Question

$show \: that \: 5 - \sqrt{3} \: is \: irrational$
​

Answer 1

Step-by-step explanation:

Given:-

5-√3

To find:-

Show that 5-√3 is an irrational number?

Solution:-

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

So it must be in the form of p/q

Where p and q are integers and q≠0

Let 5-√3 = a/b

Where a and b are co-primes.

=> -√3 = (a/b)-5

=> √3 = 5-(a/b)

=> √3 = (5b-a)/b

=> √3 is in the form of p/q

=> √3 is a rational number.

But √3 is not a rational number.

√3 is an irrational number.

This contradicts to our assumption that is 5-√3 is a rational number.

=> 5-√3 is not a rational number.

=> 5-√3 is an irrational number.

Hence, Proved.

Answer:-

5-√3 is an irrational number.

Used formula:-

• The numbers are in the form of p/q where p and q are integers and q≠0 called rational numbers.

• The numbers are not in the form of p/q where p and q are integers and q≠0 called irrational numbers.

Used Method:-

• Method of Contradiction (Indirect method).

Note:-

If a is a rational number and b is an irrational number then a+b , a-b , ab and a/b is an irrational number

Answer 2

$\underline\mathtt\green{Answer:}$

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

We can find co prime a & b ( b≠ 0 )such that 5 - √3 = a/b.

Therefore 5 - a/b = √3.

So we get 5b -a/b = √3.

Since a & b are integers, we get 5b -a/b is rational, and so √3 is rational.

But √3 is an irrational number Let us assume that 5 - √3 is a rational.

We can find co prime a & b ( b≠ 0 )such that,

∴ 5 - √3 = √3 = a/b

Therefore 5 - a/b = √3.

So we get 5b -a/b = √3.

Since a & b are integers, we get 5b -a/b is rational, and so √3 is rational.

But √3 is an irrational number, which contradicts our statement.

∴ 5 - √3 is irrational.