| CS 420 |
| 2024fall |
| ibarland |
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READING: Chpt.03
Sets, and set-operations — making new sets out of old ones.
What was the concatenation of two languages defined as, again?
For two languages A,B, we say:
Recall lexicographic order
.
Kleene star: A*; Sigma*
define: "the set A closed under f":
- ints closed under squaring (unary operator)
- ints closed under * (binary operator)
- ints *not* closed under square-rooting, nor division
- general def'n, in ENGLISH (logic as exer. below)
unary f : X→Y : for any number n, f(n)
(for a unary f, and separately for a binary f)
- Showing sets equal:
compare:
L₃ = strings over {a,b}* where every 'a' is followed by a 'b'
vs
L₃' = strings over {a,b}* which don't contain "aa"
Show/argue :
- L₃ ⊆ L₃': quick proof-by-contra
- try other dir (careful!)
- review proof-by-induction: n(n+1)(2n+1)/6
sum-of-squares formula
Want to prove:
1² + 2² + 3² + ... + (n-1)² + n² = n(n+1)(2n+1)/6
holds for all natural number n ∈ ℕ.
Proof by induction:
Let P(n) be the assertion that
1² + 2² + 3² + ... + (n-1)² + n² = n(n+1)(2n+1)/6
I. Show that P(0) holds. [check]
II. Show that for any k, P(k) → P(k+1)
So we assume P(k) is true, that is:
1² + 2² + 3² + ... + (k-1)² + k² = k(k+1)(2k+1)/6
Then we can add (k+1)² to both sides:
1² + 2² + 3² + ... + (k-1)² + k² (k+1)²
= k(k+1)(2k+1)/6 + (k+1)²
= k(2k²+k+2k+1)/6 + k²+ 2k + 1
= 2k³/6 +3k²/6 +k/6 + k²+ 2k + 1
= 2k³/6 +3k²/2 + 13k/6 +1
= (2k³ + 9k² + 13k +6)/6
= (2k³ + 3k² +4k² + 2k² + 13k + 6)/6
= (k+1)(k+2)(2(k+1)+1)/6
= P(k+1)
The rule of Proof by Induction lets us conclude:
∀n.P(n) is true
Def'n: a set S is countable if ∃f . f:ℕ→S, f is onto.1
(If S is countably infinite, it's convenient to also require f to be 1:1 — a bijection
, and hence f will be invertable.)
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