Question

given that root 5 is irrational prove that 2 root 5 minus 3 is an irrational number​

Answer 1

Answer:

Step-by-step explanation:

(Q) Given that √5 is irrational.

Prove that 2√5-3 is an irrational number.

(Sol)->

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

i.e. 2√5-3 = p/q (where p and q are integers )

Now,

2√5-3 = p/q

2√5 = p/q +3

2√5 = (p+3q)/q

√5 = (p+3q)/2q

We can see that LHS is an irrational number which can never be equal to RHS which is a rational number. Therefore, our assumption that 2√5-3 is a rational number is wrong. 2√5-3 is not a rational number, it is an irrational number.

Hence proved.

Done!!

Answer 2

$2\sqrt{5} -3$ is an irrational number.

• When we talk about the number line containing the real numbers, that is the real number line, there are numbers of the form $\frac{p}{q}$, where p and q are non-zero integers.
• These numbers, that can be represented in the fraction format are called rational numbers. The numbers, on the other hand, that cannot be represented in this manner are called irrational numbers.

Here, according to the given information, we are given that,

$\sqrt{5}$ is an irrational number.

Now, let us suppose that $2\sqrt{5} -3$ is a rational number, say, p.

Then, $p = 2\sqrt{5}-3$

Or, $\sqrt{5} =\frac{p+3}{2}$

This is a contradiction since $\sqrt{5}$ is an irrational number but the right hand side of this equation must be a rational number since p is a rational number.

Hence, it is proved that  $2\sqrt{5} -3$ is an irrational number.

Learn more here

https://brainly.in/question/6039056

#SPJ5