With an ordinary language, you are able to make good use of your knowledge if you are fluent in the language. A typical American high school student studying French will learn more French than the average French 4 year old knows. But French 4 year olds are fluent in French and can use their limited knowledge of French to communicate far more effectively than a typical American student.
I like to use the term math fluency to describe a similar ability to make use of our mathematical knowledge in flexible and creative ways. But whatever you wish to call it, this ability to make flexible and creative use of mathematics can be much more important than simply having a lot of knowledge that you can only use in very familar situations.
In all likelyhood, you would not consider yourself to be fluent in math. So how do you become fluent in mathematics? If you are studying calculus and you have never been fluent in math at any level, calculus is not the place to start to try to achieve fluency in math. Neither is algebra. The easiest way to develop fluency in math is to start at a level that is as close as possible to the ordinary language in which we are fluent--which in my case is English.
Mary is two years older than her sister. If Mary is 10, how old is her sister?
This problem can be solved with simple arithmetic. But you have to ne alert. It is a bit of a trick question. It sounds like it should be an addition problem--but it turns out to require subtraction. A large pat of what it means to be fluent in math is being able to notice when we are being given information in a form that is not the ideal form for us to use, and being able restate that information in a way we can use. In this case our restated condition is that her sister is two years younger than Mary.
Next we will consider some problems that are mathematically more advanced. A radio sells for $13 ar 35% off. What was the original price? A meal including a 15% tip comes to $46. What was the cost of the meal before the tip?
35% off mean 35% less than 100% or the amount we pay is 65% or 0.65 times the original price. To find the original price we must divide $13 by 0.65 which gives us $20 as the original price. You can check that 35% of $20 is $7 and $20 - $7 does equal $13.
For the meal we add 15% to our basic cost of 100% to get 115%. The $46 is 115% of the basic cost of 1.15 times the basic cost. To find the basic cost we divide $46 by 1.15 and get an answer of $40. You can check that 15% of $40 is $6 and 40 and 6 do add up to 46.
All of this can be expressed very efficently using algebra. As long as students are going to study algebra, why not wait and cover this material using the language of algebra? I would argue that a solution given in plain English will paint a clearer picture for most people than one expressed in the symbolic language of algebra. More importantly, when students do study algebra, what they study will be easier to understand if they have seen something similar before. Indeed, it can be very helpful begin the study of algebra by seeing how basically the same arguments can be made either with verbal reasoning or symbolically working with algebraic expressions. For most students, this will make what they do in algebra seem more understandable and meaningful.
I would like to conclude this post with a problem from the 9th grade IMP textbook called the Perils of Pauline. The girl is 3/8 of the way through when she hears a train whistle from behind her. She just has time to run back to the beginning of the tunnel and get there at the same time as the train. She also just has time to run forward and get to the end of the tunnel at the same time as the train. If the train goes 60 mph how fast doe Pauline run?
I will look at this problem in my next post--but I wanted you to see it first wiout mmy commentary. You may try to solve it if you like. Whether you try to solve it or not, I would like you to make an assessment of how difficult you think this problem would be for the average person.
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