prove that 3+5√2 is an irrational number
Answers
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.
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