
The square of a proper fraction is \[\_\_\_\_\_\_\_\_\] than the given fraction.
A smaller
B greater
C equal
D none
Answer
560.4k+ views
Hint: The given question is about fractions we had asked that the square of a proper fraction is smaller, greater, or equal to the given fraction. We will prove this sum by taking examples of proper fraction then we will square that proper fraction and then we will check that the obtained square of a proper fraction is smaller, greater, or equal to the given fraction.
Complete step-by-step answer:
In the given question, we are asked that the square of an improper fraction is smaller, greater, or equal to the given fraction which means when we square the proper fraction the resultant fraction that obtained is smaller, greater, or equal to the fraction that we took.
Firstly we will square the proper fraction whereas the proper fraction is nothing but that fraction in which the numerator is smaller than the denominator whereas in an improper fraction the numerator is greater than the denominator.
According to the given question, we have to check that the squaring of the proper fraction is smaller, greater, or equal to the fraction that is taken. Let us check this by taking the example of a proper fraction. Let the proper fraction be \[\dfrac{1}{2}\].
Now squaring of proper fraction means we have to square \[\dfrac{1}{2}\]which means \[{\left( {\dfrac{1}{2}} \right)^2} = \dfrac{{{{\left( 1 \right)}^2}}}{{{{\left( 2 \right)}^2}}} = \dfrac{1}{4}\]
It means proper fraction was \[\dfrac{1}{2}\]and squaring of proper fraction is \[\dfrac{1}{4}\]
Therefore square of proper fraction is smaller than the proper fraction which means
\[{\left( {\dfrac{1}{2}} \right)^2} < \dfrac{1}{2}\]
Therefore \[\dfrac{1}{4} < \dfrac{1}{2}\] It means the square of the proper fraction is smaller than the given fraction.
Let us check it by taking another example of a proper fraction. Now let the proper fraction be \[\dfrac{2}{3}\].
Therefore \[{\left( {\dfrac{2}{3}} \right)^2} < \dfrac{2}{3}\]
On squaring, \[\dfrac{4}{9} < \dfrac{2}{3}\]
On dividing, we get \[0.44 < 0.67\]
This means for every fraction, the square of the proper fraction is smaller than the given fraction.
Note: Let us consider an improper fraction where the numerator is greater than the denominator is \[\dfrac{3}{2}\]. Now we will check the same condition for improper fraction which means the square of the improper fraction is smaller, greater, or equal to the given fraction. So
\[ \Rightarrow {\left( {\dfrac{3}{2}} \right)^2} > \dfrac{3}{2}\]
\[ \Rightarrow \dfrac{9}{4} > \dfrac{3}{2}\]
On dividing, we get
\[2.25{\text{ }} > {\text{ }}1.5\]
This means the square of the improper fraction is greater than the given fraction.
Complete step-by-step answer:
In the given question, we are asked that the square of an improper fraction is smaller, greater, or equal to the given fraction which means when we square the proper fraction the resultant fraction that obtained is smaller, greater, or equal to the fraction that we took.
Firstly we will square the proper fraction whereas the proper fraction is nothing but that fraction in which the numerator is smaller than the denominator whereas in an improper fraction the numerator is greater than the denominator.
According to the given question, we have to check that the squaring of the proper fraction is smaller, greater, or equal to the fraction that is taken. Let us check this by taking the example of a proper fraction. Let the proper fraction be \[\dfrac{1}{2}\].
Now squaring of proper fraction means we have to square \[\dfrac{1}{2}\]which means \[{\left( {\dfrac{1}{2}} \right)^2} = \dfrac{{{{\left( 1 \right)}^2}}}{{{{\left( 2 \right)}^2}}} = \dfrac{1}{4}\]
It means proper fraction was \[\dfrac{1}{2}\]and squaring of proper fraction is \[\dfrac{1}{4}\]
Therefore square of proper fraction is smaller than the proper fraction which means
\[{\left( {\dfrac{1}{2}} \right)^2} < \dfrac{1}{2}\]
Therefore \[\dfrac{1}{4} < \dfrac{1}{2}\] It means the square of the proper fraction is smaller than the given fraction.
Let us check it by taking another example of a proper fraction. Now let the proper fraction be \[\dfrac{2}{3}\].
Therefore \[{\left( {\dfrac{2}{3}} \right)^2} < \dfrac{2}{3}\]
On squaring, \[\dfrac{4}{9} < \dfrac{2}{3}\]
On dividing, we get \[0.44 < 0.67\]
This means for every fraction, the square of the proper fraction is smaller than the given fraction.
Note: Let us consider an improper fraction where the numerator is greater than the denominator is \[\dfrac{3}{2}\]. Now we will check the same condition for improper fraction which means the square of the improper fraction is smaller, greater, or equal to the given fraction. So
\[ \Rightarrow {\left( {\dfrac{3}{2}} \right)^2} > \dfrac{3}{2}\]
\[ \Rightarrow \dfrac{9}{4} > \dfrac{3}{2}\]
On dividing, we get
\[2.25{\text{ }} > {\text{ }}1.5\]
This means the square of the improper fraction is greater than the given fraction.
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