[size=150]Numbers like -1.7, [math]\sqrt{16}[/math], and [math]\frac{5}{3}[/math] are known as [i]rational numbers.[br][/i]Numbers like [math]\sqrt{12}[/math] are [math]\sqrt{\frac{5}{9}}[/math] known as [i]irrational numbers.[/i][/size]
[size=150]Find exact solutions (not approximate solutions) to each equation and show your reasoning. Then, say whether you think each solution is rational or irrational. Be prepared to explain your reasoning.[/size][br][br][math]x^2-8=0[/math]
[math]\left(\frac{x}{4}\right)^2-5=0[/math]
[table][tr][td][math]2[/math][/td][td][math]3[/math][/td][td][math]\frac{1}{3}[/math][/td][td][math]0[/math][/td][td][math]\sqrt{2}[/math][/td][td][math]\sqrt{3}[/math][/td][td][math]\text{-}\sqrt{3}[/math][/td][td][math]\frac{1}{\sqrt{3}}[/math][/td][/tr][/table][size=150]Here are some statements about the sums and products of numbers. For each statement, decide whether it is [i]always[/i] true, true for [i]some[/i] numbers but not others, or [i]never[/i] true.[br][br]Experiment with sums and products of two numbers in the given list to help you decide.[br][br]Sums:[/size][br][list][*]The sum of two rational numbers is rational.[br][/*][/list]
[list][*]The sum of a rational number and an irrational number is irrational.[br][/*][/list]
[list][*]The sum of two irrational numbers is irrational.[/*][/list]
[size=150]Products:[/size][br][list][*]The product of two rational numbers is rational.[/*][/list]
[list][*]The product of a rational number and an irrational number is irrational.[/*][/list]
[list][*]The product of two irrational numbers is irrational.[br][/*][/list]
[size=150]It can be quite difficult to show that a number is irrational. To do so, we have to explain why the number is impossible to write as a ratio of two integers. It took mathematicians thousands of years before they were finally able to show that [math]\pi[/math] is irrational, and they still don’t know whether or not [math]\pi^{\pi}[/math] is irrational.[br][br]Here is a way we could show that [math]\sqrt{2}[/math] can’t be rational, and is therefore irrational.[br][br][list][*]Let's assume that [math]\sqrt{2}[/math] were rational and could be written as a fraction [math]\frac{a}{b}[/math], where [math]a[/math] and [math]b[/math] are non-zero integers.[/*][*]Let’s also assume that [math]a[/math] and [math]b[/math] are integers that no longer have any common factors. For example, to express 0.4 as [math]\frac{a}{b}[/math], we write [math]\frac{2}{5}[/math] instead of [math]\frac{4}{10}[/math] or [math]\frac{200}{500}[/math]. That is, we assume that [math]a[/math] and [math]b[/math] are 2 and 5, rather than 4 and 10, or 200 and 500.[/*][/list][/size][br]If [math]\sqrt{2}=\frac{a}{b}[/math], then [math]2= \dfrac{\boxed{ }}{\boxed{ }}[/math].
Explain why [math]a^2[/math] must be an even number.
Explain why if [math]a^2[/math] is an even number, then [math]a[/math] itself is also an even number. (If you get stuck, consider squaring a few different integers.)[br]
Because [math]a[/math] is an even number, then [math]a[/math] is 2 times another integer, say, [math]k[/math]. We can write [math]a=2k[/math]. Substitute [math]2k[/math] for [math]a[/math] in the equation you wrote in the first question. Then, solve for [math]b^2[/math].
Explain why the resulting equation shows that [math]b^2[/math], and therefore [math]b[/math], are also even numbers.[br]
We just arrived at the conclusion that [math]a[/math] and [math]b[/math] are even numbers, but given our assumption about [math]a[/math] and [math]b[/math], it is impossible for this to be true. Explain why this is.[br]
[size=150]If [math]a[/math] and [math]b[/math] cannot both be even, [math]\sqrt{2}[/math] must be equal to some number other than [math]\frac{a}{b}[/math].[br][br]Because our original assumption that we could write [math]\sqrt{2}[/math] as a fraction [math]\frac{a}{b}[/math] led to a false conclusion, that assumption must be wrong. In other words, we must not be able to write [math]\sqrt{2}[/math] as a fraction. This means [math]\sqrt{2}[/math] is irrational![/size]