Asked by hehee
prove that the following numbers are irrational:-
(i)4 + root 5
ii)4/root3
(i)4 + root 5
ii)4/root3
Answers
Answered by
Count Iblis
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.
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.
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