Question

Prove that \[\sqrt 3 + \sqrt 2 \] is an irrational number.

Answer 1

Hint: Here we check from the definition of irrational numbers if the numbers\[\sqrt 3 \] and \[\sqrt 2 \] satisfy the conditions of an irrational number or not. Then use the property of the sum of two irrational numbers to prove the sum as irrational.
An irrational number is a number that cannot be represented in the form \[\dfrac{p}{q}\], and the value of an irrational number has the decimal representation as non-recurring, non-repeating and non-terminating. Also, any number that is not a rational number is called an irrational number. Examples: \[\sqrt 2 ,\sqrt 3 ,\pi \]etc.

Complete step-by-step solution:
We have to prove that\[\sqrt 3 + \sqrt 2 \]is an irrational number.
We will check separately for the values of\[\sqrt 3 \] and\[\sqrt 2 \]and then check for their sum.
We know a number that can be written in the form \[\dfrac{p}{q}\] where \[q\] is non-zero and both the numerator and denominator are integers is called a rational number, i.e. the decimal representation of a rational number is either terminating or recurring.
Also, any number that is not a rational number is called an irrational number.
We know \[\sqrt 2 \] is an irrational number because it cannot be represented in the form \[\dfrac{p}{q}\], also if we try to find the square root of 2 in decimal from then, \[\sqrt 2 = 1.414213562....\] where the decimal representation is non-recurring, non-repeating and non-terminating.
Also, we know \[\sqrt 3 \] is an irrational number because it cannot be represented in the form \[\dfrac{p}{q}\], also if we try to find the square root of 3 in decimal from then, \[\sqrt 3 = 1.7320508...\] where the decimal representation is non-recurring, non-repeating and non-terminating.
We know the sum of two irrational numbers can be a rational or an irrational number depending on the two numbers. If the numbers are similar i.e. they have the same number under the square root then there is a chance that they might cancel out and then the resultant will be rational otherwise if the numbers under the square root are different (in least and simplest form)then, the sum will be irrational.
Since both the numbers \[\sqrt 3 \] and \[\sqrt 2 \] are irrational numbers and the numbers under the square root are different so there is no chance of terms to cancel each other. So, the sum of \[\sqrt 3 \] and \[\sqrt 2 \] will be an irrational number.

\[\therefore \sqrt 3 + \sqrt 2 \] is an irrational number

Note: Students are likely to make mistake while representing the values of the irrational numbers in decimal form as they tend to write the value up to two or three decimal places and don’t put continuing dots along, this causes confusion that the decimal representation of the number is terminating i.e. it is ending.