[table][tr][td]-25[/td][td]-10[/td][td]-2[/td][td]-1[/td][td]0[/td][td]5[/td][td]9[/td][td]40[/td][/tr][/table][br][size=150]Experiment with adding any two numbers from the list (or other integers of your choice). [/size][br][br]Try to find one or more examples of two integers that add up to another integer.
Try to find one or more examples of two integers that add up to a number that is [i]not[/i] an integer.
[size=150]Experiment with multiplying any two numbers from the list (or other integers of your choice).[br][br][/size]Try to find one or more examples of two integers that multiply to make another integer.
Try to find one or more examples of two integers that multiply to make a number that is [i]not[/i] an integer.
[math]4+0.175=4.175[/math][br][br]Is this sum a rational number? Be prepared to explain how you know.
[math]\frac{1}{2}+\frac{4}{5}=\frac{5}{10}+\frac{8}{10}=\frac{13}{10}[/math][br][br]Is this sum a rational number? Be prepared to explain how you know.
[math]-0.75+\frac{14}{8}=-\frac{6}{8}+\frac{14}{8}=\frac{8}{8}=1[/math][br][br]Is this sum a rational number? Be prepared to explain how you know.
[math]a[/math] is an integer: [math]\frac{2}{3}+\frac{a}{15}=\frac{10}{15}+\frac{a}{15}=\frac{10+a}{15}[/math][br][br]Is this sum a rational number? Be prepared to explain how you know.
[size=150]Suppose [math]\frac{a}{b}[/math] and [math]\frac{c}{d}[/math] are fractions. That means that [math]a,b,c,[/math] and [math]d[/math] are integers, and [math]b[/math] and [math]d[/math] are not 0.[/size][br][br]Find the sum of [math]\frac{a}{b}[/math] and [math]\frac{c}{d}[/math]. Show your reasoning.
In the sum, are the numerator and the denominator integers? How do you know?[br]
Use your responses to explain why the sum of [math]\frac{a}{b}+\frac{c}{d}[/math] is a rational number. [br]
Use the same reasoning as in the previous question to explain why the product of two rational numbers, [math]\frac{a}{b}\cdot\frac{c}{d}[/math], must be rational.[br]
[size=150]Consider numbers that are of the form [math]a+b\sqrt{5}[/math], where [math]a[/math] and [math]b[/math] are whole numbers. Let’s call such numbers [i]quintegers[/i].[br][/size][br][size=150]H[size=100]ere are some examples of quintegers:[/size][/size][size=100][br][table][tr][td][math]\bullet\quad3+4\sqrt{5}[/math] ([math]a=3[/math], [math]b=4[/math])[/td][td][math]\bullet\quad\text{-}5+\sqrt{5}[/math] ([math]a=\text{-}5[/math], [math]b=1[/math])[/td][/tr][tr][td][math]\bullet\quad7-2\sqrt{5}[/math] ([math]a=7[/math], [math]b=\text{-}2[/math])[/td][td][math]\bullet\quad3[/math] ([math]a=3[/math], [math]b=0[/math])[/td][/tr][/table][br]When we add two quintegers, will we always get another quinteger? Either prove this, or find two quintegers whose sum is not a quinteger.[/size]
When we multiply two quintegers, will we always get another quinteger? Either prove this, or find two quintegers whose product is not a quinteger.
[size=150]Here is a way to explain why [math]\sqrt{2}+\frac{1}{9}[/math] is irrational.[/size][list][size=100][*]Let [math]s[/math] be the sum of [math]\sqrt{2}[/math] and [math]\frac{1}{9}[/math], or [math]s=\sqrt{2}+\frac{1}{9}[/math].[/*][*]Suppose [math]s[/math] is rational.[/*][/size][/list][br]Would [math]s+-\frac{1}{9}[/math] be rational or irrational? Explain how you know.
Evaluate [math]s+-\frac{1}{9}[/math]. Is the sum rational or irrational?[br]
Use your responses so far to explain why [math]s[/math] cannot be a rational number, and therefore [math]\sqrt{2}+\frac{1}{9}[/math] cannot be rational.
Use the same reasoning as in the earlier question to explain why [math]\sqrt{2}\cdot\frac{1}{9}[/math] is irrational.[br]
[size=150]Consider the equation [math]4x^2+bx+9=0[/math]. [br][/size][br]Find a value of [math]b[/math] so that the equation has 2 rational solutions.
Find a value of [math]b[/math] so that the equation has 2 irrational solutions.
Find a value of [math]b[/math] so that the equation has 1 solution.
Find a value of [math]b[/math] so that the equation has no solutions.
Describe all the values of [math]b[/math] that produce 2, 1, and no solutions.[br]