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March 30, 2015

March 30, 2015

Posted by **hehee** on Tuesday, May 31, 2011 at 4:27am.

(i)4 + root 5

ii)4/root3

- maths -
**Count Iblis**, Tuesday, May 31, 2011 at 2:18pmSuppose 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.

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