Find 12 rational numbers between $ -1 $ and 2.
Answer
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Hint: We will write the given numbers in their equivalent fractions. For this, we will select a number and multiply and divide it to the given numbers. After that, we will look at the numbers that lie between the equivalent forms of the given numbers. We will make a list of 12 such numbers and conclude that these numbers lie between the given numbers.
Complete step by step answer:
The given numbers are $ -1 $ and 2. We can write these numbers as $ \dfrac{-1}{1} $ and $ \dfrac{2}{1} $ . Let us choose a number, say 5. Now we will multiply and divide the given numbers by number 5. So, we get the following equivalent fractions,
$ -1=\dfrac{-1}{1}=\dfrac{-1}{1}\times \dfrac{5}{5}=\dfrac{-5}{5} $
$ 2=\dfrac{2}{1}=\dfrac{2}{1}\times \dfrac{5}{5}=\dfrac{10}{5} $
Now, we have to find 12 rational numbers between $ \dfrac{-5}{5} $ and $ \dfrac{10}{5} $ . Rational number is a number of the form $ \dfrac{p}{q} $ where $ p $ and $ q $ are integers and $ q $ is not equal to 0. So, we can find 12 fractions that lie between $ \dfrac{-5}{5} $ and $ \dfrac{10}{5} $ . We know that the integers $ -4,-3,-2,-1,1,2,3,4,5,6,7,8,9 $ are greater than $ -5 $ and smaller than 10. We will select 12 numbers from this list and divide the numbers by 5 to obtain 12 rational numbers between $ \dfrac{-5}{5} $ and $ \dfrac{10}{5} $ . So, we have the fractions $ \dfrac{-3}{5},\dfrac{-2}{5},\dfrac{-1}{5},\dfrac{1}{5},\dfrac{2}{5},\dfrac{3}{5},\dfrac{4}{5},\dfrac{5}{5},\dfrac{6}{5},\dfrac{7}{5},\dfrac{8}{5},\dfrac{9}{5} $ that lie between $ \dfrac{-5}{5} $ and $ \dfrac{10}{5} $ . All the fractions in this list are of the form $ \dfrac{p}{q} $ where $ p $ and $ q $ are integers and $ q $ is not equal to 0. Therefore, we have found 12 rational numbers between $ -1 $ and 2.
Note:
It is important that we understand the definition of rational numbers for such type of questions. The key concept in this question is the concept of equivalent fractions. We can choose any integer to multiply and divide the given number and get an equivalent fraction. We chose 5 for convenience and no other reason. We can find different 12 rational numbers by changing this integer to 10 or 20 or any other integer with sufficient range.
Complete step by step answer:
The given numbers are $ -1 $ and 2. We can write these numbers as $ \dfrac{-1}{1} $ and $ \dfrac{2}{1} $ . Let us choose a number, say 5. Now we will multiply and divide the given numbers by number 5. So, we get the following equivalent fractions,
$ -1=\dfrac{-1}{1}=\dfrac{-1}{1}\times \dfrac{5}{5}=\dfrac{-5}{5} $
$ 2=\dfrac{2}{1}=\dfrac{2}{1}\times \dfrac{5}{5}=\dfrac{10}{5} $
Now, we have to find 12 rational numbers between $ \dfrac{-5}{5} $ and $ \dfrac{10}{5} $ . Rational number is a number of the form $ \dfrac{p}{q} $ where $ p $ and $ q $ are integers and $ q $ is not equal to 0. So, we can find 12 fractions that lie between $ \dfrac{-5}{5} $ and $ \dfrac{10}{5} $ . We know that the integers $ -4,-3,-2,-1,1,2,3,4,5,6,7,8,9 $ are greater than $ -5 $ and smaller than 10. We will select 12 numbers from this list and divide the numbers by 5 to obtain 12 rational numbers between $ \dfrac{-5}{5} $ and $ \dfrac{10}{5} $ . So, we have the fractions $ \dfrac{-3}{5},\dfrac{-2}{5},\dfrac{-1}{5},\dfrac{1}{5},\dfrac{2}{5},\dfrac{3}{5},\dfrac{4}{5},\dfrac{5}{5},\dfrac{6}{5},\dfrac{7}{5},\dfrac{8}{5},\dfrac{9}{5} $ that lie between $ \dfrac{-5}{5} $ and $ \dfrac{10}{5} $ . All the fractions in this list are of the form $ \dfrac{p}{q} $ where $ p $ and $ q $ are integers and $ q $ is not equal to 0. Therefore, we have found 12 rational numbers between $ -1 $ and 2.
Note:
It is important that we understand the definition of rational numbers for such type of questions. The key concept in this question is the concept of equivalent fractions. We can choose any integer to multiply and divide the given number and get an equivalent fraction. We chose 5 for convenience and no other reason. We can find different 12 rational numbers by changing this integer to 10 or 20 or any other integer with sufficient range.
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