Question

prove that 5-2√3 is an irrational number​

Answer 1

Answer:

Step-by-step explanation:

let, 5 - 2√3 be a rational number.

.°. 5 - 2√3 = p/q [ where p and q are integer , q ≠ 0 and q and p are co- prime number ]

= - 2√3 = p/q -5

=- 2√3 = p - 5q / q

= √3 = p - 5q / - 2q

we know that p/q is a rational number.

.°. √3 is also a rational number.

This contradicts our assumption

5 - 2√3 is an irrational number

Answer 2

Answer:

5 -2√3 is an Irrational number.

Step-by-step explanation:

Let us assume that √3 is a rational number.

⇒ √3 = $\sf{\frac{p}{q}}$

Squaring on both sides.

⇒ [√3]² = $\sf[{\frac{p}{q}}]$²

⇒ 3 = $\sf{\frac{p^{2}}{q^{2}}}$

⇒ 3q² = p²

∵ 3 divides p²

∴ 3 divides p also.  → 1

[If 'p' is a prime number and divides 'a²', then 'p' divides 'a' also where 'a' is a positive integer.]

Let p = 3c, where c is any positive integer.

Substitute this value in 3q² = p²

3q² = [3c]²

3q² = 9c²

q² = $\frac{9c^{2}}{3}$

q² = 3c²

∵ 3 divides q²

∴ 3 divides q also. → 2

[If 'p' is a prime number and divides 'a²', then 'p' divides 'a' also where 'a' is a positive integer.]

From 2 and 3, We can say that 'p' and 'q' are divisible by one another since they have 3 as their common factor. This contradicts the fact that 'p' and 'q' are co-primes. This is due to our wrong assumption that √3 is rational.

∴ √3 is an Irrational Number.

Let us assume that 5 - 2√3 is a rational number.

⇒ 5 - 2√3 can be expressed of the form p/q where p and q and co-primes and q ≠ 0.

∴  5 - 2√3 = $\sf{\frac{p}{q}}$

⇒ - 2√3 = $\sf{\frac{p}{q}}$ - $\sf{5}$

⇒ - 2√3 = $\sf{\frac{p - 5q}{q}}$

⇒ √3 = $\sf{\frac{p - 5q}{-2q \ \ }}$

⇒ $\boxed{\sf{\sqrt{3} = \sf{\frac{p - 5q}{ - 2q}}}}$

In the RHS, 'p', '5q', '-2q' are all rational numbers. Hence the RHS as a whole is Rational.

But we have proved that √3 is an Irrational number.

But Irrational ≠ Rational.

This is due to our wrong assumption that 5 - 2√3 is an rational number.

∴ 5 - 2√3 is an irrational number.

NOTE: This is a 3-4 mark solution, for the 2 mark solution, you dont have to prove that √3 is irrational. Proceed to write 'We know that √3 is irrational, and the conclusion statement given in this answer.