Algebra/Chapter 2/Real Numbers/Answers to "Why" questions

11. Take two rational numbers $$\frac{a}{b}$$ and $$\frac{c}{d}$$ the product of those two numbers is $$\frac{ac}{bd}$$, which is still rational (remember that a rational number is one that can be expressed as a ratio of two numbers). This is an example of the closure property for rational numbers.
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13. $$\sqrt{2}$$ is irrational, however, $$\sqrt{2}\sqrt{2} = 2$$ is an integer.

15. Consider two numbers $$1+\sqrt{2}$$ and $$1-\sqrt{2}$$ both of these numbers are irrational, however when you add them together, the irrational part "cancels" and you are left with a rational part. In other words, because you can always add a number and its negative to get 0 (which we consider to be rational), you can always get a rational number from an irrational number.

17. Because if $$x$$ were irrational but $$\sqrt{x}$$ were equal to $$a/b$$ for some integers $$a$$ and $$b$$, then $$x$$ would be equal to the rational number $$a^2/b^2$$, contradicting the fact that $$x$$ is irrational.

19. The product of two rational numbers is rational (see the first challenge problem) similarly the sum of two rational numbers is also rational. If $$x$$ is rational then $$x+1$$ is rational, then $$x(x+1)$$ must also be rational. Therefore, $$x$$ must be irrational.
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