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Hint: We recall the definition of rational, irrational number and surds. We observe that every irrational number cannot be expressed in the form of $\sqrt[p]{n}={{n}^{\dfrac{1}{p}}}$ where $n$ is any rational number and $p$ is a non-zero rational number. \[\]
Complete step by step answer:
We know that we call a number ration when the number can be expressed in the form of $\dfrac{p}{q}$ where $p$ is any integer and $q$ is a non-zero integer. The rational numbers are solutions of linear equations. For example $-2,0.25,\dfrac{2}{4}$ are rational numbers since we can express them in the form of $\dfrac{p}{q}$ as $\dfrac{-2}{1},\dfrac{1}{4},\dfrac{1}{2}$.
We call a number irrational when cannot express that number in the form $\dfrac{p}{q}$ for example $\sqrt{2},\sqrt[3]{4}$. \[\]
We know that we call a number surd when we can express number in the form the in the form of $\sqrt[p]{n}={{n}^{\dfrac{1}{p}}}$ where $n$ is positive rational number such that $n\ne {{x}^{p}}$ for any rational number $x$ and $p$ is a non-zero rational number. For example $\sqrt{2},\sqrt[3]{4},\sqrt[\dfrac{1}{3}]{\dfrac{1}{2}}$ are all surds. \[\]
We have numbers like $2+\sqrt{5},\dfrac{\sqrt[3]{4}}{3}$ are irrational numbers but they cannot be expressed in the form of $\sqrt[p]{n}={{n}^{\dfrac{1}{p}}}$ for any values of $n,p$. There are also transcendental numbers like $\pi ,e$ also cannot be expressed in the form of $\sqrt[p]{n}={{n}^{\dfrac{1}{p}}}$ and hence are not surds but they are irrationals. \[\]
This is why every surd is irrational, but every irrational need not be a surd. \[\]
Note: We note that the other name of surd is ${{n}^{\text{th}}}$ root since all surds are solutions to the polynomial equation ${{x}^{n}}=a$ where $a$ is any irrational number while those numbers which are solutions to polynomial equation like $a{{x}^{n}}+a{{x}^{n-1}}+....+{{a}_{0}}=0$ are called algebraic numbers and which cannot be solutions are called transcendental numbers.
Complete step by step answer:
We know that we call a number ration when the number can be expressed in the form of $\dfrac{p}{q}$ where $p$ is any integer and $q$ is a non-zero integer. The rational numbers are solutions of linear equations. For example $-2,0.25,\dfrac{2}{4}$ are rational numbers since we can express them in the form of $\dfrac{p}{q}$ as $\dfrac{-2}{1},\dfrac{1}{4},\dfrac{1}{2}$.
We call a number irrational when cannot express that number in the form $\dfrac{p}{q}$ for example $\sqrt{2},\sqrt[3]{4}$. \[\]
We know that we call a number surd when we can express number in the form the in the form of $\sqrt[p]{n}={{n}^{\dfrac{1}{p}}}$ where $n$ is positive rational number such that $n\ne {{x}^{p}}$ for any rational number $x$ and $p$ is a non-zero rational number. For example $\sqrt{2},\sqrt[3]{4},\sqrt[\dfrac{1}{3}]{\dfrac{1}{2}}$ are all surds. \[\]
We have numbers like $2+\sqrt{5},\dfrac{\sqrt[3]{4}}{3}$ are irrational numbers but they cannot be expressed in the form of $\sqrt[p]{n}={{n}^{\dfrac{1}{p}}}$ for any values of $n,p$. There are also transcendental numbers like $\pi ,e$ also cannot be expressed in the form of $\sqrt[p]{n}={{n}^{\dfrac{1}{p}}}$ and hence are not surds but they are irrationals. \[\]
This is why every surd is irrational, but every irrational need not be a surd. \[\]
Note: We note that the other name of surd is ${{n}^{\text{th}}}$ root since all surds are solutions to the polynomial equation ${{x}^{n}}=a$ where $a$ is any irrational number while those numbers which are solutions to polynomial equation like $a{{x}^{n}}+a{{x}^{n-1}}+....+{{a}_{0}}=0$ are called algebraic numbers and which cannot be solutions are called transcendental numbers.
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