Question

Proving that every positive integer is of the form $3\mathrm{q}$ or $3\mathrm{q}+1$.

Answer 1

Step 1: Determine to prove that any positive integer is in the form of $3\mathrm{q},3\mathrm{q}+1,3\mathrm{q}+2$ if $\mathrm{n}=3\mathrm{q}$

We know that from Euclid’s division lemma for $\mathrm{b}=3$

Let us assume that any positive integer $‘\mathrm{n}’$ be of the form $3\mathrm{q}$ or, $3\mathrm{q}+1$ or $3\mathrm{q}+2$.

If $\mathrm{n}=3\mathrm{q}$,

On squaring we get,

$$\begin{gathered} {\mathrm{n}}^2={\left(3\mathrm{q}\right)}^2 \\ =9{\mathrm{q}}^2 \end{gathered}$$

${\mathrm{n}}^2=3\left(3{\mathrm{q}}^2\right)$

Here, ${\mathrm{n}}^2=3\mathrm{m}$, where $\mathrm{m}$ is some integer $[\mathrm{m}=3{\mathrm{q}}^2]$.

Step 2: Determine to prove that any positive integer is in the form of $3\mathrm{q},3\mathrm{q}+1,3\mathrm{q}+2$ if $\mathrm{n}=3\mathrm{q}+1$.

If $\mathrm{n}=3\mathrm{q}+1$,

On squaring we get,

$$\begin{gathered} {\mathrm{n}}^2={(3\mathrm{q}+1)}^2 \\ =9{\mathrm{q}}^2+6\mathrm{q}+1 \end{gathered}$$ [Solved using the identity ${(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}$]

${\mathrm{n}}^2=3(3{\mathrm{q}}^2+2\mathrm{q})+1$

Here, ${\mathrm{n}}^2=3\mathrm{m}+1$, where $\mathrm{m}$ is some integer $[\mathrm{m}=3{\mathrm{q}}^2+2\mathrm{q}]$.

Step 3: Determine to prove that any positive integer is in the form of $3\mathrm{q},3\mathrm{q}+1,3\mathrm{q}+2$ if $\mathrm{n}=3\mathrm{q}+2$.

If $\mathrm{n}=3\mathrm{q}+2$,

On squaring we get,

$$\begin{gathered} {\mathrm{n}}^2={(3\mathrm{q}+2)}^2 \\ =9{\mathrm{q}}^2+12\mathrm{q}+4 \end{gathered}$$ [Solved using the identity ${(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}$]

${\mathrm{n}}^2=3(3{\mathrm{q}}^2+4\mathrm{q}+1)+1$

Here, ${\mathrm{n}}^2=3\mathrm{m}$, where $\mathrm{m}$ is some integer $[\mathrm{m}=3{\mathrm{q}}^2+4\mathrm{q}+1]$.

Therefore, the square of any positive integer is of the form $3\mathrm{q}$ or $3\mathrm{q}+1$ but not of the form $3\mathrm{q}+2$.

Hence, it is proved.