Name: Rational Numbers Between Two Rational Numbers
Uploaded: 2019-07-19
Duration: 9 min 50 s
Description: In this video, we will learn that we can find infinite rational numbers between any two rational numbers. Interested to know more? Just click!

Question

Insert $16$ rational numbers between $2.1$ and $2.2$ .

Accepted Answer

Given,

Let’s take $a = 2.1$ , $b = 2.2$ and $n = 16$

$n =$ Numbers required to be find out.

So,

$d = \frac{b - a}{n + 1}$

$= \frac{2.2 - 2.1}{16}$

$= \frac{0.1}{17}$

$= \frac{1}{170}$

$= 0.005$

Thus, $16$ rational numbers are:
$(a + d), (a + 2 d), (a + 3 d), (a + 4 d), (a + 5 d), (a + 6 d), (a + 7 d), (a + 8 d), (a + 9 d), (a + 10 d), (a + 11 d), (a + 12 d), (a + 13 d), (a + 14 d), (a + 15 d)$ and $(a + 16 d)$ .

So, $(a + d) = (2.1 + 0.005) = 2.105$

$(a + 2 d) = [2.1 + (2 \times 0.005)] = 2.110$

$(a + 3 d) = [2.1 + (3 \times 0.005)] = 2.115$

$(a + 4 d) = [2.1 + (4 \times 0.005)] = 2.120$

$(a + 5 d) = [2.1 + (5 \times 0.005)] = 2.125$

$(a + 6 d) = [2.1 + (6 \times 0.005)] = 2.130$

$(a + 7 d) = [2.1 + (7 \times 0.005)] = 2 . 135$

$(a + 8 d) = [2.1 + (8 \times 0.005)] = 2.140$

$(a + 9 d) = [2.1 + (9 \times 0.005)] = 2.145$

$(a + 10 d) = [2.1 + (10 \times 0.005)] = 2.150$

$(a + 11 d) = [2.1 + (11 \times 0.005)] = 2.155$

$(a + 12 d) = [2.1 + (12 \times 0.005)] = 2.160$

$(a + 13 d) = [2.1 + (13 \times 0.005)] = 2.165$

$(a + 14 d) = [2.1 + (14 \times 0.005)] = 2.170$

$(a + 15 d) = [2.1 + (15 \times 0.005)] = 2.175$

$(a + 16 d) = [2.1 + (16 \times 0.005)] = 2.180$

Hence, the rational numbers between $2.1$ and $2.2$ are $2.105, 2.110, 2.115, 2.120, 2.125, 2.130, 2.135, 2.140, 2.145, 2.150, 2.155, 2.160, 2.165, 2.170, 2.175, 2.180$ .

Insert $16$ rational numbers between $2.1$ and $2.2$ .

Important Questions on Number Systems

Learn and Prepare for any exam you want !

Insert 16 rational numbers between 2.1 and 2.2.

Important Questions on Number Systems

Learn and Prepare for any exam you want !

Insert $16$ rational numbers between $2.1$ and $2.2$ .