prove that the following numbers are irrational:-
(i)4 + root 5
ii)4/root3
Suppose 4 + sqrt(5) is rational:
4 + sqrt(5) = r/s ------->
sqrt(5) = r/s - 4 = another rational number
Now, we can prove that sqrt(5) is irrational, so we get a contradiction. Therefore the asumption that
4 + sqrt(5) is rational was wring and thisnumber is thus irrational.
Now, the proof that sqrt(5) is irrational is also quite simple. If
sqrt(5) is rational, then we have:
sqrt(5) = r/s
where r and s don't have divisors in common. This means that:
5 = r^2/s^2 -------->
r^2 = 5 s^2
Factor both sides into prime factors and count by multiplicity how many prime factors you have. The number r has some number of prime factors, so r^2 has twice that number, so the left hand side has an even number of prime factors.
The same is true for s^2. On the left hand side, then, the factor 5 adds one prime factor to it, so the left hand side contains an odd number of prime factors.
So, we have a contradiction, the assumption that sqrt(5) is rational is thus false.
To prove that a number is irrational, we need to show that it cannot be expressed as the ratio of two integers (i.e., it cannot be written as a fraction in the form a/b, where a and b are both integers and b is not equal to zero). Let's prove the irrationality of both numbers:
(i) 4 + √5
Assume that 4 + √5 is a rational number. This implies that it can be expressed as a fraction a/b, where a and b are integers, and b is not equal to zero.
So, we have:
4 + √5 = a/b
Now let's isolate √5:
√5 = (a/b) - 4
Square both sides of the equation to remove the square root:
5 = ((a/b) - 4)^2
5 = (a^2 / b^2) - 8(a/b) + 16
Rearrange the equation:
(a^2 / b^2) - 8(a/b) + 11 = 0
Let x = a/b, so the equation becomes:
x^2 - 8x + 11 = 0
Solve this quadratic equation using the quadratic formula:
x = (-(-8) ± √((-8)^2 - 4(1)(11))) / (2(1))
x = (8 ± √(64 - 44)) / 2
x = (8 ± √20) / 2
x = 4 ± √5
Since we assumed that a/b is equal to 4 + √5, one of the roots of the quadratic equation should match that value. However, we see that the roots are 4 + √5 and 4 - √5, both of which are not integers. This contradicts our initial assumption that 4 + √5 can be expressed as a rational number.
Therefore, 4 + √5 is an irrational number.
(ii) 4 / √3
Assume that 4 / √3 is a rational number. This implies that it can be expressed as a fraction a/b, where a and b are integers, and b is not equal to zero.
So, we have:
4 / √3 = a/b
Now, let's rationalize the denominator (√3) by multiplying both the numerator and the denominator by √3:
4 / √3 = (a/b) * (√3 / √3)
4 / √3 = (a√3) / (b√3)
Simplifying the expression:
4 / √3 = (a√3) / (b√3) = (a * √3) / (b * √3) = (a/b) * (√3 / √3) = (a/b) * 1 = a/b
We see that 4/√3 is in the form of a/b, where a and b are integers. Thus, 4/√3 is a rational number.
In conclusion:
(i) 4 + √5 is an irrational number.
(ii) 4/√3 is a rational number.