Prove that the following numbers are irrational:

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Prove that the following numbers are irrational:

(i) 5 +√2
(ii) 3 – 5√3
(iii) 2√3 – 7
(iv) √2 +√5

Solution:

(i) 5 +√2
Now, let us assume 5 + √2 be a rational number, ‘r’
So, 5 + √2 = r
r – 5 = √2
We know that, ‘r’ is rational, ‘r – 5’ is rational, so ‘√2’ is also rational.
This contradicts the statement that √2 is irrational.
So, 5 + √2 is irrational number.
(ii) 3 – 5√3
Now, let us assume 3 – 5√3 be a rational number, ‘r’
So, 3 – 5√3 = r
3 – r = 5√3
(3 – r)/5 = √3
We know that, ‘r’ is rational, ‘(3 – r)/5’ is rational, so ‘√3’ is also rational.
This contradicts the statement that √3 is irrational.
So, 3 – 5√3 is irrational number.
(iii) 2√3 – 7
Now, let us assume 2√3 – 7 be a rational number, ‘r’
So, 2√3 – 7 = r
2√3 = r + 7
√3 = (r + 7)/2
We know that, ‘r’ is rational, ‘(r + 7)/2’ is rational, so ‘√3’ is also rational.
This contradicts the statement that √3 is irrational.
So, 2√3 – 7 is irrational number.
(iv) √2 +√5
Now, let us assume √2 +√5 be a rational number, ‘r’
So, √2 +√5 = r
√5 = r – √2
Square on both sides,
(√5)2 = (r – √2)2
5 = r2 + (√2)2 – 2r√2
5 = r2 + 2 – 2√2r
5 – 2 = r2 – 2√2r
r2 – 3 = 2√2r
(r2 – 3)/2r = √2
We know that, ‘r’ is rational, ‘(r2 – 3)/2r’ is rational, so ‘√2’ is also rational.
This contradicts the statement that √2 is irrational.
So, √2 +√5 is irrational number.

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