Principles of Mathematics in Operations Research

1.3 Problems 9

Proposition 1.3.

p,qeR+3 y/pq^ V-~Y- => P hi-

proof. A: y/pq^^ and hence A: ^pq = *f*. Similarly, B: p £ q and B:

p = q. Let us assume B and go forward 2±2 = p = ^ = ^fpq. However, this

is nothing but A: ^/pq = Ey^2. Contradiction. D

1.3.4 Theorem of alternatives

If the pattern of the proposition is A => either C or (else) D is true (but not

both), we have a theorem of alternatives. In order to prove such a proposition,

we first assume A and C and try to reach D. Then, we should interchange C

and D, do the same operation.

Proposition 1.3.7 If x^2 - 5x + 6 > 0, then either x < 2 or x > 3.

Proof. Let x > 2. Then,

a;^2 - 5a; + 6 > 0 => (a; - 2)(x - 3) > 0 =» (a: - 3) > 0 => a; > 3.

Let a; < 3. Then,

x^2 - 5x + 6 > 0 => (a; - 2)(x -3)>0^(a;-2)<0^a;<2. D

Problems

1.1. Prove the following two propositions:

(a) If / and g are two functions that are continuous * at x, then the function

/ + g is also continuous at x, where (/ + g)(y) = f(y) + g(y).

(b) If / is a function of one variable that (at point a;) satisfies

3 c> 0, 5 > 0 such that Vy 3 \x - y\ < 6, \f(x) - f(y)\ <c\x-y\

then / is continuous at x.

1.2. Assume you have a chocolate bar consisting, as usual, of a number of
squares arranged in a rectangular pattern. Your task is to split the bar into
small squares (always breaking along the lines between the squares) with a
minimum number of breaks. How many will it take? Prove^2.

A function / of one variable is continuous at point x if

Ve > 0, 35 > 0 such that Vy B \x - y\ < S =^ |/(x) - f(y)\ < e.

(^2) http://www.cut-the-knot.org/proofs/chocolad.shtml