CSE390D Notes for Monday, 10/7/24
<![CDATA[
Some implicit assumptions in section 1.7 proofs that have to do with
whether integers are closed under certain operations:
addition: yes
subtraction: yes
multiplication: yes
division: no
exponentiation: yes (for y > 0), no (when y < 0 or 0^0)
-------------------------------------------------------------------------------
counterexample
prove "Every positive integer is the sum of two squares is false"
consider 3, only squares smaller are 0 and 1, can't add up to 3
-------------------------------------------------------------------------------
example: 4n + 2 is never a perfect square
(k = 4n + 2 for some n) -> ~(k = m^2 for some m)
direct doesn't help
indirect doesn't help
contradiction:
remember if t == p -> q, then we assume p ^ ~q, derive r ^ ~r
k = 4n + 2 and k = m^2
m^2 = 4n + 2 = 2(2n + 1)
m^2 is even
m is even (lemma--use indirect proof)
m is odd
m = 2x + 1
m^2 = (2x + 1)^2 = 4x^2 + 4x + 1 = 2(2x^2 + 2x) + 1
qed
m = 2x
m^2 = (2x)^2 = 4x^2
4x^2 = 4n + 2
2x^2 = 2n + 1
1 = 2(x^2 - n)
1 is even
but 1 = 2 * 0 + 1
1 is odd
contradiction
qed for original proposition
-------------------------------------------------------------------------------
Example
prove that (n + 1)^3 >= 3^n if n is a positive integer with n <= 4
proof by exhaustion (4 cases):
1: (1 + 1)^3 = 8, 3^1 = 3, 8 >= 3
2: (2 + 1)^3 = 27, 3^2 = 9, 27 >= 9
3: (3 + 1)^3 = 64, 3^3 = 27, 64 >= 27
4: (4 + 1)^3 = 125, 3^4 = 81, 125 > 81
qed
-------------------------------------------------------------------------------
Example
prove that if n is an integer, then n^2 >= n
proof by cases (3 cases)
case 1: n < 0
n <= -1 (negative) and n^2 >= 0, so n^2 >= n
case 2: n = 0
n^2 = 0, so n^2 >= n
case 3: n > 0
n >= 1
multiply by n: n^2 >= n
qed
-------------------------------------------------------------------------------
conjecture:
perfect squares: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121
prove: n is an integer then last digit of n^2 is 0, 1, 4, 5, 6, 9
let n = 10a + b for some integers a and b and 0 <= b < 10
then n^2 = (10a + b)^2 = 100a^2 + 20ab + b^2
= 10(10a^2 + 2ab) + b^2
therefore final digit of n^2 is the same as the final digit of b^2
proof by cases
b = 0, b^2 = 0
b = 1, b^2 = 1
b = 2, b^2 = 4
b = 3, b^2 = 9
b = 4, b^2 = 16
b = 5, b^2 = 25
b = 6, b^2 = 36
b = 7, b^2 = 49
b = 8, b^2 = 64
b = 9, b^2 = 81
-------------------------------------------------------------------------------
game of Chomp: guaranteed win for first player
tilings
n x n checkerboard can be tiled with dominoes iff n is even
8 x 8 checkerboard with opposite corners removed can't be tiled
with dominoes (30 white and 32 black squares or vice versa)
8 x 8 checkerboard with one corner removed can't be tiled with
triominoes (counts of three colors are unequal)
]]>
Stuart Reges
Last modified: Mon Oct 7 14:14:49 PDT 2024