Lecture 3: The real numbers

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Quiz 1:

Sketch the graph of y=\sqrt(|x|) over the domain [-4,4]. \sqrt means take the square root.

Summary of lecture. We met in the computer lab in Fretwell to learn how to access the email accounts the university provides. We also discussed how to use the web pages for this course, and had a quiz at the end of class.

This material is left from the web page for math 1100. Much of it is quite relevant, so it is left here for your benefit.

1. Representations of Real Numbers.
There are many ways to represent numbers, but one very convenient way is DECIMAL representations. We looked at what decimal representation means and how to move between the decimal representation of a rational number and the quotient of integers representation.
2. The real numbers. We constructed a venn diagram showing two ways of chopping up the real numbers: first by positive, negative and zero, and second by rational and irrational. We can classify a number as rational or irrational based on the nature of its decimal representation. If the number has terminating (ends with zeros from some point on) or repeating (the same block of digits repeats forever) decimal representation, then the number is rational.
3. Integer exponents.
4. Absolute value, and the graph of the absolute value function. How to solve equations with absolute values? There are two ways.
A. Solving by conditioning.
Solve |x|=4. First, ask `what could we say if we knew x was positive?' Of course, then x would have to be 4. Now what if we knew x was not positive? We could say that x was -4. We've just used conditioning on x to solve the equation. IE, we supposed x has some property, and made a conclusion, then we supposed that it had did not have that property and came to another conclusion.
B. Solving by geometric means
We defined the function f as follows:
f(x) = 3-2x if x satisfies x<=1, 2+x if 1< x <=5, and 2x-7 if 5 < x.
Then we asked the question, for what values of x is f(x)=6.
A. We used conditioning as follows. First, condition on x <= 1. In other words, assume the inequality is satisfied and ask if there are any of those x for which f(x) is 6. We found that -3/2 works. Then we conditioned on 1 < x <=5, and found another solution in that interval. Finally, we solved the equation 2x-7=6 and found yet another solution 13/2. We will say we conditioned on the value of x three times, one for each clause in the functions definition.
B. To solve the problem geometrically, we can sketch the graph of the function and locate the places where this graph touches the line y=6. This can be done either by hand or with the aid of a graphing calculator.
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