
Find five rational numbers that were present between 2 and 3 by mean method?
Answer
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Hint: We start solving the problem by recalling the fact that the mean of the two numbers a and b is defined as $ \dfrac{a+b}{2} $ and mean of any two numbers lies between them. We then find the mean of the given numbers 2 and 3 (say m). We then find the mean of 2 and m (say $ {{m}_{1}} $ ), also mean of m and 3 (say $ {{m}_{2}} $ ). We then find the mean of 2 and $ {{m}_{1}} $ , also mean of $ {{m}_{2}} $ and 3 to get the required answer.
Complete step by step answer:
According to the problem, we are asked to find five rational numbers that were present between 2 and 3 using the mean methods.
We know that the mean of the two numbers a and b is defined as $ \dfrac{a+b}{2} $ and we know that the mean of any two numbers lies between them.
Let us find the mean of the given numbers 2 and 3.
So, we get one of the rational numbers that lies between 2 and 3 as $ \dfrac{2+3}{2}=\dfrac{5}{2}=2.5 $ ---(1).
Now, let us find the mean of the numbers 2 and 2.5.
So, we get one of the rational numbers that lies between 2 and 3 as $ \dfrac{2+2.5}{2}=\dfrac{4.5}{2}=2.25 $ ---(2).
Now, let us find the mean of the numbers 2.5 and 3.
So, we get one of the rational numbers that lies between 2 and 3 as $ \dfrac{2.5+3}{2}=\dfrac{5.5}{2}=2.75 $ ---(3).
Now, let us find the mean of the numbers 2 and 2.25.
So, we get one of the rational numbers that lies between 2 and 3 as $ \dfrac{2+2.25}{2}=\dfrac{4.25}{2}=2.125 $ ---(4).
Now, let us find the mean of the numbers 2.75 and 3.
So, we get one of the rational numbers that lies between 2.75 and 3 as $ \dfrac{2.75+3}{2}=\dfrac{5.75}{2}=2.875 $ ---(5).
From equations (1), (2), (3), (4), and (5), we have found five rational numbers that were present between 2 and 3 as 2.125, 2.25, 2.5, 2.75, 2.825.
$\therefore$ The five rational numbers that were present between 2 and 3 as 2.125, 2.25, 2.5, 2.75, 2.825.
Note:
We can also find the rational numbers between 2 and 3 in the alternative method as shown below:
We have given numbers 2 and 3 which can be written in rational form as $ \dfrac{12}{6} $ and $ \dfrac{18}{6} $ .
We can see that the rational numbers $ \dfrac{13}{6} $ , $ \dfrac{14}{6} $ , $ \dfrac{15}{6} $ , $ \dfrac{16}{6} $ , $ \dfrac{17}{6} $ are present between these which is the required answer.
Complete step by step answer:
According to the problem, we are asked to find five rational numbers that were present between 2 and 3 using the mean methods.
We know that the mean of the two numbers a and b is defined as $ \dfrac{a+b}{2} $ and we know that the mean of any two numbers lies between them.
Let us find the mean of the given numbers 2 and 3.
So, we get one of the rational numbers that lies between 2 and 3 as $ \dfrac{2+3}{2}=\dfrac{5}{2}=2.5 $ ---(1).
Now, let us find the mean of the numbers 2 and 2.5.
So, we get one of the rational numbers that lies between 2 and 3 as $ \dfrac{2+2.5}{2}=\dfrac{4.5}{2}=2.25 $ ---(2).
Now, let us find the mean of the numbers 2.5 and 3.
So, we get one of the rational numbers that lies between 2 and 3 as $ \dfrac{2.5+3}{2}=\dfrac{5.5}{2}=2.75 $ ---(3).
Now, let us find the mean of the numbers 2 and 2.25.
So, we get one of the rational numbers that lies between 2 and 3 as $ \dfrac{2+2.25}{2}=\dfrac{4.25}{2}=2.125 $ ---(4).
Now, let us find the mean of the numbers 2.75 and 3.
So, we get one of the rational numbers that lies between 2.75 and 3 as $ \dfrac{2.75+3}{2}=\dfrac{5.75}{2}=2.875 $ ---(5).
From equations (1), (2), (3), (4), and (5), we have found five rational numbers that were present between 2 and 3 as 2.125, 2.25, 2.5, 2.75, 2.825.
$\therefore$ The five rational numbers that were present between 2 and 3 as 2.125, 2.25, 2.5, 2.75, 2.825.
Note:
We can also find the rational numbers between 2 and 3 in the alternative method as shown below:
We have given numbers 2 and 3 which can be written in rational form as $ \dfrac{12}{6} $ and $ \dfrac{18}{6} $ .
We can see that the rational numbers $ \dfrac{13}{6} $ , $ \dfrac{14}{6} $ , $ \dfrac{15}{6} $ , $ \dfrac{16}{6} $ , $ \dfrac{17}{6} $ are present between these which is the required answer.
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