# Ex.1.1 Q2 Real Numbers Solution - NCERT Maths Class 10

## Question

Show that any positive odd integer is of the form \(6q+1\) or \(6q+3,\) or \(6q+5,\) where *\(q\)* is some integer.

## Text Solution

**What is known?**

Some integer \(q.\)

**What is unknown?**

That any positive odd integer is of the form \(6q+1,\) or \(6q+3,\) or \(6q+5,\) where *\(q\) *is some integer.

**Reasoning:**

To solve this question, first think about the Euclid’s division algorithm. Suppose there is any positive integer '\(a\)' and it is of the form \(6q+ r\). where \(q\) is some integer. This means that \(0 \le r < 6\) i.e. \(r=0\) or \(1\) or \(2\) or \(3\) or \(4\) or \(5\) but it can’t be \(6 \) because \(r\) is smaller than \(6.\). So, by Euclid’s division lemma possible values for '\(a\)' can be \(6q\) or \(6q+1\)or \(6q+2\) or \(6q+3\) or \( 6q+4\) or \(6q+5.\)

**Steps:**

Let '\(a\)' be any positive integer. Then according to Euclid’s algorithm,

\(a = 6q + r\) for some integer \(q \ge 0\)

and \(r= 0,1,2,3,4,5\) because \(0\le r < 6\)

Therefore,

\(a = 6q + 0\) or \(6q + 1\) or \(6q + 2\) or \(6q + 3\) or \(6q + 4\) or \(6q + 5.\)

Now, \(6q + 1 = 2 \times 3q + 1 = 2 {k_1}+1\)

(where \(k_1\) is a positive integer)

\(\begin{align}6q+3 &= 6q+2 +1\\&= 2 (3q + 1)+1 \\&=2{k_2}+1\end{align}\)

(where \(k_2\) is a positive integer)

\(\begin{align}6q+5&=6q+4+1\\&=2 (3q+2)+1\\&=2{k_3}+1\end{align}\)

(where \(k_3\) is a positive integer)

Clearly, \(6q+1,6q+3\) and \(6q+5\) are of the form \(2k+1,\) where \(k\) is an integer.

Therefore, \(6q+1, 6q+3\) and \(6q+5\) are not exactly divisible by \(2.\)

Hence, these expressions of numbers are odd numbers and therefore any odd integers can be expressed in the form \(6q+1\) or \(6q+3\) or \(6q+5.\)