Using Euclid's division lemma, show that the square of any positive
integer is either of the form 3m or (3m +1) for some integer m.
Answers
Let 'a' be any positive integer.
On dividing it by 3 , let 'q' be the quotient and 'r' be the remainder.
Such that ,
a = 3q + r , where r = 0 ,1 , 2
When, r = 0
∴ a = 3q
When, r = 1
∴ a = 3q + 1
When, r = 2
∴ a = 3q + 2
When , a = 3q
On squaring both the sides,
\begin{gathered} {a}^{2} = 9 {q}^{2} \\ {a}^{2} = 3 \times (3 {q}^{2} ) \\ {a}^{2} = 3 \\ where \: m = 3 {q}^{2} \end{gathered}
a
2
=9q
2
a
2
=3×(3q
2
)
a
2
=3
wherem=3q
2
When, a = 3q + 1
On squaring both the sides ,
\begin{gathered} {a}^{2} = (3q + 1)^{2} \\ {a}^{2} = 9 {q}^{2} + 2 \times 3q \times 1 + {1}^{2} \\ {a}^{2} = 9 {q}^{2} + 6q + 1 \\ {a}^{2} = 3(3 {q}^{2} + 2q) + 1 \\ {a}^{2} = 3m + 1 \\ where \: m \: = 3 {q}^{2} + 2q\end{gathered}
a
2
=(3q+1)
2
a
2
=9q
2
+2×3q×1+1
2
a
2
=9q
2
+6q+1
a
2
=3(3q
2
+2q)+1
a
2
=3m+1
wherem=3q
2
+2q
When, a = 3q + 2
On squaring both the sides,
\begin{gathered} {a}^{2} = (3q + 2)^{2} \\ {a}^{2} = 3 {q}^{2} + 2 \times 3q \times 2 + {2}^{2} \\ {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\end{gathered}
a
2
=(3q+2)
2
a
2
=3q
2
+2×3q×2+2
2
a
2
=9q
2
+12q+4
a
2
=(9q
2
+12q+3)+1
a
2
=3(3q
2
+4q+1)+1
a
2
=3m+1
wherem=3q
2
+4q+1
Therefore , the square of any positive integer is either of the form 3m or 3m+1.
THIS IS THE BEST EVER ANSWER OF YOUR QUESTION
Done by my ownself in 30 minutes only for you
Answer:
It is the correct answer.
Step-by-step explanation:
Hope this attachment helps you.
