tag:blogger.com,1999:blog-80290897687252072872018-05-28T20:19:41.073-07:00Math is fun but hardHMMnoreply@blogger.comBlogger4125tag:blogger.com,1999:blog-8029089768725207287.post-27204338507226023842011-03-15T04:44:00.000-07:002011-03-15T07:33:27.659-07:00Some thoughts on the math testsI heard this little story about math tests from my former math professor. His math teacher friend had made a little experiment on the math tests. His philosophy concerning the tests was that the first question of every test should be easy enough to be solved by everybody in the course.<br /><br />So slowly and carefully he chose every time little easier first question than last time. However, it seemed like that making of the first question easier was useless, as there was every time almost an equal number of wrong answers.<br /><br />Finally he got fed up with the whole experiment of his and decided to end it once and for all. In the next test he put the first question as follows: <i>How much do we get if we add 1 to 1? </i>Surprisingly not everyone were able to get even this right. Some of the students thought it was a riddle and answered something else than <i>two</i>.<br /><br />Anyway, I think it's not too rare that students face a question that isn't completely unambiguous or clear, and the students are forced to make a "sophisticated guess" about the interpretation of the problem. I have here a test question from my probability course (maybe 6 years ago) where I had a little problem:<br /><br /><b>Question: </b>Five persons are going to be elected in a committee. There are eight candidates for the committee, 5 women and 3 men. What is the probability that the <i>minority</i> of the committee will be men, if five persons from eight are chosen by random?<br /><br />Okay, at first let's think about all the possible combinations for the committee:<br />{f, f, f, f, f} = A<br />{f, f, f, f, m} = B<br />{f, f, f, m, m} = C<br />{f, f, m, m, m} = D, where f=female, m=male.<br /><br /><b>The problem is basically this</b>: are men the minority in the set A even thought there's no men?<br /><br />This is one of the first definitions for minority I found: <i> </i><br /><i>Minority - the smaller in number of two groups forming a whole.</i><br /><br />Some might think that the minority in A is an empty set, others that the minority doesn't exist in A. I think it doesn't actually matter, as there's anyway no such a subset of A that would contain a male as an element. So males can't be the minority in the set A.<br /><br />The right interpretation according to my math teacher was that the sets A, B and C were cases where men were the minority. Fortunately enough I had let all my pedantic pride go in the test and answered interpreting A as a minority set like the teacher meant.<br /><br />I guess the moral of the story is to save your clever remarks for your blog.<img src="http://feeds.feedburner.com/~r/MathIsFunButHard/~4/hv6fQGxT6WM" height="1" width="1" alt=""/>HMMnoreply@blogger.com9http://mathisfunbuthard.blogspot.com/2011/03/some-thoughts-on-math-tests.htmltag:blogger.com,1999:blog-8029089768725207287.post-3687871238901332702011-03-13T06:55:00.000-07:002011-03-13T06:55:40.250-07:00A fun fact about elementary number theoryEver wondered what kind of rational numbers (so called "fractions", like 3/4) have a finite decimal representation? At first let's take a look the usual base ten numeral system with the following examples. <br /><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://lh5.googleusercontent.com/-DVnXxsMs0e4/TXzJt6WV7UI/AAAAAAAAABE/IuRpEEzmLyU/s1600/numbertheory1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="212" src="https://lh5.googleusercontent.com/-DVnXxsMs0e4/TXzJt6WV7UI/AAAAAAAAABE/IuRpEEzmLyU/s320/numbertheory1.png" width="320" /></a></div><br /><br />I'm not sure if I chose the examples carefully enough so that it would be easy to see what's common with all the finite fractions. Anyway let's make another kind of representation for the finite ones:<br /><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://lh3.googleusercontent.com/-gQULwlAMUiw/TXzKAUFxbII/AAAAAAAAABQ/ExQY2MpiL_I/s1600/numbertheory2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="132" src="https://lh3.googleusercontent.com/-gQULwlAMUiw/TXzKAUFxbII/AAAAAAAAABQ/ExQY2MpiL_I/s320/numbertheory2.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><br /><br />Well, the common thing is that in the finite cases the denominator is a product of only numbers 2 or 5. Where does these two numbers come from? It's well known fact, that every positive integer (>1) is either a prime number (can be divided only with 1 or itself) or a product of prime numbers (Euclid proved this). The number system we were studing was the base ten and the prime factors of 10 are 2 and 5.<br /><br />So the answer to the original question is: all the fractions where the denominator can be presented as a product of 2's and 5's, have a finite decimal representation. More mathematically we could say that the denominator should be:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://lh6.googleusercontent.com/-KWZUHLEZMJ4/TXzJyPctUfI/AAAAAAAAABM/jARBNF2MHfU/s1600/numbertheory3.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="28" src="https://lh6.googleusercontent.com/-KWZUHLEZMJ4/TXzJyPctUfI/AAAAAAAAABM/jARBNF2MHfU/s320/numbertheory3.png" width="320" /></a></div>where i and j are non-negative integers.<br /><br />In other the numeral systems than the base ten, the answer is quite similar. The allowed numbers in the denominator always depens on the prime factors of the numeral systems base number. In base eight it could only contain 2 (8=2*2*2), in base twelve 2 and 3 (12=2*2*3), in base nine just 3 (9=3*3) and so on.<img src="http://feeds.feedburner.com/~r/MathIsFunButHard/~4/sVOITpiCvi0" height="1" width="1" alt=""/>HMMnoreply@blogger.com10http://mathisfunbuthard.blogspot.com/2011/03/fun-fact-about-elementary-number-theory.htmltag:blogger.com,1999:blog-8029089768725207287.post-91899712370847767332011-03-12T10:24:00.000-08:002011-03-12T10:28:51.144-08:00Some basic combinatorics<a href="http://en.wikipedia.org/wiki/Pigeonhole_principle">Pigeonhole principle</a> is known in mathematics as a theorem that states: if n items are put in a m pigeonholes and n > m, then at least two items must be in the same pigeonhole. It sounds quite a simple and it is, but still it can solve some seemingly more complex problems.<br /><br />Let's think about the following problem; how many <b>rooks</b> can be placed in a chess board so that none of the rooks threatens another?<br /><br />Quite a quickly it becomes obvious that because only one rook can be placed on the same row or column, the maximum number of the rooks will be equivalent with the number of squares on the diagonal (see the picture). <br /><br /><div class="separator" style="clear: both; text-align: center;"></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://lh3.googleusercontent.com/-vHmEaTnq0aM/TXuzTvIFEqI/AAAAAAAAAA0/PPj94Zaq9W0/s1600/combinatorics02.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="155" src="https://lh3.googleusercontent.com/-vHmEaTnq0aM/TXuzTvIFEqI/AAAAAAAAAA0/PPj94Zaq9W0/s400/combinatorics02.png" width="400" /> </a></div><div class="separator" style="clear: both; text-align: center;"> </div>So now we have been able to place eight rooks on the board, but how do we prove that the ninth one cna't be placed there? Well as you probably guessed already, with pigeonhole principle. The right board is split in eight boxes (pigeonholes) and it becames quite obvious that it's impossible to add a rook in anywhere on the board withouth being placed in a box which already contains a rook (so it will threathen).<br /><br /><br />What about <b>kings</b>?<br /><br />Well it's quite easy to start looking for the maximum set up by placing the first king in some corner, and then adding the next king not in the next but following square. That is the closest place so that the kings don't threaten one another. We can fit four kings on the same row this way, next row none and four again to the next row etc.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://lh5.googleusercontent.com/-B1bjpIOQ1Ws/TXu2f2UCbvI/AAAAAAAAAA4/s1Wxlz95KeU/s1600/combinatorics_kings.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="163" src="https://lh5.googleusercontent.com/-B1bjpIOQ1Ws/TXu2f2UCbvI/AAAAAAAAAA4/s1Wxlz95KeU/s400/combinatorics_kings.png" width="400" /></a></div><br />When proving that the 17th king can't be placed, we proceed the same way as before, the board is just divided in the pigeonholes little different.<br /><br /><b>Knights</b>.<br /><br />I think this is already a little more complex problem. I'm not sure if this one can be solved as the king problem. At least it's not that clear that if we place the first knight in some corner, will the nearest square be the best place for the next one? Actually the answer is no (or is it?). Let's first look at the movement of the knight (picture on the left). The key is to notice that if the knight is placed on the white square, it threatens only places on the board that are black. Now it's not very hard to realise that we can place knights on every white (or black) square on the board without any threat.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://lh6.googleusercontent.com/-JScn9w8NrT4/TXu7WRatr_I/AAAAAAAAABA/rxwxXkly2sI/s1600/combinatorics_knights.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="165" src="https://lh6.googleusercontent.com/-JScn9w8NrT4/TXu7WRatr_I/AAAAAAAAABA/rxwxXkly2sI/s400/combinatorics_knights.png" width="400" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"></div><br />The boxing in this case is little different. It can't be done by placing only one knight to a box but four. It's quite easy to se this way, that five knights can't be placed in the same box without conflict.<img src="http://feeds.feedburner.com/~r/MathIsFunButHard/~4/7YZxEsB2hg0" height="1" width="1" alt=""/>HMMnoreply@blogger.com4http://mathisfunbuthard.blogspot.com/2011/03/some-basic-combinatorics.htmltag:blogger.com,1999:blog-8029089768725207287.post-23800792543553847632011-03-12T06:30:00.000-08:002011-03-12T09:08:25.516-08:00About transcendential functions<script src="http://mathjax.connectmv.com/MathJax.js" type="text/javascript"> MathJax.Hub.Config({ extensions: ["tex2jax.js","TeX/AMSmath.js","TeX/AMSsymbols.js"], jax: ["input/TeX", "output/HTML-CSS"], tex2jax: { inlineMath: [ ['$','$'], ["\\(","\\)"] ], displayMath: [ ['$$','$$'], ["\\[","\\]"] ], }, "HTML-CSS": { availableFonts: ["TeX"] } }); </script><br />I have been studing abstract algebra for a while now and I'm familiar with the concept of elements being algebraic or transcendential over certain field (or a ring?). Well anyway, yesterday I faced this interesting post regarding the solving of the equation:<br />\[ <br />x^x = 25<br />\]<br />I have never really given a thought about $x^x$ before, but I discovered that it is a so called transcendential function, meaning "<i>it cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication, and root extraction</i>." (wikipedia).<br /><br />Why this is so interesting? It is because this equation mentioned earlier can't be solved by using algebra, only with numerical methods. Here's couple of other interesting examples of transcendential functions:<br />$x^{\pi}$<br />$x^{1/x}$<br />$c^x$, where $c$ not in ${0, 1}$<img src="http://feeds.feedburner.com/~r/MathIsFunButHard/~4/HH-l_WPm9_c" height="1" width="1" alt=""/>HMMnoreply@blogger.com0http://mathisfunbuthard.blogspot.com/2011/03/about-transcendential-functions.html