Advertisement Remove all ads

Advertisement Remove all ads

Advertisement Remove all ads

Sum

** **Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

Advertisement Remove all ads

#### Solution

Let **a** is **b** be two positive integers in which **a** is greater than **b**.

According to Euclid's division algorithm; **a** and **b** can be expressed as a = bq + r, where q is quotient and r is remainder and 0 ≤ r < b.

Taking b = 4, we get: a = 4q + r, where 0 ≤ r < 4 i.e., r = 0, 1, 2 or 3

**r = 0** ⇒ a = 4q, which is divisible by 2 and so is **even**.

**r = 1** ⇒ a = 4q + 1, which is not divisible by 2 and so is **odd**.

**r = 2** ⇒ q = 4q + 2, which is divisible by 2 and so is **even**. and

**r = 3** Þ q = 4q + 3, which is not divisible by 2 and so is **odd**.

**Any positive odd integer is of the form**

**4q + 1 or 4q + 3; **where q is an integer.

**Hence the required result.**

Concept: Euclid’s Division Lemma

Is there an error in this question or solution?