The teenager fond of mathematics does not have to share the nostalgia of the teachers. So take this picture only as an invitation to visit my dokuwiki (Valentina did me the favor of staying in your virtual space). I have it a bit forgotten, but ... the cyber-readers visit it by clicking here
I followed these days by reading the book 500 challenges (see below) and I can recommend without reservation. The solution to the following problem (with comments inserted) is an example of the art of reasoning for clues (clues that only experience in problem solving taught to interpret). Previous comments
A logical principle of common sense, formalized as a technique for solving problems in Combinatorics and Discrete Mathematics, Box is called Pigeon Hole Principle or Principle (which can be translated as pigeonhole principle and the principle of pichoneras , respectively).
The principle says that if there are n pichoneras (nests) and n +1 pigeons (pigeons), then one of the nests will sleep 2 pigeons. (Formerly otherwise arise but for solution of problems with this approach is sufficient folk.) Let me
share an example from the Barbeau and Klamkin (E. Barbeau, M. Klamkin, W. Moser, Five Hundred Mathematical Challenges, MAA, 1995) to illustrate the fact that, although the principle is a truism, which is not obvious is how to build pichoneras do the trick of the show asked.
Problem. Any subset of size 7 drawn from {1,2 ,..., 126} can choose 2 elements b such that b / a is greater than 1 but not greater than 2. Are asked to prove this statement (proposition). Previous comments
2.
The first thing you should do the cognizable teenager who aspires to be a superstar in troubleshooting is experience exemplifying. And this in order to reach a full understanding of the statement. (Well, I think before that should be the mood and willpower - and fighting spirit - that immunity to the scarecrows - that is, cognizable should not panic.)
Tip:
Can you put the condition otherwise but equivalent - but more readable, ie, more understandable to you? Can you verbalize the equivalent condition? Problem
partner:
Trying to build a subset of size 7 with numbers 1 to 126 and that does not meet the given condition. (First it may be worth taking a subset of any size 7 and see if it meets the condition.)
short
Solution:
pichoneras The 6 {1.2}, {3,4,5,6}, {7 ,..., 14}, {15, .. ., 30}, {31 ,..., 62}, {63 ,..., 126} do the trick. (Note that each pichonera has the property that contains an x \u200b\u200bsuch that 2x is also on the pichonera. Moreover, they form the union of {1,2 ,..., 126}.)
Note 1: The
condition imposed on the problem can be put another way to multiply by the double inequality 1 < b/a < (o igual)2. Es decir, la condición sería equivalente a a < b < (o igual)2a. La cual ya es más fácil de verbalizar: "existen a y b tales que b está entre a y 2a (y, si acaso, b=2a)"
Note 2:
If we try to build a subset of size 7 to comply with the condition sooner or later we'll get a {1 , 3,7,15,31,63,127}. And this is illuminating and, as you can see that any element of the set has a double in the subset. From here you may eventually get the idea that
sets that meet the condition would be those in which there is an element that has its double in the set. (The pichoneras built into the solution meets this condition as adjusted - ie have only one element of that type.)
Note 3:
In problems of this kind - where the principle of the pichoneras trivially solved - how the wording suggests that pichoneras use. (Of course, to achieve capture the suggestion from the cognizable statement requires the expertise have resolved about 20 basic problems of this kind - through the experience acquired most of the codes.)
Note 4:
These problems can be frustrating due to the fact that you know the assertion is proved is true - for example, having built (in this case) the whole note 2, the revelation is accomplished ... "Of course! Not be otherwise, because would lack elements (127), but from there to prove it formally is a considerable distance.
0 comments:
Post a Comment