+1
x^2=3 (3q^2+2q)+1
x^2=3q+1 (let 3q^2+2q be q)
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x =3q+2
x^2=(3q+2)^2
x^2=9q^2+ 12q+4
x^2=3 (3q^2+4q+1)+1
x^2=3q+1 (let 3q^2+4q+1 be q)
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it proves q^2 in the form of 3q and 3q+1 but not in the form of 3q+2
lorengray:
why are you squaring it.???
Answered by
2
let,n be the positive integre on dividing by 3 q is the quotient and r will be the remainder .by euclid division lema
n=3q+r , where r=0,1,2
case 1 when r=0
n=3q+r
n=3q+0=3q
case2 when r=1
n=3q+r
n=3q+1
case 3 when r=2
n=3q+r
n=3q+2
n=3q+r , where r=0,1,2
case 1 when r=0
n=3q+r
n=3q+0=3q
case2 when r=1
n=3q+r
n=3q+1
case 3 when r=2
n=3q+r
n=3q+2
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