PROBLEM V. If two Post-Boys A and B, at 59 Miles Distance from one another, set out in the Morning to meet. And A rides 7 Miles in two Hours, and B 8 Miles in three Hours, and B begins his Journey an Hour later than A; to find what Number of Miles A will ride before he meets B.

Math teachers would probably find the substance of this word problem, if not its expression, to be comfortably familiar. They may be surprised to learn, however, that this particular problem comes from the 1769 edition of an English translation of a textbook written in Latin by Isaac Newton (1643–1727).

Celebrated as the inventor of calculus and the author of Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy), Newton also prepared lecture notes during the time that he was at the University of Cambridge as the Lucasian Professor of Mathematics. All undergraduates were required to attend the Lucasian lectures, starting in their third year.

In 1669, Newton had made extensive notes on an algebra textbook written by Gerard Kinkhuysen and published in 1661. Like other 17th-century textbooks on solving equations, Kinkhuysen's book built on traditional material from the previous century, with the addition of some new ideas from RenĂ© Descartes and Frans van Schooten.

Newton used these notes later when he compiled his Lucasian lectures, organizing the material in much the same way that Kinkhuysen had in his book. It seems unlikely that Newton ever delivered the lectures, but William Whiston (1667–1752), Newton's successor as Lucasian Professor, edited and published the notes in 1707 as the book Arithmetica universalis.

Newton himself was unhappy with the publication of his notes and refused to let his name appear on the book. When the book was printed, he apparently threatened to buy all the copies so that he could destroy them. The book was later translated into English by Joseph Raphson (1648–1715) and published in 1720 as Universal arithmetick.

A second, amended edition of Arithmetica universalis was published in 1722. Newton's name first appears on the 1761 edition of Arithmetica universalis, printed well after his death.

Arithmetica universalis covers a wide range of elementary topics, including algebraic notation, arithmetic, the relationship between geometry and algebra, limits and infinite series, the binomial theorem, and the solution of equations, including the extraction of roots.

Because Newton's book closely follows the structure and content of Kinkhuysen's Algebra, it has "the appearance of being far more elementary than in fact it is," writes Jacqueline A. Stedall of The Queen's College, Oxford. "To the traditional ideas expounded by Kinkhuysen, Newton added some far-reaching insights of his own."

Nonetheless, she adds, Newton often "described rules or procedures without explanation, giving rise to extensive commentary later on."

Today's high school teachers could profit from pondering the way Newton handled fundamental operations and questions of arithmetic and the wide array of examples and problems that he posed, says mathematician Michel Helfgott of East Tennessee State University.

Helfgott and George Baloglou of the State University of New York at Oswego themselves went back to Newton's Universal arithmetick when they recently investigated relations among a triangle's area, perimeter, and angles. They generalized Newton's derivation of the formula x = P/2 – 2A/P, expressing a right triangle's hypotenuse, x, in terms of its area, A, and perimeter, P. The results are in their paper "Angles, Area, and Perimeter Caught in a Cubic," published earlier this year in Forum Geometricorum.

Here's Newton's solution to the problem posed at the beginning of this article, as published in the 1769 edition of Universal arithmetick.

## March 11, 2008

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## 2 comments:

Newton's method is almost surely the method any algebra text would

follow: make the goal a variable to be solved for, constrain it, and find the space of solutions to that set of constraints.

But if there had been two questions, the first being how far did A travel before B started, one might be tempted to bypass algebra and just say 7/2 miles, this being how far A travels in an hour.

Had one taken this digression and then asked when A and B will meet, one might decide to divide the remaining distance of 59 - 7/2 = 111/2 miles by the combined velocity of 7/2 + 8/3 = 37/6 miles per hour to arrive at

111/2 * 6/37 = 9 hours after B started, which is 10 hours after A's start. But then it is obvious how far A travels, namely this time multiplied by A's velocity, 10 * 7/2 = 35 miles.

So this particular problem might not the ideal advertisement for algebra, since this approach entailed no variables at all.

One then wonders, what test could Newton have applied to recognize that his problem had this shortcoming? While I don't propose to answer it, it seems like a nice question for the producers of algebra problems, if not for their consumers in the trenches who must assign them to their students.

On the other hand problems with this "weakness" might be a great way of picking out the real mathematicians in the class as those who saw the variable-free approach as being easier than following the rules.

Unfortunately it may be smarter yet for the student to follow the rules so as not to risk losing points for a nonstandard solution even when the alternative is obvious. It is a rare teacher that can convey when it is ok to paint outside the lines in mathematics, and rarer yet one that can not only recognize when a student is smarter than the teacher but can still tell when that student is right or wrong.

Yes, I feel many algebra problems can be reduced to arithmetic.

You essentially solved and reasoned through

this algebraic equation [(7/2)+(3/8)]x + 7/2 = 59

I feel the standard way would be

r t =d

A's rate is (7/2). So a travelled for x+1 hours for a distance of that product.

B's rate is (8/3) so he travelled for x hours for a distance of that product.

Then (7/2)x+1 +(8/3)x = 159

This can be reduced to logic and arithmetic also. After A travels for 1 hour,

they are 159 - 7/2 miles apart. A now goes 7/2 miles per hour for the same number

of hours as b goes 8/3 miles per hour. SO together they go (7/2 + 8/3) miles every hour for

the same number of hours. Know it is essentially the rule of three which is taught in arithmetic.

(7/2+8/3) ; 1 hours as 159-7/2 is to the number of hours b travelled. And so on. I'm sure there are

many other ways to solve this problem too.

x=9 so (7/2) (X+1) = 35

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