Question

$\huge\mathfrak\red{\underline{\underline{Answer}}}$
Prove that 3-√5 is irrational
No spam
​

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

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 2

$\huge\bf{Question:-}$

Prove that 3+√5 is irrational

$\huge\bf{Answer:-}$

Answer:

Given

3 + √5

To prove:

3 + √5 is an irrational number.

Proof:

Letus assume that 3 + √5 is a rational number.

So it can be written in the form a/b

3 + √5 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving

3 + √5 = a/b

we get,

=>√5 = a/b – 3

=>√5 = (a-3b)/b

=>√5 = (a-3b)/b

This shows (a-3b)/b is a rational number.

But we know that √5 is an irrational number, it is contradictsour to our assumption.

Our assumption 3 + √5 is a rational number is incorrect.

3 + √5 is an irrational number

Hence proved.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬