Question

prove that 3+√2 is an irrational​

Answer 1

Step-by-step explanation:

Root of any prime number is always irrational

Answer 2

ᘜIᐯᗴᑎ :-

$\rm{ a \: irrational \: number \: 3 + \sqrt{2} }$

TO ᑭᖇOᐯᗴ :-

$\rm{ \: that \: 3 + \sqrt{2} \: \: is \: irrational}$

ՏOᒪᑌTIOᑎ :-

Let us assume, to the contrary, that

$3 + \sqrt{2}$ is rational. Then, there exist co-prime positive integers a and b such that

$\implies \rm{ \: 3 + \sqrt{2} = \dfrac{a}{b} }$

$\implies \rm{ \: \sqrt{2} = \dfrac{a}{b} \: - 3 }$

$\implies \rm{ \ \sqrt{2} = \dfrac{ \: a \: - 3b \: }{b} }$

$\implies \: \rm{ \sqrt{2 \: } \: \: \: is \: \: \: rational}$

[ 3,a and b are integers ∴

$\rm{\dfrac{a - 3b}{b} }$ is a rational number]

This contradicts the fact that $\sqrt{2}$ is irrational. So, our assumption is not correct. Hence, $\rm{ \: 3 + \sqrt{2} }$ is an irrational number.

OTᕼᗴᖇ IᑎᖴOᖇᗰᗩTIOᑎ :-

IRRATIONAL :

• A number is called rational if it cannot be expressed in the form p/q where p and q are integers ( q> 0) Example : √3,√2,√5,p etc.

• Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Using this theorem we can represent the irrational numbers on the number line.

• They have non-terminating and non-repeating decimal expression

• The sum, difference, multiplication and division of irrational numbers are not always irrational. Irrational numbers do not satisfy the closure property under addition, subtraction, multiplication and division