Prove that $\dfrac{1}{\sqrt{2}}$ is an irrational number.
Answer
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Hint: Try to recall the definition of rational and irrational numbers. You need to use the method of contradiction to prove that $\dfrac{1}{\sqrt{2}}$ is an irrational number. Also, focus on the point that the square root of an integer which is not a perfect square, is irrational.
Complete step-by-step answer:
Before moving to the options, let us talk about the definitions of rational numbers followed by irrational numbers.
So, rational numbers are those real numbers that can be written in the form of $\dfrac{p}{q}$ such that both p and q are integers and $q\ne0$ . In other words, we can say that the numbers which are either terminating or recurring when converted to decimal form are called rational numbers. All the integers fall under this category.
Now, moving to irrational numbers.
Those real numbers which are non-terminating and non-recurring are termed as irrational numbers.
The roots of the numbers which are not perfect squares fall under the category of irrational numbers. $\pi \text{ and }e$ are also the standard examples of irrational numbers.
Now let us move to the solution to the above question.
We take $\dfrac{1}{\sqrt{2}}$ to be a rational number. Therefore, using the definition of rational number we can say that:
$\dfrac{1}{\sqrt{2}}=\dfrac{p}{q}$
Where p and q both are integers. Now if we take the reciprocal of both the sides of the equation, we get
$\sqrt{2}=\dfrac{q}{p}$
And we know that reciprocal of all rational numbers except 0 must be rational. But $\sqrt{2}$ is not a rational number. Hence, it contradicts our assumption that $\dfrac{1}{\sqrt{2}}$ is a rational number. So, $\dfrac{1}{\sqrt{2}}$ is irrational.
Note: Be careful with the rational number zero, as if you use the above method for zero, you would end up getting 0 to be an irrational number, but in actuality 0 is rational. Also, you can remember a result that the root of the multiplicative inverse of an integer is irrational, provided the integer is not a perfect square.
Complete step-by-step answer:
Before moving to the options, let us talk about the definitions of rational numbers followed by irrational numbers.
So, rational numbers are those real numbers that can be written in the form of $\dfrac{p}{q}$ such that both p and q are integers and $q\ne0$ . In other words, we can say that the numbers which are either terminating or recurring when converted to decimal form are called rational numbers. All the integers fall under this category.
Now, moving to irrational numbers.
Those real numbers which are non-terminating and non-recurring are termed as irrational numbers.
The roots of the numbers which are not perfect squares fall under the category of irrational numbers. $\pi \text{ and }e$ are also the standard examples of irrational numbers.
Now let us move to the solution to the above question.
We take $\dfrac{1}{\sqrt{2}}$ to be a rational number. Therefore, using the definition of rational number we can say that:
$\dfrac{1}{\sqrt{2}}=\dfrac{p}{q}$
Where p and q both are integers. Now if we take the reciprocal of both the sides of the equation, we get
$\sqrt{2}=\dfrac{q}{p}$
And we know that reciprocal of all rational numbers except 0 must be rational. But $\sqrt{2}$ is not a rational number. Hence, it contradicts our assumption that $\dfrac{1}{\sqrt{2}}$ is a rational number. So, $\dfrac{1}{\sqrt{2}}$ is irrational.
Note: Be careful with the rational number zero, as if you use the above method for zero, you would end up getting 0 to be an irrational number, but in actuality 0 is rational. Also, you can remember a result that the root of the multiplicative inverse of an integer is irrational, provided the integer is not a perfect square.
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