Math, asked by RohitRSB, 1 year ago

use Euclid division lemma to show that the square of any positive integer is either of the form 3m or 3m +1 for some integer m​

Answers

Answered by pushpakala086
2

As per Euclid's Division Lemma

If a & b are 2 positive integers, then

a = bq + r

where \: 0 \leqslant r < b

Let positive integer be a

And b = 3

hence \: a = 3q + r

where(0 \leqslant r < 3)

r is an integer greater than or equal to 0 and less than 3

hence, r can be either 0, 1 or 2.

CASE 1:-

if \: r = 0 \\ our \: equation \: becomes \\ a = 3q + r \\ a = 3q + 0 \\ a = 3q \\ squaring \: both \: the \: sides \\ {a}^{2} = {(3q)}^{2} \\ {a}^{2} = 9 {q}^{2} \\ {a}^{2} = 3(3 {q}^{2} ) \\ {a}^{2} = 3m \\ where \: m = 3 {q}^{2}  

CASE 2:-

if \: r = 1 \\ our \: equation \: becomes \\ a = 3q + r \\ a = 3q + 1 \\ squaring \: both \: the \: sides \\ {a}^{2} = {(3q + 1)}^{2} \\ {a}^{2} = {(3q)}^{2} + {1}^{2} + 2(3q) \\ {a}^{2} = 9 {q}^{2} + 6q + 1 \\ {a}^{2} = 3 ({3q}^{2} + 2q) + 1 \\ {a }^{2} = 3m + 1 \\ where \: m \: = 3 {q}^{2} + 2q

CASE 3:-

if \: r \: = 2 \\ our \: equation \: becomes \\ a = 3q + r \\ a = 3q + 2 \\ squaring \: both \: the \: sides \\ {a}^{2} = {(3q + 2)}^{2} \\ {a}^{2} = {(3q)}^{2} + {2}^{2} + 2(2)(3q) \\ {a}^{2} = 9 {q}^{2} + 12q + 4 \\ {a}^{2} = 9 {q}^{2} + 12q + 3 + 1 \\ {a}^{2} = 3(3 {q}^{2} + 4q + 1) + 1 \\ {a}^{2} = 3m + 1 \\ where \: m = 3 {q}^{2} + 4q + 1

Hence, square of any positive number can be expressed of the form 3m or 3m + 1  

HENCE PROVED

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Answered by ItsMysteriousGirl
6

\huge\bf\blue {Hey\:Mate!!!}

As per Euclid's Division Lemma ,

if a and b are 2 positive integers then,a = bq + r \\ where \: 0 \leqslant r &lt; b

Let positive integer be a and b = 3.

hence \: a =   3q + r \\ (where \: 0 \leqslant r &lt; 3)

r is an integer greater than or equal to 0 and less than 3.

Hence ,r can be either 0, 1 or 2.

________________________________

Case - 1

if \: r = 0 \\ our \: equation \: becomes \\ a = 3q + r \\ a = 3q + 0 \\ a = 3q \\ squaring \: both \: the \: sides \\  {a}^{2}  = {(3q)}^{2}  \\  {a}^{2}  =  {9q}^{2}  \\ a = 3(3 {q}^{2} ) \\  {a}^{2}  = 3m \\ where \: m  = 3 {q}^{2}

________________________________

Case - 2

if \: r = 1 \\ our \: equation \: becomes \\ a = 3q + r \\ a = 3q + 1 \\ squaring \: both \: the \: sides \\  {a}^{2}  = {(3q + 1)}^{2}  \\  {a}^{2}  =  {3q}^{2} +  {1}^{2} + 2(3q) \\  {a}^{2}  = 9 {q}^{2}  + 6q + 1\\  {a}^{2}  = 3(3 {q}^{2} + 2q) + 1  \\  {a}^{2} =3m + 1 \\   where \: m  = 3(3 {q}^{2}   + 2q)

________________________________

Case - 3

if \: r = 2\\ our \: equation \: becomes \\ a = 3q + r \\ a = 3q + 2\\ squaring \: both \: the \: sides \\  {a}^{2}  = {(3q + 2)}^{2}  \\  {a}^{2}  =  {(3q)}^{2} +  {2}^{2} + 2(2)(3q) \\  {a}^{2}  = 9 {q}^{2}  + 12q + 4\\  {a}^{2}  = 9 {q}^{2} + 12q + 3 + 1   \\   {a}^{2}   = 3(3 {q}^{2} +  4q + 1) + 1\\ {a}^{2} =3m + 1 \\   where \: m  = 3 {q}^{2}   + 4q + 1

Square of any positive integer can be expressed in the form of 3m or 3m + 1.

Hence Proved

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