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

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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.

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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