CSE390D Notes for Friday, 10/4/24
<![CDATA[
Terminology of proofs:
theorem a major statement that can be shown to be true
axioms statements we assume to be true
proofs a valid argument that establishes the truth of a theorem
lemmas a less important theorem
corrolary a theorem that follows from another proof
conjecture a statement that is proposed to be true
qed quod erat demonstrandum, "which was to be demonstrated"
proofs often have an implied application of a universal quantifier:
If x > y, where x and y are positive real numbers, then x^2 > y^2
really means
For all positive real numbers x and y, if x > y, then x^2 > y^2
and often involve an implied application of universal generalization
as long as the choice of values like x and y is arbitrary
definitions:
The integer n is even if there exists an integer k such that n =
2k, and n is odd if there exists an integer k such that n = 2k +
1. (Note that an integer is either even or odd, and no integer is
both even and odd).
The real number r is rational if there exist integers p and q with
q != 0 such that r = p/q. A real number that is not rational is
called irrational.
-------------------------------------------------------------------------------
Methods of proof (think of it as a set of tools to consider)
direct
indirect (contraposition)
contradiction
equivalence
counterexample
-------------------------------------------------------------------------------
direct proof:
prove that p -> q
assume p
derive q
example: prove that n is odd only if n^2 is odd
use of only: other direction satisfied if n is odd and n^2 even
(n odd) -> (n^2 odd)
assume n is odd
then n = 2k + 1 for some k
and n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1
therefore n^2 is odd
qed
-------------------------------------------------------------------------------
indirect (contraposition)
prove that p -> q
assume ~q
derive ~p
example: prove that if 3n + 2 is odd then n is odd
(3n + 2 odd) -> (n odd)
direct approach leads to dead end:
3n + 2 = 2k + 1
3n = 2k - 1
not much to work with, so try to prove the contrapositive (~q -> ~p)
assume n is not odd and prove 3n + 2 is not odd
assume n is not odd
therefore n is even
then n = 2k
and 3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1)
thus, 3n + 2 is even, 3n + 2 is not odd
qed
-------------------------------------------------------------------------------
contradiction
given some overall proposition p
assume ~p
derive ~r ^ r
~p -> (r ^ ~r) is logically equivalent to p
what if it's of this form:
prove t: p -> q
assume not: ~t == ~(p -> q) == ~(~p v q) == (p ^ ~q)
example: prove that n is odd when 3n + 2 is odd
(3n + 2 odd) -> (n odd)
assume 3n + 2 is odd and n is even
then 3n + 2 = 2j + 1 for some j
and n = 2k for some k
then 3(2k) + 2 = 2j + 1
6k + 2 = 2j + 1
6k + 1 = 2j
1 = 2 (j - 3k)
but 1 is odd
this is a contradiction
therefore original statement is true
qed
-------------------------------------------------------------------------------
equivalence
prove p <--> q by showing p -> q ^ q -> p
more generally, show that p1, p2, p3, ..., pn equivalent by showing:
p1 -> p2
p2 -> p3
p3 -> p4
...
p(n-1) -> pn
pn -> p1
example: prove these are equivalent:
n is even
n - 1 is odd
n^2 is even
1) n is even -> n - 1 is odd
direct:
assume n is even
then n = 2k for some k
then n - 1 = 2k - 1 = 2(k - 1) + 1
therefore n - 1 is odd
qed
2) n - 1 odd -> n^2 even
direct:
assume n - 1 is odd
then n - 1 = 2k + 1 for some k
then n = 2k + 2
and n^2 = (2k + 2)^2 = 4k^2 + 8k + 4 = 2(2k^2 + 4k + 2)
therefore n^2 is even
qed
3) n^2 even -> n even
indirect: prove n not even -> n^2 not even
assume n is not even, which means it's odd
then n = 2k + 1
and n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(k^2 + 2k) + 1
therefore n^2 odd
qed
-------------------------------------------------------------------------------
example of an incorrect proof:
step reason
1. a = b given
2. a^2 = ab multiply both sides of (1) by a
3. a^2 - b^2 = ab - b^2 subtract b^2 from both sides of (2)
4. (a - b)(a + b) = b(a - b) factor both sides of (3)
5. a + b = b divide both sides of (4) by a - b
6. 2b = b replace a by b in (5) because a = b
7. 2 = 1 divide both sides of (6) by b
step 5 is the problem (division by 0)
-------------------------------------------------------------------------------
example: prove that given real numbers r1 and r2, r1 and r2 are
rational only if (r1)(r2) is rational
r1 and r2 rational -> (r1)(r2) rational
direct proof:
assume r1 and r2 are rational
then r1 = a1/b1 and r2 = a2/b2 where a1, b1, a2, b2 are integers
then (r1)(r2) = (a1/b1)(a2/b2) = (a1 * a2)/(b1 * b2)
a1 * a2 and b1 * b2 are integers
therefore (r1)(r2) is rational
qed
]]>
Stuart Reges
Last modified: Fri Oct 4 12:29:04 PDT 2024