Clinton and Obama in Selma, socalled selfish athletes, levity
Workout notes 3600 yards of swimming; 500 fist in 9:05, 10 x (drill/free), 10 x (25 fly, free, back, free) on the 2 (1:421:45 each, mostly 1:431:44), 10 x 50 free, 25 back (form) on the 1:30, 500 pull in 8:11, 5 x 25 kick, 25 free (fins).
The pool was somewhat crowded today; luckily I shared a lane with someone who could swim in a straight line.
(click here for a larger image) This is kind of what my swimming is like; I’ve done a few masters meets and have gotten my doors blown off. ðŸ™‚
I’ve been lapped in a 400 meter race in a 50 meter pool!!!
Social commentary: if you think that professional athletes are selfish brutes, read this.
Basically, one ex NFL football player (Everson Walls, formerly of the Cowboys and Giants) gave his friend (Ron Springs, formerly of the Cowboys) one of his kidneys. Now THAT is being a friend!!!
More commentary: we are the proud owners of a Prius. Leave it to Julie Larson to get us to laugh at ourselves:
(click here for a larger version)
Politics: Obama and Clinton attend a civil rights event in Selma, Alabama.
Democratic rivals Barack Obama (news, bio, voting record) and
Hillary Rodham Clinton, joined by former President
Bill Clinton, courted black voters at a hallowed civil rights shrine on Sunday and said the movement’s leaders had set the stage for their White House bids.
ADVERTISEMENTAt a daylong series of events in the small town of Selma, Alabama, the trio of political stars commemorated the 42nd anniversary of the 1965 civil rights march, paying homage to the heroes of “Bloody Sunday” and calling for a new generation to take up the fight.
“I stand on the shoulders of giants,” Obama, who hopes to become the first black president, said at a packed ceremony in the AME church used as a headquarters by civil rights leader the Rev. Martin Luther King. “I’m here because somebody marched for freedom.”
Clinton, in a simultaneous speech in a packed Baptist church less than a block away, said the voting rights won after the march had made possible her campaign to be the first woman president, as well as the runs by Obama and New Mexico Gov. Bill Richardson, who would be the first Hispanic president.
“I know where my chance came from, and I am grateful,” Clinton said. “The people of Selma understood that voting matters.”
So far so good, but again, I am bothered by the fact that this was viewed as pure posturing for votes by Senator Obama and Senator Clinton. Yes, I know that they want votes and will work for them. But this is a cause that I truly believe that both of the candidates believed in, from the start of their adult years. Neither is a “Johnny come lately” to this cause!
Here they are starting across the bridge.
Senator Obama giving a speech.
Senator Clinton was there too, but even though I am a staunch Obama backer, I believe that she was there out of deep personal conviction.
A former President and a future one. Both are very intellegent, intellectually curious, and both know how to connect to the common person.
Mathematical Induction: how and why it works
I am writing this post because I’ve seen that there is some misunderstanding of what mathematical induction is and why it works.
 What is mathematical induction? It is a common proof technique. Basically, if one wants to show that a statement is true in generality and that one can index the set of statements via the integers (or by some other appropriate index set), then one can use induction.
Here is a common example: suppose one wants to show that
1 + 2 + 3 + ….+ k = (1/2)*(k)*(k+1) for all positive integers k
(for example, 1 + 2 + 3 + 4 + 5 = (1/2)*(5)*(5+1) = 15).Initial step: 1 = (1/2)*(1)*(1+1)= 1 so the statement is true for k = 1.
Inductive step: assume that the formula holds for some integer k.
Finish the proof: show that if the forumla holds for some integer k, then it holds for k+1 as well.So 1 + 2 + 3 +….+ k + (k+1) = (1 + 2 + …+ k) + (k+1) =
(1/2)*(k)*(k+1) + (k + 1) (why? because we assumed that k was an integer for which
1 + 2 + 3 + ….+ k = (1/2)*(k)*(k+1). )so 1 + 2 + 3 +….+ k + (k+1) = (1/2)*(k)*(k+1) + (k + 1) = ((1/2)*k + 1)*(k+1) (factor out a k+1 term)
= (1/2)*2*((1/2)*k + 1)*(k+1) = (1/2)*(k+2)*(k+1)=(1/2)*(k+1)*(k+2)=
(1/2)*(k+1)*(k+1 + 1)
which is what we needed to show. So the proof would be done.  Why does induction “prove” anything? Mathematical induction is equivalent to the so called “least positive integer” principle in mathematics.
 What is the least positive integer principle? It says this: “any nonempty set of positive integers has a smallest element”. That statement is taken as an axiom; that is, it isn’t something that can be proved.
Notice that this statement is false if we change some conditions. For example, is is NOT true that, say, any set of positive numbers (or even rational numbers) has a smallest element. For example, the set of all numbers between 0 and 1 (exclusive; 0 is not included) does NOT have a least element. Why? Let “b” be a candidate to be the least element. Then “b” is between 0 and 1. But then (1/2)*b is greater than zero but is less than “b”; hence “b” could not have been the least element. Neither could any other number.Note that the set of negative integers has no least element; hence we need the condition that the integers are positive.
Notice also that there could be groups of positive integers with no greatest element. For example, Let x be the largest element in the set of all even integers. But then 2*x is also even and is bigger than x. Hence it is impossible to have a largest one.
 What does this principle have to do with induction? This is what: an induction proof is nothing more than a least integer argument in disguise. Lets return to our previous example for a demonstation; that is, our proof that 1 + 2 + ….+ n = (1/2)*n*(n+1).
We start by labeling our statements: 1 = (1/2)*(1)*(1+1) is statement P(1),
1 + 2 = (1/2)*(2)*(2+1) is statement P(2), …1+2+…+ 5 = (1/2)*(5)*(5+1) is statement P(5) and so on.We assume that the statement is false for some integer. The set of integers for which the statement is false has a least element by the least element priciple for positive integers.
We assume that the first integer for which the statement is false is “k+1”. We can always do this, because we proved that the statement is true for k = 1, so the first possible false statement is k = 2 or some larger integer, and these integers can always be written in the form k + 1.
That is why the anchor statement (the beginning) is so important.
We now can assume that the statement is true for n = 1,….n = k since k+1 is the first time the statement fails.
Now when whe show “if statement P(k) is true then P(k+1) is also true (this is where we did the algebra to add up 1 + 2 + …..+ k + (k+1) = ((1/2)*k*(k+1)) + (k+1) ). This contradicts that statement P(k+1) is false.
Hence the statement cannot be false for ANY positive integer k.
 Weak versus strong induction. As you can see, the least positive integer principle supposes that the statement is true for all statements P(1) through P(k), so in fact there is no difference (when inducting on the set of positive integers) between weak induction (which assumes the induction hypothesis for some integer k) and strong induction (which assumes the induction hypothesis for n = 1 through n = k).
 Other index sets: any index set that one has to induct on has to have the “least element principle” to its subsets. Also, if there is a cardinal w that has no immediate predecessor, then one must “reanchor” the induction hypothesis as k = w prior to proceeding.

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