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Show that the following sets form a field when combined with real number addition and multiplication:
$Q(\sqrt{3}) = \{x = a + b\sqrt{3} \ | \ a,b \in \mathbb{Q}\}$
$Q(\sqrt{7}) = \{x = a + b\sqrt{7} \ | \ a,b \in \mathbb{Q}\}$
The following sets also form fields when combined with real number addition and multiplication. Show that the elements of this field are closed under these two operations.
$S_1 = \{x = a + b\sqrt{2} + c\sqrt{5} + d\sqrt{10} \ | \ a,b,c,d \in \mathbb{Q}\}$
$S_2 = \{x = a + b\sqrt[4]{2} + c\sqrt[4]{4} + d\sqrt[4]{8} \ | \ a,b,c,d \in \mathbb{Q}\}$
For each, write an equivalent simplified expression where the denominator has been rationalized. Common factors in the numerator and denominator should be eliminated, but only upon noting any restrictions on the values of variables in the final expression related to doing so.
$\cfrac{x(y+2)}{y\sqrt{y+2}}$
$\cfrac{x^2 (4x^2 - 6x + 9)}{\sqrt[3]{x} \sqrt[4]{2x-3}}$
$\cfrac{2x^2 - 8x + 8}{\sqrt{x} + \sqrt{2}}$
$\cfrac{x^3 - 1}{1 - \sqrt{x}}$
$\cfrac{x^3 + 1}{1 + \sqrt[3]{x}}$
Recall from the exercises on rational expressions the difference quotient encountered in calculus for a function $f$ at some given $x$ is given by $$\frac{f(x+h) - f(x)}{h}$$ For each function given below, find the corresponding difference quotient at the indicated value of $x$. What value of $h$ causes this difference quotient to be undefined? Under an assumption that we avoid using this value of $h$, simplify the difference quotient. Hint: you may wish to consider rationalizing numerators as you attempt to simplify things! What is the value of the simplified expression at the same value of $h$ that previously caused a problem?
$f(x) = \sqrt{x}$
$f(x) = \sqrt{9-x}$
$f(x) = \cfrac{1}{\sqrt{x}}$
$f(x) = x^{3/2}$
$f(x) = \sqrt[3]{x}$
$f(x) = x + \sqrt{x}$