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2014-08-27T17:34:26+05:30

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Rational number is a real number which can be expressed as a ratio of two integer numbers.

For example, \\ \frac{4}{1}, \frac{4}{2}, \frac{989}{3456}, \frac{9987}{-987}, \frac{-76}{67}, \frac{1}{2},\ 5\frac{1}{2}, - 2\frac{1}{10}.

Certain numbers like √2 is not expressible in the form of ratio of integers. So it is not a rational number.All fractional numbers and integers are rational numbers.

All rational numbers can be expressed as ratio of two integers that are co-prime (relatively prime to each other). That means that, the integers do not have common factors.

When we have to prove  x is a rational number, then assume that x = \frac{p}{q}. Then after squaring or repeated multiplication or arithmetic operations or rationalization, if you can prove that p and q are not integers or some other rule about factorization is violated, then our assumption about\ x =\frac{p}{q} is wrong.

This is the contradiction method. You assume and arrive at a contradiction.

Let\ x = 3 + 2 \sqrt{5} = \frac{p}{q},\ \ such\ \ that\ p\ and\ q\ are\ co-prime. \\ \\ q = \frac{p}{3+2\sqrt{5}} ,\\ \\ Rationalize\ by\ multiplying\ with\ 3-2\sqrt{5}\ in\ numerator\ and\ denom \\ \\ q = \frac{p(3-2\sqrt{5})}{(3-2\sqrt{5})(3+2\sqrt{5})} = \frac{p(3-2\sqrt{5})}{3^2 - 2^2*5} = \frac{p(3-2\sqrt{5})}{-11} \\ \\ = \frac{p (2\sqrt{5} - 3)}{11} \\ \\ 11 q = p(2\sqrt{5} - 3) \\ \\ This\ means\ that\ p\ has\ a\ factor\ 11\ \ and\ \ q\ has\ a\ factor\ (2\sqrt{5} - 3).\\ \\

If q has a factor(2√5-3), then it is not an integer.


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