Prove that 3-√5 is irrationalNo spam

1.	Prove that 3-√5 is irrationalNo spam
Answer» $\large\underline{\sf{Solution-}}$ $\sf \: Let \: us \: suppose \: that \: 3 - \sqrt{5} \: is \: not \: irrational.$ $\rm :\implies\:3 - \sqrt{5} \: is \: rational.$ $\sf \: Let \: 3 - \sqrt{5} = \dfrac{a}{b} - - (1) \: where \: a \: and \: b \: are \: positive \: integers \:$ $\sf \: and \: a \: and \: b \: are \: co - prime \: integers.$ $\rm :\longmapsto\:3 - \dfrac{a}{b} = \sqrt{5}$ $\rm :\longmapsto\: \sqrt{5} = \dfrac{3b - a}{b}$ $\sf \: as \: a \: and \: b \: are \: integers \: \rm :\implies\:\dfrac{3b - a}{b} \: is \: rational$ $\bf\implies \: \sqrt{5} \: is \: rational$ $\sf \: which \: is \: contradiction \: as \: \sqrt{5} \: is \: irrational.$ $\sf \: Hence, \: our \: assumption \: is \: wrong.$ $\bf\implies \:3 - \sqrt{5} \: is \: irrational.$ Additional INFORMATION :- Rational number A rational number, in Mathematics, can be DEFINED as any number which can be represented in the FORM of p/q where q ≠ 0. The results are ALWAYS a rational number if we multiply, add, or subtract any two rational numbers. A rational number remains the same if we divide or multiply both the numerator and denominator with the same factor. If we add zero to a rational number then we will get the same number itself. Rational numbers are closed under addition, subtraction, and multiplication. Rational numbers are terminating DECIMALS or non - terminating but repeating. Irrational number An irrational number is a number that cannot be expressed as a fraction for any integers and. . Irrational numbers have decimal expansions that neither terminate nor repeating. Multiplication of any irrational number with any nonzero rational number results in an irrational number. The addition of an irrational number and a rational number gives an irrational number

Answer»

$\large\underline{\sf{Solution-}}$

$\sf \: Let \: us \: suppose \: that \: 3 - \sqrt{5} \: is \: not \: irrational.$

$\rm :\implies\:3 - \sqrt{5} \: is \: rational.$

$\sf \: Let \: 3 - \sqrt{5} = \dfrac{a}{b} - - (1) \: where \: a \: and \: b \: are \: positive \: integers \:$

$\sf \: and \: a \: and \: b \: are \: co - prime \: integers.$

$\rm :\longmapsto\:3 - \dfrac{a}{b} = \sqrt{5}$

$\rm :\longmapsto\: \sqrt{5} = \dfrac{3b - a}{b}$

$\sf \: as \: a \: and \: b \: are \: integers \: \rm :\implies\:\dfrac{3b - a}{b} \: is \: rational$

$\bf\implies \: \sqrt{5} \: is \: rational$

$\sf \: which \: is \: contradiction \: as \: \sqrt{5} \: is \: irrational.$

$\sf \: Hence, \: our \: assumption \: is \: wrong.$

$\bf\implies \:3 - \sqrt{5} \: is \: irrational.$

Additional INFORMATION :-

Rational number

A rational number, in Mathematics, can be DEFINED as any number which can be represented in the FORM of p/q where q ≠ 0.

The results are ALWAYS a rational number if we multiply, add, or subtract any two rational numbers.

A rational number remains the same if we divide or multiply both the numerator and denominator with the same factor.

If we add zero to a rational number then we will get the same number itself.

Rational numbers are closed under addition, subtraction, and multiplication.

Rational numbers are terminating DECIMALS or non - terminating but repeating.

Irrational number

An irrational number is a number that cannot be expressed as a fraction for any integers and. . Irrational numbers have decimal expansions that neither terminate nor repeating.

Multiplication of any irrational number with any nonzero rational number results in an irrational number.

The addition of an irrational number and a rational number gives an irrational number