m101m - Helpful Math Tutor

Solution a27264a - British Gentleman.

2011-12-28T00:50:00.000-08:00

Problem: A British gentleman and his Canadian grandson were renting a car in the United States and were told that the car averages 26 miles (mi) per gallon (US gal). Grandpa wanted to know how much this is in miles per imperial gallon (imp gal), and the grandson wanted to know the fuel consumption in liters (L) per 100 km. Find the figures each are looking for, using these facts:

Solution:

To get the conversion the other way around, just equation solve on each of the facts given above:

To get the miles per imperial gallon, using the conversion factors we were given, we need to go via liters:

For the fuel consumption we need to have volume per distance, which is the reciprocal of fuel economy, or in other words, the fraction up side down:

So, the answers are that the 26 mi/US gal is approximately the same as 31 mi/Imp gal or 9 L/100km.

Comment:

We note that the units (US gal, Imp gal, L, mi, km) can be treated as variables for the purposes of equation solving, simplifications, substitutions, and calculations, in order to convert from one unit to another.

Problem a27264a - British Gentleman.

2011-12-27T16:42:00.000-08:00

A British gentleman and his Canadian grandson were renting a car in the United States and were told that the car averages 26 miles (mi) per gallon (US gal). Grandpa wanted to know how much this is in miles per imperial gallon (imp gal), and the grandson wanted to know the fuel consumption in liters (L) per 100 km. Find the figures each are looking for, using these facts:

Factor Table.

2011-12-08T20:26:00.001-08:00

The Factor Table, is the multiplication tables backwards, and more!

Knowing the multiplication tables makes the numbers your friends, and you start to be familiar with characteristics of the most commonly used numbers.

The 36 items, highlighted in green in the factor table below are from the single- digit multiplication tables and they are useful to know by heart. The numbers highlighted in yellow are the prime numbers, and those with red highlight are the squares.

The multiplication entries without highlighting are not required knowledge, but are still useful and actually easily determined using the single digit multiplication tables forwards and backwards repeatedly.

When faced with 29/42 for example, we know that since 29 does not appear in any of the single digit multiplication tables, it must be a prime number. And since 42 is not a multiple of 29, we can approximate with 28 instead for an easy estimate. 28/42 reduces immediately to 4/6 since both 28 and 42 are in the 7 multiplication table. 4/6 reduces to 2/3 since both 4 and 6 are in the 2 multiplication table. So we can conclude that 29/42 is approximately two thirds, which gives us a much better understanding of its value.

Alternatively we could have estimated 29 with 30 and looked at 30/42. Since both 30 and 42 are in the 6 multiplication table, we reduce it to 5/7, which is a little harder than two thirds to grasp. If we on the other hand just halved 30 and 42 to get 15/21 we could then estimate 21 with 20 to get 15/20 = 3/4, which is easy to understand. However, here we did two approximations along the way, and we are therefore probably not very accurate, so the two thirds still is the preferred guess, and actually the most correct one as well.

Here is a colorful Factor Table for the fridge door, or for the bulletin board by your desk. A PDF version for printing is available by clicking here.

Multiplication Table.

2011-12-01T23:00:00.001-08:00

The multiplication table, is it really needed now when every student has a calculator?

It is a matter of speed, estimates, thinking on your feet, and independence!

Knowing the 36 items, framed in green in the single-digit multiplication tables shown below, is as necessary to function in daily life as the ability to speak ones own language.

One should not need to bring out the calculator to understand that you can get five apples at 59 cents each for around three dollars. Estimate 59 with 60, 6 times 5 is 30, and the total price follows.

Knowing the multiplication tables also gives the answer to 24*7/8 instantly, because 24*7=7*24 and you have 7*24/8=7*3=21, since 3*8=24, so 24/8=3.

Calculators can be useful for comparing prices per liter for example, to see which is cheaper, 5 fl oz, (US or Imperial?, they are different!), for $1.09, or 235 mL for $1.77. Yes, there is a place and time for using calculators.

But as with anything mechanical or electronic, the calculator can break, lose power, or be lost altogether, when needed the most. You do not want to be too dependent on it for daily chores and conversations.

Here is a colorful multiplication table for the fridge door, or for the bulletin board by your desk. A PDF version for printing is available here.

Solution a27237a - Arithmetic Sum.

2011-02-27T23:08:00.002-08:00

Problem: Find the sum of the sequence of terms less than one thousand, and starting with 37 + 50 + 63 + 76 + ... .

Solution:

This is a typical arithmetic sequence problem, and the formulas to use are easily derived by starting a table of the terms, with a the first one, and d the difference between the terms, as follows:

In our case we see that a = 37 and d = 13 so we have:

To find n, the number of terms to include, we solve the inequality:

Since n is the largest integer less than 75.1 we see that n = 75 and we get:

Hence our final answer is that the Arithmetic Sum is 38850.

Comment:

One could wonder where this formula for summing up the first consecutive natural numbers come from:

Well, if we go to n instead of n − 1 it might look a little bit more familiar:

To justify it conceptually, just add the first and the last, the second and the next to the last, etc, numbers:

Of course this only works if n is an even number. But if n is odd, start by zero instead:

In the first case, for even n, we have half of n rows all summing to n + 1 each, so the total clearly is:

In the second case, for odd n, we have added zero to make it an even number of rows, so we now have half of n + 1 rows, that is the one zero row plus half of n − 1 rows, all summing to, in this case, n each. The total therefore is again:

For those who want a more serious mathematical proof for this formula, we can use the Principle of Mathematical Induction as follows:

Step 1:

For n = 1 the formula is true since:

Step 2:

For any natural number n > 1 we assume that the formula is true for n − 1, that is, as we saw above:

We then add n to see that we get the same formula with n − 1 replaced by n, and consequently n replaced by n + 1:

Step 3:

Since the formula according to Step 1 is true for n = 1, and since according to Step 2, if it is true for some n − 1 (for example n − 1 = 1 when n = 2) then it is also true for n, the Principle of Mathematical Induction gives that it is also true for all natural numbers n. This proves the formula.

Problem a27237a - Arithmetic Sum.

2011-02-27T16:28:00.000-08:00

Find the sum of the sequence of terms less than one thousand, and starting with 37 + 50 + 63 + 76 + ... .

Solution a27235a - Fraction Equation.

2011-02-08T00:22:00.000-08:00

Problem: Solve for x the equation:

Solution:

Before starting to manipulate the equation, we take a look at what we have. Noticing that x is in a denominator, we know that divide by zero is not defined, so x can not be zero.

In addition to noticing that x is not zero, we should also notice the bigger denominator, although less clearly visible, that:

We make a note of the two values that x can not be. Otherwise any value of x we might get from solving the equation should be OK. We start by handling the fractions to simplify the equation:

Here we see that x = 0 would actually solve the equation correctly, making both sides zero and maintaining the equality. However, we have already noted that x cannot be zero, so this would not do.

On the other hand, since x is not zero, we can divide by x on both sides of the equation, and continue solving it:

We are done, and the solution is:

Comment:

To double check that we have the correct answer, we put the answer in for x in the original equation, simplify, and verify that we have equality between the left and right sides of the equation.

Since both left and right sides of the equation agree, we can be reasonably confident that our solution to the equation is correct.

Middle school students may not notice, the invalid values of x due to the requirement that division by zero is undefined. While the emphasis in middle school is on correct manipulation of the fractions, and equation solving in general, the check by substitution of the answer(s) into the original equation is very important to ensure correctness. In high school, more emphasis is put on the determination of the admissible domain of the x values, which when done correctly can save some work and time while still ensuring correctness.

Problem a27235a - Fraction Equation.

2011-02-07T23:48:00.000-08:00

Solve for x the equation:

Solution a27229a - Nested Fractions.

2011-02-07T15:27:00.000-08:00

Problem: Simplify to an improper fraction in lowest terms:

Solution:

To simplify this correctly, repeated application of the fraction rules is needed, starting from the inside and working out one operation at a time:

The question now remains if this fraction is in lowest terms, or if it can be simplified further. To verify this we need to check if 233 and 151 have any common factors. With 151 being the smaller number we need to see if we can factor 151 into any smaller prime factors:

2 is not a factor of 151 because 151 is an odd number.

3 is not a factor of 151 because its digit sum 1+5+1 = 7 is not divisible by 3.

(4 can not be a factor if 2 is not.)

5 is not a factor of 151 because 151 does not end in 0 or 5.

(6 can not be a factor if 2 is not.)

7 is not a factor of 151 because 151/7 = 21 4/7.

(8 can not be a factor if 2 is not.)

(9 can not be a factor if 3 is not.)

(10 can not be a factor if 2 is not.)

11 is not a factor of 151 because 151/11 = 13 8/11.

(12 can not be a factor if 2 is not.)

13 can not be a factor because

so if 13 was a factor of 151, the other factor must be smaller, and we have already seen that 151 has no smaller factors.

Therefore we can conclude that 151 is a prime number, and no further simplification is possible.

Comment:

When checking for prime factors we only need to test the prime numbers. The numbers in brackets above are not prime, so they could have been omitted.

The fraction rules are as follows, where a, b, c, and d, are any integers, except zero when in the denominator:

Problem a27229a - Nested Fractions.

2011-02-07T00:10:00.000-08:00

Simplify to an improper fraction in lowest terms:

Solution a27227a - Wind Speed.

2011-02-06T16:28:00.000-08:00

Problem: An aircraft flies from Vancouver to Montreal and back. The trip to Montreal is with the wind and the flight time is only four and a half hours. The trip back to Vancouver against the wind took one hour longer. If the normal aircraft speed without wind is 900 km/h, calculate the wind speed in meters per second (m/s).

Solution:

This is a typical speed problem, and the formulas to use are given away by the unit used for speed, namely km/h. This means that by equation solving on this we get all three formulas we might need:

To organize the data and see what can be done we make up a table as follows:

The times are given in the problem text in hours. The distance d is not given, but it is clear that the distance to and from must be the same. The speed must be different though, as in one case we are traveling with the wind, and in the other case, struggling against it. The aircraft speed is given as 900 km/h so we introduce a variable for the unknown wind speed w, and add or subtract the wind speed to adjust the trip speeds.

Looking at this table we search for an equation that allows us to solve for the unknown variable w. The three formulas above tell us how each column can be calculated from the other two. Since we have two lines, and two unknowns, d and w, we could perhaps expect to get two equations that allows us to solve for both unknowns. This may be so, but the problem question is only asking for the wind speed, so we see that there is an easier opportunity to calculate the same d in both lines in two different ways and setting them equal, thus producing an equation with only the one unknown w:

Thus the wind speed is 90 km/h. However, the question asked for the speed in m/s instead, so using the fact that 1 km = 1000 m and 1 hour = 3600 seconds, we convert the units:

So, the final answer is that the wind speed is 25 m/s.

Problem a27227a - Wind Speed.

2011-02-06T15:45:00.000-08:00

An aircraft flies from Vancouver to Montreal and back. The trip to Montreal is with the wind and the flight time is only four and a half hours. The trip back to Vancouver against the wind took one hour longer. If the normal aircraft speed without wind is 900 km/h, calculate the wind speed in meters per second (m/s).

Math Notes

2011-01-22T21:40:00.000-08:00

Links to general notes about math practices, notation, happenings, or opinions, etc.

Multiplication Table

Factor Table

Calculator Usage

Mixed Fractions Confusion

2011-01-22T21:30:00.000-08:00

Mixed fractions as learned in elementary school can be a cause for confusion in high school. The notation is inconsistent. With numbers, mixed fractions assume addition, as in:

In high school however, students learn algebra, and the same expression using letters instead of digits, suddenly has a different meaning. With letters, mixed fractions now assume multiplication, as in:

For a = 2, b = 3, and c = 4 the answer is now:

This is just one example of math learned at an early age, which later has to be adjusted a little to accommodate more general advanced concepts. Another example of this is documented in the comment at the end of

Solution a14889a – Radical Equation.

Solution a27197a - Intersecting Curves.

2011-01-19T01:10:00.000-08:00

Problem:

For positive real number a, the two curves

intersect in point M. Determine b, as a function of a, such that the area between the x-axis and the curves below M, and the area between the line y = b and the curves above M are equal. Also find a such that the curves intersect at right angles at M.

Solution:

To find the point M we note that the two curves are symmetric around the y-axis, wherefore M should be a point on the y-axis, where x = 0. Hence

and

For the area between the x-axis and the two curves we have, because of the symmetry, twice the area under the left, or blue, curve:

To find where the blue curve intersects the line b, we solve:

To get the area above the curves between M and b we subtract the integral which gives the area under the curve, from the area of the rectangle enclosing it:

To get b to make the areas below and above the curves equal we set the areas equal and solve for b:

To make the curves cross at right angles we need to make sure that the product of the slopes of their tangent lines, given by the derivatives, equal −1, which is the same as saying that the slopes are negative reciprocals. We have the derivatives:

Since the curves intersect on the y-axis we have x = 0 and the perpendicular requirement gives the equation:

The value of a when used in the results above gives the complete answers:

Problem a27197a - Intersecting Curves.

2011-01-19T00:58:00.000-08:00

For positive real number a, the two curves

Solution a15105a - Square in Circle.

2011-01-14T00:49:00.000-08:00

Problem: A square is inscribed in a circle with its four vertices on the circle. Calculate the ratio of the square perimeter to the circle circumference as a percentage rounded to a whole number. Do the same for the ratio of the square area to the circle area.

Solution:

First we draw a picture to better understand the situation.

The diameter of the circle is the diagonal of the square, so the circle radius r is half the square diagonal.

Since no numbers are given, we need to only use the formulas. If r is the circle radius, and s is the square side length, we have:

To get to a percentage for an answer, we need to somehow get rid of the variables r and s. But as we are looking at a ratio, they hopefully may cancel out. This would only work if the square and the circle are using the same variable though.

But the square is composed of two halves that are right triangles, if the square is split along its diagonal, which also is the circle diameter.

Hence we can use the Pythagorean Theorem to relate r and s:

Here we can of course discard the negative expression for s as we are dealing with positive distances only.

Replacing

we can then calculate the ratios asked for:

We double check the formulas and calculations made above. To see if the answers are reasonable, we note that since the square is smaller than the circle it does make sense that the ratios are less than one.

The fact that the variables cancel out also makes sense since we are comparing objects measured in the same units, namely length in one case, and area in the other.

Comment:

We see that approximating the circle circumference and area with a square does not give very accurate values, although the circumference approximation is significantly better than the area approximation.

Problem a15105a - Square in Circle.

2011-01-14T00:48:00.000-08:00

A square is inscribed in a circle with its four vertices on the circle. Calculate the ratio of the square perimeter to the circle circumference as a percentage rounded to a whole number. Do the same for the ratio of the square area to the circle area.

Calculator Usage

2011-01-04T20:31:00.001-08:00

In the debate about calculator usage, we have many saying that with calculators there is no need for mental calculation skills, and the multiplication tables in particular. The counter argument is that learning mental calculation is still useful.

My take on this is somewhere in the middle. Calculators, and mathematical computer software, are certainly needed for more time-consuming calculations, like for example:

On the other hand, what we do NOT need the calculator for, are simple things like:

What we definitely do not need the calculator for are howlers like these that I have actually seen high school students perform using their calculators:

The issue with these last two examples is that it should be a lot faster to do them mentally than it would be to type them into the calculator.

The trade off point between mental calculation speed and typing into calculator speed, will of course remain slightly different for different people. I would suggest though that most people would benefit from knowing the multiplication tables and simple operations with plus, minus, and fractions.

A math tutor's language lesson.

2010-12-07T15:27:00.001-08:00

Merry Christmas!

If you can guess what language this is, let me know.

If you are a math nerd, let me know if you can figure out the equation for the curved edge at the bottom of the picture.

3x3 Magic Squares

2010-11-29T23:53:00.000-08:00

A 3x3 magic square with row, column, and diagonal, sum 3(N+4M) is easily constructed using this pattern:

N+7M	N	N+5M
N+2M	N+4M	N+6M
N+3M	N+8M	N+M

Here are the first 16 examples, with N blue, M green, the magic square yellow, and the sums in black and white:

Starting up!

2010-11-25T01:10:00.000-08:00

Solution a14889a - Radical Equation.

2010-11-19T00:41:00.000-08:00

Problem: Solve for real numbers x the equation

Solution:

This is a standard radical equation problem where an extraneous root may be introduced unless we pay careful attention to the domain for each step in the equation solving process.

Looking at this equation we make note of the domain restrictions imposed by the radicals:

Since both sides of the equation clearly are positive, we can square both sides without introducing extraneous solutions:

Here we again look at the situation and notice that the right side radical is not negative and we make note of this domain restriction:

With this additional domain restriction we can safely square both sides without introducing extraneous solutions:

The first root is clearly outside the domain for x which was determined above to be

and can be discarded. The answer therefore is:

Left and right sides with solution point.

Comment:

Had we not paid attention to the domain restrictions due to the square roots requiring non-negative contents on the inside and being non-negative on the outside, we might have included the extraneous root −5.8 in the answer. However, looking at the graph, this would have made no sense at all.

We might ask why the radicals have this non-negative requirements when for example

This does of course not apply to the square roots for which the non-negative argument works well. But should we treat cube roots and square roots differently when solving equations?

When it comes to generalizing the root concept we get into the area of exponential functions via the realization that:

Using fractional exponents eventually leads to this seriously questionable result:

This is one of the reasons why the convention has been generally adopted that, when working in the real numbers domain, any radicals are non-negative both inside and out.

Problem a14889a - Radical Equation.

2010-11-18T23:37:00.000-08:00

Solve for real numbers x the equation

Solution a27196a - Parabola Points.

2010-11-18T20:06:00.003-08:00

Problem: Find the general quadratic equation, the vertex, all axis intercepts, domain and range, and sketch the graph, for a parabola that goes through the points (2,−2), (−3, 3), and (5, 7).

Solution:

The general quadratic equation has the form:

Substitution of the coordinates for the three points (x, y) give three equations:

Here we have three simultaneous equations with the three unknowns a, b, and c. We solve them by first eliminating c from two different combinations of the three equations:

Using the new equations (11) and (12) we can now either use substitution of either a or b and solve for the other, or we can use the elimination method again. We choose to eliminate b and solve for a:

With a resolved we can now get b from equation (12) and finally c from equation (1):

The general quadratic equation asked for is then:

With the general form of the equation it is easy to get the y-intercept, which occurs when x = 0, so we have:

To find the vertex we need to complete the square using the pattern:

With our particular numbers we get:

Comparing this to the standard form of the quadratic equation

where the vertex is (p, q) we have with our numbers that the vertex is:

Once we know the vertex we also have the domain and range. The general form of the equation clearly has no limitations on x so the domain is all real numbers. The standard form of the equation gives us the range, since the vertex is either the minimum or maximum, in this case the minimum. The range therefore is all real numbers y, such that

To sketch a graph we now only need the x-intercepts which we easily get from the standard form of the equation by solving for x when y = 0:

To summarize, the answers are:

Problem a27196a - Parabola Points.

2010-11-17T16:40:00.000-08:00

Find the general quadratic equation, the vertex, all axis intercepts, domain and range, and sketch the graph, for a parabola that goes through the points (2,−2), (−3, 3), and (5, 7).

Solution a27193a - Family Ages.

2010-11-12T18:20:00.006-08:00

Problem: Brother’s and sister’s ages add up to half of their mother’s age. Their father is five times as old as sister, and there are the same differences in ages between father and mother as between brother and sister. If the mother is 42 years old, how old are the others in the family?

Solution:

Reading the text carefully several times we realize that we need four unknowns:
   b = brother’s age in years
   s = sisters’s age in years
   m = mother’s age in years
   f = father’s age in years

The problem text also gives four equations:

With m = 42 already given, we can substitute 42 for m wherever m appears. We also notice that the first equation (i) can be changed to b = 42/2 − s. This means that we can put everything into the third equation and solve it for s:

Using s = 9 to calculate the other ages we easily get the answers:
  sister: 9,
  brother: 21 − 9 = 12,
  mother: 42,
  father: 5 * 9 = 45.

Comment:

Four equations seems rather complex, but a closer look shows that these particular equations actually make substitution quite easy. Middle school students learn to make substitutions, while simultaneous equation solving by the different method of elimination is a high school topic.

Math Problems with Solutions

2010-11-12T01:35:00.017-08:00

These problems are provided to demonstrate standard solution techniques, and in some cases a couple of different solutions have been included, as well as some extra comments. The grade level classifications are very approximate and may vary from place to place, but should give an indication of the relative difficulties of the selected problems.

Students are encouraged to try solving the problems first, and then after a successful solution, or at least a serious attempt, check the solution here.

Parents can use these problems, solutions, and comments, to help their children with math at home.

These problems and solutions are my own. Please let me know via Post Comment if you have any suggestions for improvements, corrections, or requests for other problems. Thank you.

[ this is a work in progress - more problems will be added shortly ]

Grade 7:

What Number	Solution 1	Solution 2
Family Ages	Solution
Nested Fractions	Solution

Grade 8:

Fraction Equation	Solution
Square in Circle	Solution

Grade 9:

Wind Speed

Solution

Grade 10:

Arithmetic Sum	Solution
British Gentleman	Solution

Grade 11:

Parabola Points	Solution
Radical Equation	Solution

Grade 12:

pre-calculus:
calculus:	Intersecting Curves	Solution

Problem a27192a - What Number.

2010-11-12T01:28:00.008-08:00

A quarter of twice a number is three less than the number. What is the number?

Problem a27193a - Family Ages.

2010-11-12T01:28:00.007-08:00

Brother’s and sister’s ages add up to half of their mother’s age. Their father is five times as old as sister, and there are the same differences in ages between father and mother as between brother and sister. If the mother is 42 years old, how old are the others in the family?

Solution a27192a 1 - What Number.

2010-11-12T01:28:00.001-08:00

Problem: A quarter of twice a number is three less than the number. What is the number?

Solution 1:
This first solution method is by trial and error.

We do not want to try big numbers, so we look for the smallest possible. We also assume that we are dealing with positive whole numbers. If something with the number is 3 less than the the number, the minimum possible number seems to be around 4 or 5. Trying 5, twice it will be 10, and a quarter of 10 is not a whole number, which is no good. Instead trying 6, twice it is 12, and a quarter of it is 3. Now if 3 is 3 less than the number, the number must be 6, and we have a solution!

The number is 6.

Comment 1:
Trial and error solutions can sometimes quickly give the right answer and with many problems at this level this is all that is needed. On the other hand it may never give an answer, if we had for example started at 7 and gone up from there. To be sure to get the correct answers, and to know whether there is only one or many answers, the trial and error solution method typically includes making up a table of all the possibilities:

We have found one solution. What if there are more?
Looking at this table we see that we do not need to try odd numbers because then we end up with a fraction. Had we started by calculating N −3 we could have eliminated the first three rows, assuming we want positive numbers.

Well, it does not look very promising for finding more solutions - the quarter
seems to lag further behind N−3 the higher we go. We note that N−3 is increasing
in steps of 2, and is already ahead of the quarter, which only increases by steps of
1. This allows us to conclude that N = 6 is the only possible solution.

Solution a27192a 2 - What Number.

2010-11-12T01:28:00.000-08:00

Problem: A quarter of twice a number is three less than the number. What is the number?

Solution 2:
This second solution method is by equation solving.

We translate the problem text to an equation and solve the equation:

The only possible answer is therefore that the number is 6.

Comment 2:
The equation solving method has the advantage of directly computing the correct answer, and also to guarantee that this is the only answer.

In middle school, the trial and error method is typically used in the lower grades while equation solving is part of the curriculum for the upper grades.

Problem solving activities often lead successful students to the discovery of equation solving techniques, before these are formally taught.

Untitled

2010-11-10T23:20:00.002-08:00

About me

2010-11-10T23:05:00.005-08:00

Born in Stockholm, Sweden, 1951. Studied math at the University of Lund to a level equivalent to a B.Sc. with a major in math.

Married Marita, and immigrated to Canada 1974. Worked as a methods analyst and systems programmer in Montreal for four years.

Moved to Vancouver 1978 to work as a systems analyst for the local phone company. Two wonderful children were born here 1984 and 1988 respectively.

Ventured out as an independent systems consultant in 1993. In addition to some minor ones, worked on three major projects as database administrator in Surrey, systems manager in a TRIUMF research group, and as a systems consultant in ICICS at UBC.

Finally in 2004, I decided to return to my first love, mathematics, and become a full time math tutor. This is perhaps one of my happiest work decisions. There is nothing more satisfying than being able to work with what you love, and at the same time help kids to a better future.

m101m.org

More about me:
A math tutor's favorite sunset.
A math tutor's breakfast read.
A math tutor's lunch, edible fractals.
A math tutor's promenade, inedible fractals.
A math tutor's water bottle.
A math tutor's rock pile.
A math tutor's evening snack.
A math tutor's source of oxygen.
A math tutor's dandelion delight.

Testimony

2010-11-10T22:54:00.003-08:00

Magnus Birkner tutored me through Calculus a few years ago. I found Magnus to be an enthusiastic and knowledgeable tutor. He was very patient in his explanation of calculus procedures. Under his teaching, I passed the course with a First Class mark.

Dawn Stronstad, Burns Lake, B.C.

Memorable quotes

2010-11-10T22:45:00.007-08:00

Here is a collection of some of the more memorable quotes I have heard from my students and their parents during math tutoring.

The quotes have been neutralized to protect the identity of the speakers.

grade 11 student:
"Pythagorean theorem? – I have never heard of it"

grade 10 student:
"geometry is so hard!"

grade 11 student:
"geometry is easy – it is just logic"

grade 8 student:
"I never do any homework"

grade 10 student:
"I have done all the problems in the math book"

parent of grade 9:
"my student must pass the course"

parent of grade 11:
"my student needs an A+"

grade 9 student:
"answer is wrong? – who cares if it is wrong or right – it does not matter"

grade 4 student:
"your pink mechanical pencil is so nice – can I have it?"

grade 11 student:
"does your son have a girlfriend?"

Math links

2010-11-10T22:31:00.002-08:00

Interactive Math:
Wolfram Alpha

Mathematica-based computational knowledge engine

Education:
UBC Math
SFU Math

University of British Columbia Mathematics Department
Simon Fraser University Mathematics Department

Canadian Organizations:
   CMS
   Fields
   PIMS
   CRM
   CAIMS

Canadian Mathematical Society
Fields Institute for Research in Mathematical Sciences
Pacific Institute for the Mathematical Sciences
Centre de Recherches Mathématiques
Canadian Applied and Industrial Mathematics Society

American Organizations:
   AMS
   MAA
   CMI
   AIM
   NCTM

American Mathematical Society
Mathematical Association of America
Clay Mathematics Institute
American Institute of Mathematics
National Council of Teachers of Mathematics

International:
   IMU
   LMS
   SMF

International Mathematical Union
London Mathematical Society
Société Mathématique de France

Why have a math tutor?

2010-11-10T21:52:00.004-08:00

Individualized teaching.

Why have a math tutor when we have math teachers in school?

Schoolteachers are forced to provide a general teaching level that fits most kids in the class.

Kids with special challenges need more individual attention and help from a math tutor that understands their personal situation.

Kids with exceptional abilities need the individual stimulation and extra stretch provided by a private math tutor.

Encouragement.

Why have a math tutor when there are plenty of math books to read for free at the library?
... or there are lots of good math books available in the nearest bookstore?
... or with a computer and internet access, there are lots of web sites and software that teach math to anyone?

Well, who uses these resources?

There are people who could buy a math book and learn its contents by reading it.

But most people, even if they could, are not doing it!

The math tutor encourages people to learn new things, helps them get started, and helps them keep going.

A math tutor can also save valuable time by providing hints and guidance in the right direction.

Clear sailing ...

Why have a math tutor if I have problems with English or Science?

Math is not just adding numbers together and other computations of this sort.

Math provides training in general logical thinking, and few subjects can be done well without thinking well.

Math includes:

thinking about how to solve a problem
planning the solution technique
coming through with the correct solution
presenting the solution for others

Clear sailing in math therefore smoothes the path in other subjects as well.

... expanded horizon ...

Why have a math tutor if I do not need one?

This is like asking: why see a movie, is it necessary?

Or, like the boy who at first did not want to go to summer camp because he could play at home.
Camp was hilarious and he got lots of new fun friends that bettered his life for years.

A math tutor can show you new perspectives, and show you new ways of doing things that will make a difference in all areas of your life.

A math tutor can also introduce you to a wonderful world of fascinating concepts to explore and play with to your hearts content.

Why say math tutor?

Try cognitive coach - coco!

How to get ahead in math.

2010-11-10T21:33:00.001-08:00

How to get ahead in math and
love math!

Math starts with natural ideas and builds on these step by step. By taking each step carefully, making sure to see every feature, great views will be discovered up high.

Why do some kids
hate math?

They were brought along too quickly, trying to skip steps. Now they are tripping and falling. This is no fun.

Math tutoring can
help!

Post a Comment to this blog, which will instead send me an email, to book an appointment for math tutoring, or just to find out more about it.

Discover the fun in math, learn techniques to solve math problems, and enter a new exciting world of math exploration!

Welcome to my blog and web site, all in one!

2010-11-10T20:52:00.005-08:00

This blog is totally moderated by me, and I do not publish any comments posted here.

Posting a comment on any page will just send me an email. This way I can then phone you back, or email you back, provided you have included your contact information.

If you are interested in math tutoring, please specify the school grade level and any specifics regarding the math needs of the student. I tutor only mathematics, and I accept new students in any of the grades 7 to 11.

Typically, I will come to see the student one hour per week, as long as you live in Vancouver. I find that we can usually go through any difficulties during this session, and then discuss the study work to be done by the student for the rest of the week until next time.

Online tutoring is an option that I am interested in exploring, and I will work with anybody who would like tutoring via email, messenger, or even Skype. Since this will be a new experiment, I will not charge anything for this in the beginning, until things are working well.

Important note: my emphasis will always be to make sure that the student learns to do mathematics herself or himself. I will not do assignments for students. Rather, I will teach students how to do their assignments.

[ this blog was tested using Google Chrome - let me know if you see any formatting problems ]

A math tutor's source of oxygen.

2010-11-03T20:34:00.002-07:00

While the black bear is using some, the mighty trees produce enough oxygen for all of us. Even the chemistry tutor who might need some extra oxygen for oxidation experiments gets some.

Long live the trees!!!!

A math tutor's rock pile.

2010-10-29T16:17:00.003-07:00

When climbing Mount Baker, Washington, a couple of weeks ago, I decided to build a monument as a memory of my October 2010 visit. I call it my 8 sl8 m8.

no advice for young math students!

2010-09-28T18:31:00.003-07:00

One of the most famous contemporary research mathematicians has no advice for young math students! See: http://www.math.ubc.ca/Dept/Newsletters/Robert_Langlands_interview_2010.pdf where this question and answer sequence appears:

Farzin Barekat: What advice would you give to aspiring mathematics students?

Professor Robert Langlands: None. In so far as I accomplished anything of value, I did it by following my own inclinations. I hope they do so too.

What does this mean? Does the professor not care about students?

It seems to me that he definitely cares a lot about students, and he wants to give them the maximum benefit of total academic freedom. In addition, I believe there is a deeper insight implied here as well: those who are truly capable of doing outstanding original research will do it, whether told to or not, as long as they are not substantially hindered by outside circumstances.

What about regular students, who are not research prodigies? We can all use some encouragement, I think, and especially students struggling with a busy schedule and demanding courses. Gentle guidance as what to focus on, as well as how to proceed through difficult parts, are almost requirements to succeed, and make the best use of time, in our competitive society.

A math tutor's favorite sunset.

2010-06-18T01:57:00.001-07:00

A math tutor's evening snack.

2010-06-07T23:29:00.002-07:00

A math tutor's breakfast read.

2010-06-06T17:28:00.002-07:00

A math tutor's water bottle.

2010-06-04T13:49:00.002-07:00

The Origin.

2010-06-03T00:30:00.002-07:00

Some people think of the Origin merely as just a dot, a point, or an intersection of the axes. However, a closer look may reveal more details.

Infinite rows of organically grown Local Maxima.

2010-06-02T23:17:00.001-07:00

A math tutor's promenade, inedible fractals.

2009-03-22T23:48:00.002-07:00

A math tutor's lunch, edible fractals.

2009-02-10T13:59:00.001-08:00

enough math books?

2009-02-01T20:02:00.003-08:00

Why are there so many math books? There are at least three main reasons for so many books; 1) adventurers (aka math students) need to take their own notes to learn math well, and some eventually publish their notes in books, 2) math writers come up with new math, or better ways of explaining existing math, 3) publishers need to maintain their textbook market share when courses change.

New Beginnings!

2009-01-05T19:26:00.000-08:00

Happy New Year! Searching for "Introduction to" on Amazon.com gives no less than 733,199 books on all subjects. "Introduction to Calculus" finds 3,800 books! Typing “Introduction to Mathematics” brings up 20,355 books. No shortage of study materials for the interested adventurer!

adventure land

2008-11-18T00:30:00.000-08:00

Math students are adventurers. Unless they seek to complete a specific education, or seek the truth, in which case they could be called seekers. Otherwise, math is a largely uncharted adventure land where people can get lost, find fame, or just have a good time.

discoverers

2008-11-17T23:54:00.000-08:00

Here is what I think; mathematicians, and any student of math actually, are discoverers, rather than inventors. The world of math facts is out there for anyone to discover. And it exists whether anybody knows about it or not. The idea that math is created as a figment of the mind, as some philosophers of mathematics imagine, is in my opinion rather absurd. The universal agreement around the world of what is true mathematics is a strong argument for the independent existence theory.

2nd question:

2008-11-17T12:35:00.001-08:00

Are math students seekers or adventurers?

Question to ponder:

2008-10-26T22:26:00.000-07:00

Are mathematicians discoverers or inventors?

No time ?

2006-08-24T00:47:00.000-07:00

Once, Twice, Thrice, ?