Category: 11 Mathematics HL

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11 HL: Proof

Complete the questions below before you return from the November Break. (Do some tonight, but you don’t have to complete them all right now.)

Exercise 9B questions 1, 2, 3, 5, 9
Exercise 9C questions 1, 3, 4, 6, 8
Exercise 9D questions 2b, 4, 7
Exercise 9E questions 3, 6, 7
Exercise 9F questions 1, 2, 4

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11 HL: Combinations

Complete the following questions before our next class.

Exercise 7D.1 questions 1, 3, 6, 7, 10, 12, 18
Exercise 7D.2 questions 3, 5
Exercise 7E 1, 2, 6, 7, 8, 9, 11, 14, 20, 24

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11 HL: Rational Functions

Complete the following questions before our next class. (Most of this should be review from last year, but if you find any unfamiliar material we can look through that in our next class.)

Exercise 6D.1 questions 2adgj, 4

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11 HL: Absolute Values in Equalities

Complete the following questions before our next class.

Exercise 6C.3 1be, 2ade, 3bc, 4afg

If you’re interested in looking in more detail at the \(\leq\) relation, here’s a fun puzzle to consider.

The real numbers are totally ordered by the \(\leq\) relation, that is, \[\text{(1) for any } a,b \in \mathbb{R},\text{ we have } a \leq b \text{ or } b \leq a.\] We also have the following properties, which are true for any \(a,b,c \in \mathbb{R}\)

\[\text{(2) if } 0 \leq a \text{ and } b\leq c, \text{ then } ab\leq ac, \text{ and}\]

\[\text{(3) if } b\leq c, \text{ then } b+ a\leq c+a.\]

What happens if we try to extend the \(\leq\) relation to the complex numbers? If we can, we would want to do so without “breaking” properties 1–3. Can we extend our ordering relation to the complex numbers without running into trouble in this way?

Hint
If our ordering is total, then we must have either \(0\leq i\) or \(i\leq 0\). Show that both of these options lead to a problem. (Property (2) might be particularly useful here.)