Question

$show\;that\;5 - \sqrt3\;is\;irrational$

Answer 1

To Show :

• 5 - √3 is irrational.

Solution :

We have to prove that 5 - √3 is irrational. So let's assume the opposite that 5 - √3 is rational. Hence if 5 - √3 is a rational number then it can be written in the form of p/q where q ≠ 0.

Writing in the form of p/q :

⠀⟶⠀5 - √3 = p/q

⠀⟶⠀- √3 = p/q - 5

⠀⟶⠀- √3 = p - 5/q

⠀⟶⠀- √3 = p - 5q/q

⠀⟶⠀√3 = - (p - 5q/q)

⠀⟶⠀√3 = 5q - p/q

⠀⟶⠀Irrational = Rational

Hence, 5q - p/q is a rational number but √3 is irrational.

Since, Rational ≠ Irrational

⛬ 5 - √3 is irrational.

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★ Additional Info :

• A number which is not a perfect square, its square root is also irrational. eg :- √2 , √3 , √5 , √7 and so on.
• The the sum or difference of a rational and irrational is irrational.
• The product aur question of a known zero rational number with a irrational number is irrational.
• Sum / Difference / Product / Quotient of two irrational number may be rational or irrational.

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Answer 2

Answer:

$5\;-\;\sqrt3$ is irrational

Step-by-step explanation:

Given :-

$5\;-\;\sqrt3$

To find :-

To prove that $5\;-\;\sqrt3$ is irrational.

Solution :-

Assume $5\;-\;\sqrt3$ is rational.

So,

$5\;-\;\sqrt3\;=\;\frac{a}{b}$

Transposing,

$5\;-\;\frac{a}{b}\;=\;\sqrt3$

$\sqrt3\;=\;5\;-\;\frac{a}{b}$

$\sqrt3\;=\;\frac{5b-a}{b}$

Since a and b are integers, $5\;-\;\frac{a}{b}$ is rational, and we know $\sqrt3$ is irrational.

We also know that rational number cannot be equal to an irrational number.

So our assumption is incorrect, so we proved that $5\;-\;\sqrt3$ is irrational.