Question

How do you find two fractions between $2$ and $3$?

Answer 1

Hint: In the question above, we have two numbers $2$ and $3$, and we are supposed two find two fractions between them. We can do so using the mathematical formula that exists for it. One of the following formulas requires two rational numbers. So, you have two rational numbers, $x$ and $y$, the number $\dfrac{{x + y}}{2}$ is rational and satisfies $x < \dfrac{{x + y}}{2} < y$

Complete step-by-step solution:
We have two numbers $2$ and $3$, and we have to find their two fractions that exist between them. For finding these fractions, we will have to start by substituting the numbers in the existing formulas, after satisfying the required condition.
We know that the numbers 2 and 3 are rational,
 So, automatically the number
\[ \Rightarrow \dfrac{{2 + 3}}{2} = \dfrac{5}{2}\]
Should also be rational and between 2 and 3.
But again, the number
\[ \Rightarrow \dfrac{{2 + \left( {\dfrac{5}{2}} \right)}}{2}\;\]
Solving the above calculation with basic methods,
\[ \Rightarrow \dfrac{{\dfrac{{4 + 5}}{2}}}{2}\]
Adding the numerators, in the numerator, we get,
\[ \Rightarrow \dfrac{{\dfrac{9}{2}}}{2}\]
Shifting the denominator to the denominator and multiplying, we get
$ \Rightarrow \dfrac{9}{4}$
Which is between $2$and $\dfrac{5}{2}$,
So now, we have
\[ \Rightarrow 2 < \dfrac{9}{4} < \dfrac{5}{2} < 3\],
And the two fractions that we find between the two numbers are,
$ \Rightarrow \dfrac{9}{4}$ and $\dfrac{5}{2}$

So, for any two numbers $2$ and $3$, the fractional number between them will be $\dfrac{9}{4}$ and $\dfrac{5}{2}$.

Note: A fraction is a value in two parts; each part, the numerator or denominator, is an integer. The numerator is the fraction's top number, while the denominator is its bottom number. Lower-order fractional math like addition and subtraction requires that the denominators of the involved fractions be the same value. When finding a fraction that comes between two others, you ignore normal fractional math in favour of a simpler method.