Question

Insert one rational number between 3 and 4

Answer 1

Hint: If a and b are any two real numbers with a < b, then $a < \dfrac{a+b}{2} < b$. Also, the sum of two rational numbers is a rational number, and the ratio of two rational numbers is also a rational number. Hence if a and b are rational, then $\dfrac{a+b}{2}$ is also rational.

Complete step-by-step answer:
Rational Numbers: Numbers which can be expressed in the form of $\dfrac{p}{q}$ where “p” and “q” are integers and q is non-zero, are called rational numbers. Every integer n is a rational number since it can be expressed in the form $\dfrac{n}{1}$, where both n and 1 are integers, and 1 is non-zero. The sum of two rational numbers is always a rational number, and the ratio of two rational numbers is also a rational number where the denominator is non-zero.
We know that if a and b are any two real numbers with a < b, then $a < \dfrac{a+b}{2} < b$.
Hence if a = 3 and b = 4.
We have $3 < \dfrac{3+4}{2} < 4\Rightarrow 3 < \dfrac{7}{2} < 4$
Hence 3 < 3.5 < 4.
Hence a rational number between 3 and 4 is 3.5.

Note: We can insert n rational numbers between a and b by making use of properties of arithmetic mean.
If we insert n A.Ms between a and b, then $a,{{A}_{1}},{{A}_{2}},...,{{A}_{n}},b$ form an A.P.
Since a < b, we have d > 0. So $a < {{A}_{1}} < {{A}_{2}} < \cdots < {{A}_{n}} < b$
Also, $b=a+\left( n+1 \right)d$
Hence $d=\dfrac{b-a}{n+1}$
Hence the n rational numbers between two rational numbers a and b are
$a+\dfrac{b-a}{n+1},a+2\dfrac{b-a}{n+1},\cdots ,a+n\dfrac{b-a}{n+1}$
Consider the case of inserting one rational number between 3 and 4.
We have a = 3, b=4 and n =1.
Hene the number is $3+\dfrac{4-3}{1+1}=3+\dfrac{1}{2}=3.5$
Hence a rational number between 3 and 4 is 3.5