Question

Prove that 3+2√5 is irrational such that it is provided that√5 is irrational.​

Answer 1

Answer:

Hey buddy!! Here's the answer

Step-by-step explanation: To prove this we need to use the contradiction method.

ABOUT CONTRADICTION :-

In this process we assume something and solve it. We just try to prove that whatever we assumed was wrong.

So, the given situation was wrong and ultimately the actual thing will be converse of it.

So, let's solve!!

TO PROVE :- 3+2root 5 is irrational.

PROOF :- Assume that the given real number is rational. This means that the number can be expressed in the form p/q where p and q belong to integers as well as are co-prime.

So,

3 + 2root5 = p/q

Or,

2root5 = p/q - 3 = (p -3q)/q Or,

Root5 = (p-3q)/2q ....... (i)

Now, (p-3q)/2q is a rational.

So,

irrational number ≠ rational number.

This means root5 is rational.

But, root5 is an irrational.

How??

Assume root5 as rational. So, Root5 = a/b

Where a and b are integers and co-primes.

So,

Squaring both sides:-

5 = p²/q²

So, p² =5q² ...... (ii)

So, p² has 5 as a factor. So, p also has 5 as its factor for some integer c.

Now,

p =5c

Or, p² =25c²

Putting it in (ii)

5q² =25c²

Or, q² = 5c²

So, q² is a multiple of 5 So, q is also a multiple of 5.

Now, Both p and q have a common factor 5 This means they are not co-primes but it is given that they are co-primes.

Hence, it's a contradiction which has risen because of taking root5 as rational.

So, √5 is irrational.

Now,

Back to the question. From (i) :-

√5 = (p-3q)/2q

So, This is not possible as√5 is irrational and RHS of the equation is rational.

As irrational ≠ rational.

Hence, it is a contradiction.

This has risen because of taking the given number (3 + 2√5) as rational number.

This implies that 3 + 2√5 is an irrational number.

Hope it helps:)

#Stay Safe buddy:)

Answer 2

AnswEr :

⠀⠀⠀⠀⠀⠀⠀☯ Let's assume that 3 + 2√5 is an rational number. So, It can be written in the form of p/q. where, p is not equal to zero.

$:\implies\sf 3 + 2 \sqrt{5} = \dfrac{a}{b} \\\\\\:\implies\sf 2\sqrt{5} = \dfrac{a}{b} - 3 \\\\\\:\implies\sf 2\sqrt{5} = \dfrac{ a - 3b}{b} \\\\\\:\implies\sf \sqrt{5} \ \bf{{}^{\text {1}}\!/{}_{\text{2}}} \Bigg( \dfrac{a - 3b}{b} \Bigg) \\\\\\:\implies\sf \sqrt{5} = \dfrac{a - 3b}{2b}$

$\star \ \underline{\frak{Here, \ \sqrt{5} \ is \ an \ irrational \ number \ \& \ \dfrac{a - 3b}{2b} \ is \ rational. }}$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀

It arises contradiction because of our wrong assumption [Rational ≠ Irrational].

Thus, 3 + 2√5 is an irrational Number.

⠀⠀⠀⠀⠀⠀⠀Hence Proved!