SUNY Geneseo Department of Mathematics
Wednesday, February 22
Math 239 01
Spring 2017
Prof. Doug Baldwin
Sedar Ngoma, Auburn University (and candidate for a faculty position here)
“Recovering a Diffusion Coefficient in a Parabolic Equation Arising in Geochronology”
Thursday, Feb. 23, 2:30 PM, Newton 203
Extra credit for writing a paragraph or so summarizing connections you make between the talk and your own interests.
What assumptions can you make in proofs?
Those given to you by the proposition you’re proving.
Since many propositions are conditionals “if P then Q,” this rule often means you can assume P. Note that the conditionals are sometimes implied by universal quantifiers, e.g., “for all real numbers x ...” is often treated as “if x is a real number ...” for purposes of proofs,
Proof by cases gives you additional assumptions by breaking P into several parts, each of the form “P and something else.”
Section 3.4
Define f(x) by
Prove that for all real numbers x, f(x) ≥ x.
Relevant ideas from the reading (or questions about it)
Proof: There are two cases, namely x < 0 and x ≥ 0. In the first case, x < 0, let x be any negative real number, and note that x2 is positive, i.e., greater than 0. Since x is negative, f(x) is greater than x. In the second case, x ≥ 0, we see immediately that 2x ≥ x. Since it is true in every case, we have thus shown than f(x) ≥ x. QED.
This proof illustrates a couple of common features of case-based proofs, in particular that they are very useful for proving things about functions defined piecewise, and that the proofs of the various cases can be quite different from each other.
Define max(x,y) to be the largest of x and y, i.e., max(x,y) = x if x ≥ y, max(x,y) = y if x < y.
For example, max(3,2) = 3, max(2,7) = 7, max(7,7) = 7.
Proposition: for all real numbers a, b, and c, max( a, max(b,c) ) = max( max(a,b), c )
Proof: We prove that for all real numbers a, b, and c, max( a, max(b,c) ) = max( max(a,b), c ) in three cases, according to which of a, b, or c is the maximum.
Case 1: a ≥ b and a ≥ c. In this case max(b,c) is either b or c but in either case a is larger, so max( a, max(b,c) ) must be a. On the right side of the equation, max(a,b) = a and so max( max(a,b), c ) = max(a,c) = a. So if a ≥ b and a ≥ c, then max( a, max(b,c) ) = max( max(a,b), c )
Case 2: b ≥ a and b ≥ c. In this case max(b,c) is b, and so max( a, max(b,c) ) must be b. On the right side of the equation, max(a,b) = b and so max( max(a,b), c ) = max(b,c) = b. So if b ≥ a and b ≥ c, then max( a, max(b,c) ) = max( max(a,b), c ).
Case 3: c ≥ a and c ≥ b. In this case max(b,c) = c, so max( a, max(b,c) ) must be c. On the right side of the equation, max(a,b) is either a or b, but in either case c is larger so max( max(a,b), c ) = c. So if c ≥ a and c ≥ b, then max( a, max(b,c) ) = max( max(a,b), c ).
Since in all cases, max( a, max(b,c) ) = max( max(a,b), c ) the proposition is proven. QED.
This proof illustrates a situation in which the right cases to use aren’t necessarily obvious. It also involves “sub-cases” of the first and third cases, but those sub-cases can be dealt with in just a few words and so probably aren’t worth explicitly identifying as cases. Finally, the proofs of each case are very similar, and in fact most publications would probably just show the proof for the first case and then say “the remaining cases are similar.”
See handout for details
The division algorithm and congruence
Read section 3.5