# Ex.1.1 Q4 Real Numbers Solution - NCERT Maths Class 10

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## Question

Use Euclid’s division lemma to show that the square of any positive integer is either of the form $$3\,\rm{m}$$ or $$3\,\rm{m} + 1$$ for some integer $$\rm{m.}$$

[Hint: Let $$x$$ be any positive integer then it is of the form $$3q,\, 3q + 1$$ or $$3q + 2.$$ Now square each of these and show that they can be rewritten in the form $$3\,\rm{m}$$ or $$3\,\rm{m} + 1.$$]

Video Solution
Real Numbers
Ex 1.1 | Question 4

## Text Solution

To Prove:

The square of any positive integer is either of the form $$3\,\rm{m}$$ or $$3\,\rm{m} + 1$$ for some integer $$\rm{m}$$ (using the Euclid’s division lemma).

Reasoning:

Suppose that there is a positive integer ‘$$a$$’. By Euclid’s lemma, we know that for positive integers $$a$$ and $$b,$$ there exist unique integers $$q$$ and $$r,$$ such that $$a=bq+r,\,0\le r\lt b$$

If we keep the value of $$b = 3,$$ then $$0 ≤ r < 3$$ i.e. $$r = 0$$ or $$1$$ or $$2$$ but it can’t be $$3$$ because $$r$$ is smaller than $$3.$$ So, the possible values for $$a = 3q$$ or $$3q + 1$$ or $$3q + 2.$$ Now, find the square of all the possible values of $$a.$$ If $$q$$ is any positive integer then its square (let’s call it as “$$\rm{m}$$”) will also be a positive integer. Now, observe carefully that the square of all the positive integers is either of the form $$3\,\rm{m}$$ or $$3\,\rm{m} + 1$$ for some integer $$\rm{m.}$$

Steps:

Let $${“a”}$$be any positive integer and $$b=3.$$

Then, $$a=3q+r$$ for some integer $$q\ge 0$$ and $$r = 0, 1 , 2$$ because $$0\le r <3.$$

Therefore,

\begin{align} a&=3q \;\text{ or }\;3q+1\;\text{ or }\;3q+2\text{ or}\;\\ {{a}^{2}}&={ }{{\left( 3q \right)}^{2}}\;\text{ or }\;{{\left( 3q{ }+{ }1 \right)}^{2}}\;\text{or }\;{{\left( 3q{ }+{ }2 \right)}^{2}} \\ {{a}^{2}}&={ }3{{(3q)}^{2}}\;\text {or }\;(9{{q}^{2}}+6q+{ }1{ })\;\text{or }\;(9{{q}^{2}}+{ }12q{ }+{{4}^{{}}}) \\{{a}^{2}}&={ }3{ }\left( 3{{q}^{2}} \right)\;\text{or }3\left( 3{{q}^{2}}+2q \right)+{ }1\\&\qquad \text{ or}\;3\left( 3{{q}^{2}}+{ }4q{ }+1 \right){ }+1 \\&=\rm{m}\; \text{or} \;3\rm{m}+1 \\\end{align}

Where $$\rm{m}$$ is any positive integer.

Hence it can be said that the square of any positive integer is either of the form \begin{align}{3\,\rm{m} \;\rm{or}\; 3\,\rm{m} + 1.} \end{align}

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