Question

given tht √2 is irrational prove that (5+3√2) is an irrational number​

Answer 1

Answer:

Prove that 3+2√5 is irrational?

To prove this we need to use the contradiction method.

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, root5 is irrational.

Now,

Back to the question. From (i) :-

Root5 = (p-3q)/2q

So, This is not possible as root5 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 + 2root5) as rational number.

This implies that 3 + 2root5 is an irrational number.

Answer 2

Answer:

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Step-by-step explanation:

Let us assume the contrary.

i.e; 5 + 3√2 is rational

∴ 5 + 3√2 = abab, where ‘a’ and ‘b’ are coprime integers and b ≠ 0

3√2 = abab – 5

3√2 = a−5bba−5bb

Or √2 = a−5b3ba−5b3b

Because ‘a’ and ‘b’ are integers a−5b3ba−5b3b is rational

That contradicts the fact that √2 is irrational.

The contradiction is because of the incorrect assumption that (5 + 3√2) is rational.

So, 5 + 3√2 is irrational.