Marveling At The Historical

Math Oldies But Goodies

  • About This Blog

    This blog is mostly about math procedures in textbooks dated from about 1825-1900. I’m writing about them because some of the procedures are exquisite and much more powerful, and simpler, than some of the procedures in current text books. Really!

    I update this blog as frequently as possible ... every 2-3 days. And, if you are a lover of old texts and unique procedures, you might want to talk to me about them, at I’m not an antiquarian; the books I have are dusty, musty, brown-paged scribbled-in texts written by authors with insights into how math works. Unfortunately, most of their procedures have vanished. They’ve been overcome by more traditional perspectives, but you have to realize that at that time, they were teaching the traditional methods.

Archive for the ‘Historical Math’ Category

Bringing historical math into the modern

Concrete to Abstract

Posted by mark schwartz on April 16, 2017


When presenting operations with signed numbers, an instructor must deal with the issue of notation as well, to allow for the plus and minus having two different meanings; this has to be addressed. I spent a long time playing with this until I found a way for students to ‘see’ the difference. Given a number line which I call a road and a car which they can drive on this number-line-road, the car can be put into drive or reverse, So the direction of moving is with reference to the car. When the car is put on the number-line-road, it can face forward (toward the positive) or back (toward the negative), so the facing is with reference to the number line. Students twiddle with this a little but eventually get it. It seems clear that facing (positive or negative) is different from moving (positive or negative).

I wrote a piece in this blog (Driving the Integer Road), a somewhat long detailed almost lesson plan which describes how all this works but I never present the traditional notation until I’ve gone through a bunch of exercises which I call facing and movement. What’s interesting about it is that students seem to appreciate and understand the differences now between plus and minus in terms of the operation or the sign of the number. Although other instructors may use something like this I haven’t seen it presented anywhere. I’d never seen it in my collection of 1800s texts either … until recently. Here’s how Durell and Robbins, in a very similar way had people walk a number line. They omitted some detailed explanation but I will discuss this.

The Story

In Durell and Robbins 1898 School Algebra Complete (pgs. 20-21) they have the students visualizing walking on a number line. The students don’t actually walk the line but only visualize it. Before this ‘exercise’, the authors point out “ … the signs + and – are employed for two purposes – first, to express positive and negative quantity; and second, to indicate the operations of addition and subtraction.” This prompts the students to pay attention to the notation in a problem. I’m now going to paraphrase what the authors did to show students the relationship between walking on a number line and the traditional +/- notation.

On a number line with A at zero, B at + and C at – , a person walking from A toward B a distance of 5 units and then walking back toward A a distance of 3 units, has in total walked a distance of 2 positive units from A, or zero. In notation, the authors write “+ 5 + (- 3)”, essentially the sum of a positive and negative quantity. They point out that this is symbolically what was done on the number line. They discuss this in terms of positive and negative distance and demonstrate that “Hence, we see that adding negative quantity is the same in effect as subtracting positive quantity; therefore in the expression 5 – 3 the minus sign used may be considered either a sign of the quantity of 3, or as a sign of operation to be performed on 3.” That’s a very powerful statement and hopefully when instructors used this text, they emphasized this point because what this really does is show that + 5 + (– 3) = +5 (+ 3) = 5 – 3. The authors don’t detail this expression; they just state it … but it can be shown by walking on their number lines

Given that the authors point out “ … the signs + and – are employed for two purpose…” when they wrote the activity as + 5 + (– 3), were consistent with their own schema by designating the – as the sign of the number and the + as the sign of the operation. That’s interesting because the ‘walker’ is facing in the negative direction with reference to the number line, while walking forward with reference to his own movement and the authors don’t mention this. Their expression of the walking activity could have been written as + 5 –  ( + 3), meaning that the walker walked in a negative direction with reference to the number line, while facing in a positive direction with reference to his own movement. This is a subtle difference but consistent with their schema.

Look at the expression + 5 – (+ 3). With the ‘walker’ starting at A and taking 5 steps toward B, this is +5. If the walker does not change the direction he’s facing, then he could – with reference to his own movement – step backwards 3 units. This is the operation and thus the – outside the parentheses. So, facing forward with reference to the number line is the sign of the number, thus +3.

“With reference to” becomes a critical phrase in parsing one’s way through this demonstration of what the authors have done. Perhaps it’s too subtle for a class discussion but from my experience, this subtlety seems to make an appearance when students talked about the fuzziness in all the operations with signed numbers.

In summary, Durell and Robbins in 1898 captured the core elements of my “Driving the Integer Road” but didn’t explore the subtleties of the notation when talking about the sign of the number and the sign of the operation. I would urge instructors to explore ways of making concrete the ‘abstract’ use of – and + for this as well as other math relationships.

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One 1873 View of Percent

Posted by mark schwartz on February 18, 2017


I’m writing this short piece to give you the flavor of what students had to do when they studied percent using Rev. York’s 1873 The Man of Business and Railroad Calculator. Today’s texts typically present one formula for percent and then discuss variations on it, like percent increase, decrease, percent proportion or finding values given a percent. Also, typically, the problems are very similar to the examples given and rarely, if ever, include fractional values. Rev. York presents a much more demanding idea.

The Story

In his book he discusses percent across 11 pages, making 13 different conditions (like ‘given x, find y’) and ends the discussion with a presentation of 8 formulae. In essence, these 8 formulae are simple variations on ‘percent = part/whole’ but his presentation gives the appearance that these 8 formulae are to be used depending on the nature of the problem. In addition to the typical presentation of percent in today’s text, you can see my concept of percent proportion in this blog (see Percent Proportion). At no point does he state the basic, simple relationship algebraically.

The best way to show what students had to do is to list the kinds of problems he presented. I’ve included the answers as well. I’m not going to list his 8 statements. Let me remind you that students in the 1870s had no calculators and that the work Rev. York presented suggests the importance of mastering fractions. At that time, units of measurement weren’t as standardized and a lot of conversion between systems involved fractional relationship.

The problems as he presented them are below; the answers are at the bottom, in the event you want to play with the problems.

  1. What percent is 1/4 of 2/5?
  2. If a merchant sell calico at 12 1/2 cents per yard and makes 12 1/2 percent. ; what did it cost per yard?
  3. If I sell an article for $250, and make 125 percent; what did it cost me?
  4. One of the stockholders of a rail road company owns 19 shares of $50 each; the dividend is declared to be 7 1/2 per cent premium; what ought he to receive?
  5. If I sell 4/7 of an article for as much as I paid for 2/3 of it; what percent did I make?




  1. 62 1/2 percent
  2. 11 1/2 cents
  3. $111 1/9
  4. $67.50
  5. 14 2/7 percent

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Recognize x∧2 – x – 1 = 0?

Posted by mark schwartz on February 12, 2017


If you recognize the equation in the title then you are already familiar with the golden ratio of 1.618. There are a number of books that explore the occurrence of the golden ratio in nature, architecture, the pyramids, art, aesthetics and a variety of other suspected places. I’d like to explore something about 1.618 that I’ve not seen in any books; just something to add to the mythology of it. If you aren’t that familiar with it, just Google it and have fun.

The Story

When I was teaching a pre-service teacher’s class about math concepts, I introduced the golden ratio because it was a nice vehicle for demonstrating how a pattern can show up in a lot of places. I talked a lot about teaching students to explore for patterns because math is saturated with patterns and relationships.

I asked if anyone in the class was involved in mysticism, the occult and if they would feel comfortable talking to us about it. There was one person who spoke up. I directed the conversation to focus on the pentagram (a five pointed star inscribed in a circle) because I knew that it was a significant symbol. I asked her to talk about the significance of the pentagram and of all the stuff she talked about, the golden ratio wasn’t mentioned. I mention it because it’s a somewhat invisible yet significant characteristic of the pentagram. Let me start with the star.

From here on, you will be asked to do a lot of ‘construction’ but it doesn’t have to be elegant or perfect and hopefully my directions can be easily followed. I just believe that your participating – as I had my students do – will be entertaining and educational, not drudgery. There is some geometry and trigonometry involved so if you don’t recall stuff, I’ll provide the information needed to do the work.

You can ‘construct’ a five-pointed star, starting by sketching a circle. Don’t make it too small because some of the points and lines will need room to be labeled. When you get done, you will have constructed a pentagram.

Starting at 12 o’clock, place 5 equidistant points on the circle. Starting with the point at 12 o’clock, label that one A and then going clockwise label the others B, C, D, and E. Now, draw a straight line from A to C, then a line from B to D, and continue with C to E, D to A and E to B. You should have a somewhat respectable looking 5 point star. I want to highlight that this pentagram has a pentagon in the middle and each face of the pentagon has a triangle on it. Take the triangle at the top that has the vertex A and label the other two vertices in that triangle F and G. Still with me?

Now if your pentagram were a perfectly crafted one, the pentagon and each triangle would appear to be equilateral. The pentagon would be but not the triangles! What? – they look equilateral. How can we determine that they are not?

To do this we need to know something about the angles in triangle AFG. Allowing that AF and AG are equal to 1 (we could pick any arbitrary value and the outcome of this exercise wouldn’t change; it’s just that 1 simplifies the calculations). Geometry allows us to determine that angle A equals 36 degrees. Now drop a perpendicular from vertex A to side FG, labeling that point on FG as H. Why? Well, we just formed a right triangle AFH which allows us to use some trigonometry to determine the length of FH. The perpendicular also bisects angle A, so angle FAH is 18 degrees. The sine of this angle is FH over 1, or just FH. The sine of 18 degrees is .809 – the length of FH – so the length of FG is .618, thus demonstrating that this triangle and all the triangles in the pentagram are not equilateral.

Given this, the perimeter of triangle AFG is 1 + 1 + .618, or 2.618. If you’re not familiar with the golden ratio, let me note a few relationships. First, the square root of 2.618 is 1.618 and clearly 1.618-squared is 2.618. And, as trivial as it seems, you can see that 1 + 1.618 equals 2.618.

And now the point of this exercise! Recall that the title of this article is x2 – x – 1 = 0. If you solve this equation, you get 1.618. But substituting 2.618 for x2, 1.618 for x you now have 2.618 – 1.618 – 1 = 0, which is a true statement for the perimeter of the triangle. So, triangle AFG and all the other triangles in this pentagram contain the golden ratio. But, there’s more.

Look at the pentagon in the pentagram. I won’t ask you to do all the construction (you can do it if you like) but I’ll step through it to show that the pentagon also contains the golden ratio. Briefly, starting with point F and moving clockwise, the vertices of the pentagon are G, I, J, K. Drawing a straight line from K to G, forming triangle KFG and assigning KF and FG a value of 1, what is the value of KG? As with the star construction, geometry lets us know that angle KFG is 108 degrees, and a perpendicular dropped from F to KG (label this point M) bisects the angle, and results in a right triangle KFM with angle KFM equal to 54 degrees. Again, trigonometry lets us determine that the length of KM is .809, and thus the length of KG is 1.618.

So, what is true for each of the triangles is also true for the pentagon, although it’s not the perimeter of the pentagon, but a construct (a ‘chord’ of the circle too) within the pentagon.

And one more thing. All of this analysis came about because I was watching the National Geographic Channel and the show was all about creatures of the sea. One of the creatures shown was the star fish. If you connect the tips of the starfish arms you get a pentagon!

So, in addition to the pentagram having a powerful influence in the occult, perhaps reinforced by the presence of the golden ratio, nature again (using a little imagination) provides another possible example of the ‘hidden’ sway of 1.618.

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A Short, Short Discourse on Digit Sum

Posted by mark schwartz on January 16, 2017


My daughter and I play with numerology; it’s just play, nothing serious. We play with calendar dates, prime numbers, birthdays, etc., looking for patterns and such. Most recently she texted me that she added the digits in my wife’s birthday and continued to add them until there was a single digit, the result was 4. She wasn’t aware of digit sum nor casting-out-9s. So, I played back.

The Story

I first took a look through some of my old math texts, dating from about 1850 to 1900. I looked there because in those days, having a way to check your work was important and digit sum was popular, but not noted in all texts. What I don’t know is whether instructors may have taught it although it wasn’t in the text. I did find a few (I have about 75 old texts) that actually demonstrated how addition can be checked using digit sums. Oddly though, none of those that presented how to check addition indicated that digit sum can be used to check subtraction, multiplication and division as well. Yes, it can.

But, let’s take a look at why digit sum works. It’s based on what is called casting-out-9s. In essence, given a number – 23 – if you cast out 9s (which can be done by subtracting 9 until you have a single digit), you get 23 – 9 = 14, then 14 – 9 = 5. Notice that if you simply added the 2 and 3 in the number 23, you also get 5, so what simplifies getting a digit sum is simply add the digits repeatedly until you get a single digit. For the number 268, at first you get 16, then 7.

Why does this work? Let’s get basic. Using 23 again, this is really 2(10) + 3(1). Rewriting this in what I call ‘slow motion’ math, it becomes 1(10) + 1(10) + 3(1), then 1(9) + 1(1) + 1(9) + 1(1) + 3(1). If the ‘1(9)s’ are now ‘cast out’, the result is 1(1) + 1(1) + 3(1), giving 5(1).

This of course is not a rigid proof but rather a demonstration of casting-out-9s.

For example, 235 + 568 = 803. The digit sum for 235 is 1; the digit sum for 568 is 1, and the sum of these is 2. This equals the digit sum of 803, so it checks. I realize that in today’s technical world, this procedure isn’t likely to be taught nor used but in olden days without calculators, it seemed reasonable to check your work.

Now back to what I sent back to my daughter. I generated the digit sum for the birthdays for all the members of our family and then generated the digit sum for the sum of them and lo and behold the result was 1! Of course, the family is unity!

Told you it was a short discourse, but couldn’t resist sharing it. Can’t wait to see what she comes up with next.


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Yet Another Subtraction Algorithm!

Posted by mark schwartz on November 4, 2016


I recently posted Revisiting Mr. Stoddard’s 1852 Subtraction. In that posting I modified Mr. Stoddard’s idea by introducing a procedure which allows for subtraction without borrowing. This posting modifies that modification.

The Story

I’ll use a simple subtraction example to demonstrate the procedure, but I have examined much more sophisticated problems such as 20801 ̶ 278 and the procedure is still good.

Basically, treating ‘ab’ as a 2-digit number and ‘c’ as a single digit number, in the problem “ab ̶ c”, if c > b, the answer to ‘b ̶ c’ is 10 ̶ ( c ̶ b ) and then add 1 to the 10s place value in the subtrahend. For example, 12 ̶ 8 gives 10 ̶ (8 ̶ 2), or 4, then add 1 to the 10s place value in the subtrahend, giving 1 ̶ 1 or 0, which isn’t written.

What I didn’t note clearly are two things. First, if in that example, b > c, then write down that value as the answer. Do not add 1 to the next place value in the subtrahend. However, if c > b, then the algorithm as noted is to be used. And here’s the modification – continue with this algorithm!

Here’s an example in slow-motion math. Using the problem 7234 ̶ 567 as a traditional ‘vertical’ problem, we hav


In the 1s column, 7 is greater than 4, so the answer is 10 ̶ ( 7 ̶ 4) which is 7. Add 1 to the 6 in the subtrahend 10s column. Then in the tens column, 7 is greater than 3, so the answer is 10 ̶ ( 7 ̶ 3), which is 6. Add 1 to the 5 in the subtrahend 100s column. Then in the 100s column, 6 is greater than 2, so the answer is 10 ̶ ( 6 ̶ 2), which is 6. Add 1 to the zero in the subtrahend 1000s column. Then in the 1000s column, 7 is greater than 1, so the answer is simply the difference of 6. The solution looks like this:

– 567

There are many subtraction algorithms posted in this blog and most of them focus on avoiding the need to borrow, so if you feel like trolling through the entire blog and compiling them, you might find one you like.

Posted in basic math operations, Historical Math, math instruction, mathematics, remedial/developmental math, subtraction | Tagged: , , , | 1 Comment »

Heron’s Area of a Triangle

Posted by mark schwartz on October 24, 2016


This is a brief story about a fun event that almost always happens when discussing the area of a triangle. The formula for the area is A = 1/2 bh, where ‘b’ is the base and ‘h’ is the height.

The Story

Simple enough if it’s a right triangle and the base and height can readily be seen or calculated from a2 + b2 = c2. But what happens when it’s not a right triangle? Well, one has to wiggle around a bit and do a few more calculations to determine the height, but it can be found.

But being an instructor that likes to stretch students thinking and imaginations, I draw a very scalene triangle with sides of 4, 8, and 10 and present them the task of finding the area. As I roam the room watching them work and listening to their grumblings, I ultimately have them stop and present to them Heron’s formula. I don’t bother with the derivation and for those that are interested in knowing it, I recommend Googling it.

The formula is A = 1/4 the square root of (a + b + c)( ̶ a + b + c)( a ̶ b + c)( a + b ̶ c).

This is a really nifty formula because of the pattern in it – add all the sides together, then in the next three parentheses, just negate each side, one at a time, in order. And, given that there are 4 parentheses, just divide the square root of this product by 4.

In class, I first apply the formula to a 3, 4, 5 right triangle to demonstrate how it works. Then we play with a few other right triangles to get comfortable with the formula. Then we return to that weird scalene triangle. Using Heron’s formula, we get an area of 15.1987 … and now the fun begins!

Someone typically asks “how do you know this is correct?” So, this sets us up to explore how to find the area if we didn’t have Heron’s formula; so we do all the Algebra necessary to demonstrate that indeed 15.1987 is correct. In one class, this whole presentation got applause and I take that as their having had fun with it.

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A 1st Day Handout to Students

Posted by mark schwartz on October 17, 2016


Author’s Note: The following is literally a handout given to students the first day of class. I give them time to read it and then we talk about it. The discussion set the tone for their learning and the idea of freedom was a surprising but satisfactory idea, although scary to some who expected this class to be like all previous math classes. What follows is the handout.

In the 1960s, a book titled “Freedom, not License” hit the bookstores. Briefly, it’s a story of the core philosophy of a school named Summerhill in England. The title refers to a subtle distinction between two conditions: freedom – being able to determine your own behaviors, live with the consequences, be self-determining, guided by your own internal discipline and control; and license – interpreting the circumstances in which you are allowed, permitted and “controlled” by an external authority. Actually, it’s misinterpreting the freedom as license, whereby the misinterpretation leads one to rely on external events, rather than understand the freedom to govern one’s own behavior and actions. License also is interfering with other’s freedom.

I give you freedom to succeed but it has to be your success, not driven by external rewards and punishments. I will teach well and you have to learn to learn well. Don’t rely on me to chase you down the hall demanding that you get assignments done on time. That’s your responsibility. Don’t rely on me to threaten you with loss of grade if you don’t attend class. Attendance is your responsibility. Don’t rely on me to control the classroom as is done in elementary school; hushing the noisy, punishing the “unruly”. It’s your responsibility to respect the classroom environment and not disrupt my teaching or the learning of others.

Freedom is a little scary if you’ve never experienced it in a classroom. But consider it a responsibility just like driving. You’re responsible for your car – for its maintenance and performance; for driving responsibly within the wide legal constraints of the speed limit, parking areas, passing, not drinking while driving, etc.

According to the Oxford English Dictionary, “education” is derived from its Latin root, “educare”.  Educare means “to rear or to bring up”.  Educare itself can be traced to the Latin root words, “e” and “ducere”.  Together, “e-ducere” means to “pull out” or “to lead forth”.  Hence we use the word “educare” to communicate the teaching method through which children and adults are encouraged to “think” and “draw out” information from within.

Notice the last three words: “information from within”. It is within you to learn well and to learn any subject well. I can help you draw it out, but the “you” is the important word in that sentence. You have to attend class, do the assignments, and act respectfully toward yourself and all others in the classroom.

Let me repeat – freedom is scary if you’ve never experienced it in the classroom. I will not check your classwork to see if you’ve done it and it is correct; answers are in the text. I will work with you if your answers are incorrect. You’re responsible for that and it will be hard for you to accept that responsibility because it will be tempting to leave class early and not do it because math makes you uncomfortable and anxious. But I can help you address the lack of math skills that lead you to feel that way.

My teaching doesn’t automatically lead to your learning. But take the freedom offered and use it; don’t let it become license that interferes with your learning.

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A Student’s Aha Moment

Posted by mark schwartz on October 12, 2016


One more example of how an aggregation of student’s imaginations led the class, including the instructor, to apply Algebra to Algebra. A type of problem which frequents texts in the U.S. is the ‘mixture’ problem. Mixing solutions of different concentrations to get a third with a desired concentration; mixing different kinds of candy or nuts to get a mixture to sell at a certain price, or getting a return on investing in two accounts at different interest rates (and other applications).

The Story

Gus has on hand a 5% alcohol solution and a 20% alcohol solution. He needs 30 liters of a 10% alcohol solution. How many liters of each solution should he mix together to obtain the 30 liters?

The classical solution is to write the equation .05x + .20(30 – x) = .10 (30). Solving the equation, the outcome is 20 liters at 5% and 10 liters at 20%.

As this type of problem was discussed, one student asked “can 15+x and 15-x be used, since they add to the total of 30 and this seems easier?” I wasn’t certain if it would work nor why it seemed easier, but we explored the idea. The student presented the problem on the board as 5(15 + x) + 20(15 – x) = 10 (30). Not only did he re-craft the unknowns, but he used whole numbers, not decimals. When asked why, he simply stated that we would be getting rid of the decimals anyway. I noted that ‘getting rid of’ is not a mathematical operation, but clearly it works. The answer to this equation is -5, and some students believed this solution to be awkward for two reasons. First because x = -5, and a negative quantity doesn’t make sense and second, because finding x doesn’t finish the problem. Recall that the unknowns in the equation are 15+x and 15–x, so another step is required to come to the correct answer of 20 liters at 5% and 10 liters at 20%. Then the question came: “Isn’t there another way to do this?”

At this point, I introduced “Alligation”, a procedure described in detail in a post in this blog titled Mixing it up with Alligation. I won’t go into detail about the procedure, so look it up if you’re interested. It’s a very different approach which was popular in the 1800s but doesn’t seem to be in any current texts.

We did discuss several other ideas and it was an enjoyable session in which the class actually reported having fun doing Algebra!

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Revisiting Mr. Stoddard’s 1852 Subtraction

Posted by mark schwartz on September 29, 2016


In this blog is a posting Mr. Stoddard Subtracts in 1852. If you haven’t read it, you don’t need to (but of course you can!). Mr. Stoddard presents an idea in subtraction which avoided the need for “borrowing”. For some reason, I was playing with a subtraction idea and after I had written out the entire algorithm, I realized that I basically had modified Mr. Stoddard’s; thus the title.

The Story

I’ll use a simple subtraction example to demonstrate the procedure, but I have examined much more sophisticated problems such as 20801 ̶ 278 and the procedure is still good.

Basically, treating ‘ab’ as a 2-digit number and ‘c’ as a single digit number, in the problem “ab ̶ c”, if c > b, the answer to ‘b ̶ c’ is 10  ̶  ( c ̶ b ) and then add 1 to the 10s place value in the subtrahend. For example, 12 ̶ 8 gives 10 ̶ (8 ̶ 2), or 4, then add 1 to the 10s place value in the subtrahend, giving 1 ̶ 1 or 0, which isn’t written.

Here’s why it works. In essence, it could be said that borrowing has happened but it’s hidden as well as not written!

In essence, 10  ̶  ( c ̶ b ) is borrowing, but it’s hidden. The ‘10’ in the 10 ̶ (8 ̶ 2) could be said to have been borrowed from the 10s column in the minuend. Given that, that ‘10’ can be said to have been subtracted from the10s column in the minuend. It’s known that if the same value is subtracted (or added) from both the minuend and subtrahend of a subtraction problem, the answer will be the same. Thus, adding a 1 to the next place value in the subtrahend adds a value which will be subtracted.

There it is. It’s a mild modification to Mr. Stoddard, but my ‘aha’ moment with 10 ̶  ( c ̶ b ) may well have been his idea incubating all this time. Try it with other problems – like 20801 ̶ 278 and after a while it becomes as automatic as doing the problem using borrowing.

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Vedic Version of a Line From Two Points

Posted by mark schwartz on September 25, 2016

In Vedic Mathematics (revised edition, 1992) a very interesting algorithm is presented. It allows one to find the equation of a line in standard form by visually examining the values of the two points, doing a little mental calculation, and writing down the equation! One need not use the slope-intercept or the point-slope formula.

Given two points (a,b) and (c,d), the vedic version (pg. 343) is: x(b-d) – y(a-c) = bc – ad

A slight notation change gives the standard form (ax+by =c), thus (b-d)x – (a-c)y= bc – ad

For example, using the vedic version with (9,7) and (5,2) the equation is:

(7 – 2)x – (9 – 5)y = 7⦁5 – 9⦁2, giving 5x – 4y = 17.

I was curious about this because it looked familiar; basically, the difference in the y-values is the x-coefficient and the difference in the x-values is the y-coefficient. The constant is the ‘inner’ minus the ‘outer’, if you are familiar with FOIL. As I played with this, I realized that the vedic algorithm could be derived from combining the slope-intercept and the point-slope formulae. Starting with the point-slope formula, one gets:

(y – y1) = m(x – x1

(y – y1) = ((y2 ̶ y1)/(x2 ̶ x1)) (x – x1)

(x2 – x1) (y – y1) = (y2 – y1)(x – x1)

(x2 – x1)y – (x2 – x1)y1 = (y2 – y1)x – (y2 – y1)x1

– (y2 – y1)x + (x2 – x1)y = (x2 – x1)y1 – (y2 – y1)x1

– (y2 – y1)x + (x2 – x1)y = x2y1 – x1y1 – x1y2 + x1y1

 – (y2 – y1)x + (x2 – x1)y = x2y1 – x1y2

 (y2 – y1)x – (x2 – x1)y = x1y2 – x2y1

 -1(y1 – y2)x – (-1)(x1 – x2)y = (-1)(x2y1 – x1y2)

(y1 – y2)x – (x1 – x2)y = x2y1 – x1y2

This form (y1 – y2)x – (x1 – x2)y = x2y1 – x1y2 is the vedic form (b-d)x – (a-c)y = bc – ad.

Furthermore, this vedic form allows one to generate the equation of the line if given the slope and a point, or a point with a line perpendicular or parallel to a given line because a second point can be found from the given point and the slope.

Using the same example as above, if presented the point (9,7) and the slope 5/4, the second point is (9 + 4, 7 + 5), or (13,12), as well as (9 – 4, 7 – 5), or (5,2). Consider using this vedic version.

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