Question

Prove that 5+ √5 is an irrational number.​

Answer 1

Answer:

Let's prove this by the method of contradiction-

Say, √5 is a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.

⇒√5=p/q

⇒5=p²/q² {Squaring both the sides}

⇒5q²=p² (1)

⇒p² is a multiple of 5. {Euclid's Division Lemma}

⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}

⇒p=5m

⇒p²=25m² (2)

From equations (1) and (2), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5. {Euclid's Division Lemma}

⇒q is a multiple of 5.{Fundamental Theorm of Arithmetic}

Hence, p,q have a common factor 5. this contradicts that they are co-primes. Therefore, p/q is not a rational number. This proves that √5 is an irrational number.

For the second query, as we've proved √5 irrational. Therefore 2-√5 is also irrational because difference of a rational and an irrational number is always an irrational number.

Answer 2

• To prove the given condition let us assume that $5 + \sqrt{5}$ is any rational integer .

• We know that every rational integer can be expressed in the form of $\frac{x}{y}$
• This basically implies that $(5 + \sqrt{5} ) = \frac{x}{y}$
• So on transposing 5 we get, $\sqrt{5} = \frac{x}{y} + 5$
• This implies that √5 is rational but we already know that √5 is an irrational integer
• Hence , our assumption is wrong.
• Therefore, $5 + \sqrt{5}$ is an irrational number.

hope this helps you

mark brainliest

and

follow me