1/19
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ordered field
you can add, subtract, divide, multiply and you can tell that x<y for some x,y in Q
how do we get R
we start with Q and build a bigger ordered field around R
triangle inequality
∀ x,y in R, |x+y| ≤ |x|+|y|
triangle inequality corollary
look up
Let E be a subset of R, E is bounded above if
∀x∈E, x≤β, for some β∈R
Let E be a subset of R, E is bounded below if
∀x∈E, α≤x, for some α∈R
Let E be a subset of R, E is bounded (words only intuition)
if E is bounded above AND below
Let E be a subset of R, E is bounded (math language)
iff ∃m≥0 s.t. ∀x∈E, |x|≤m
definition of absolute value
if x∈R, then |x| = { [x if x>0] [-x if x<0] }
∀x,y ∈R, |xy|=
|x||y|
∀x∈R
-|x| ≤ x ≤ |x|
|a|≤b IFF
-b≤a≤b
Q has holes theorem
add from lecture 8/19
axiom of completeness
every nonempty set of R that is bounded above, has a supremum (least upper bound) in R
R has AOC but
Q does not
lower bound aoc
8/21 pg2
archimedean principle intuition
you can always make a number bigger by multiplying
archimedean principle (math language)
∀x>0, y ∈ R, ∃n∈N s.t. nx>y
density of Q
8/24