New math of the 70-ies. Yikes, or nostalgy ?

[Fredrik]

My kids are being taught "lattice multiplication", which is new for me- so I guess that's the new new math :).

I had to google it, and found this:
http://www.coolmath4kids.com/times-tables/times-tables-lesson-lattice-multiplication-3.html

It is "standard" multiplication, but turned over 45 degrees

The math-education folks I talked to about it uniformly love lattice multiplication-

I enjoyed reading about this, mostly because when I read it I realized that I have no idea how I was taught to multiply numbers in school. I have completely forgotten it. When I need to multiply 14 and 56 (the example from the page vanesch linked to), I certainly don't use "lattice multiplication" or whatever algorithm I was taught in school. I just do this in my head: 14\cdot 56=\underbrace{10\cdot 56}_{=560}+4\cdot 56=560+\underbrace{4\cdot 50+4\cdot 6}_{=200+24=224}=560+224=784 Isn't this how every math nerd does it? Lattice multiplication and similar algorithms seem pretty useless to me, at least as long as we're talking about two-digit numbers. Maybe they're more efficient than my method when we're dealing with four-digit numbers, but I'm not convinced that this is a great reason to teach these algorithms. The method I use for four-digit numbers is to type them into the Google search box in the upper right corner of my Firefox. I'm satisfied knowing that I could do it with my method if I wanted to.

 

[mathwonk]

the reason i prefer the way i showed is that it explains every step in the formula, i.e. why there is a 4 [because the difference between (r+s)^2 and (r-s)^2 is 4rs], why there is a square root [because you are finding the squared difference of r and s], why you add and subtract the square root [because you have r+s and r-s and want to eliminate r or s], why you divide by 2 [because by adding and subtracting you are getting 2r and 2s],...


tiny tim, why do you prefer the usual completing the square method? That was the method i was taught in high school and it always made it seem unmemorable, because to me it is unmotivated. i.e. it is clever but why do you do those things? i.e. how do you think of it? but i was never strong at algebra.

there are also constructions in euclidean geometry where the completing the square method is easier to translate into a geometric construction. such as euclid's construction of a regular pentagon, where he completes the square geometrically to solve the equation

X^2 = R(R-X). I.e. Euclid solves this equation geometrically by completing the square,

i.e. by writing it as X^2 + RX = R^2, he completes the square by bisecting segment R and writing it as (X+ R/2)^2 = X^2 + RX + (R/2)^2 = R^2 + (R/2)^2, then he solves it by constructing the right triangle with sides R and R/2, since then the hypotenuse must be (X + R/2), so we get X by subtracting R/2.

This is quite easy to do geometrically in a circle of radius R by constructing two perpendicular diameters, bisecting one radius, and connecting its midpoint to the point where the other diameter meets the circle. I.e. this segment is X + R/2. Then one easily copies R/2 onto this hypotenuse and copies the remainder as a secant X of the circle.

Then later he proves that X is the side of a regular decagon inscribed in a circle of radius R.

 

[tiny-tim]

hi mathwonk!

the reason i prefer the way i showed is that it explains every step in the formula, i.e. why there is a 4 [because the difference between (r+s)^2 and (r-s)^2 is 4rs], why there is a square root [because you are finding the squared difference of r and s], why you add and subtract the square root [because you have r+s and r-s and want to eliminate r or s], why you divide by 2 [because by adding and subtracting you are getting 2r and 2s],...


tiny tim, why do you prefer the usual completing the square method? That was the method i was taught in high school and it always made it seem unmemorable, because to me it is unmotivated. i.e. it is clever but why do you do those things? i.e. how do you think of it? but i was never strong at algebra.

because it's a direct proof, involving only a b and c, while yours introduces r and s and proves a lemma about r and s before finally abandoning them and returning to a b and c

also because it makes it obvious where the ± comes from

also because (although the general proof works fine) you can use it with the given numbers almost without thinking …

eg x² + 58x +337 = 0

(x + 29)² +337 - 841 = 0

x = -29 ± √504 ​

 

[symbolipoint]

Mathwonk,

We could only imagine that you learned or were taught your r & s expressions for the Quadratic Formula when you were young, either in high school, or early in your college education, or both. Most of us were taught about Completing the Square to find the general solution to a quadratic equation as well as to memorize the general solution. We were also commonly taught a relationship between the two solutions to a quadratic equation and were given maybe three or four homework exercises with this relationship. This is why most of us are unfamiliar with those r & s formulas which you are comfortable with.

We know that we can derive the general solution by comparing with a general quadratic equation and a quadratic trinomial (so I say, but I forgot just how to do it). Most of us may remember choosing a symbolic term in the form of a square and adding this to both sides of an equation, starting from a quadratic expression. Also, some of us found or maybe were actually shown a visual method to demonstrate what Completing the Square means, using cutting and rearranging pieces from a rectangle to show a missing square corner. All the lengths of sides can be labeled with variables and we can derive what this missing square area piece is. So you see, we have good development in learning about completing the square to solve a quadratic equation. We can also use fairly simple algebra to convert between standard form and general form of the quadratic equation for our purposes. In standard form, we can easily graph the equation (that is, when we have a quadratic function).

 

[mathwonk]

a formula you can use without thinking makes me uncomfortable. I was taught the same completing the square way you were. I just didn't get it. I was never happy until I finally learned what the connection was between the roots r,s that we are looking for and the coefficients a,b,c.The mystery was cleared up for me when I finally realized that every quadratic can be factored as (X-r)(X-s) = X^2 - (r+s)X + rs = X^2 -bX +c, and that's what b and c mean. they are sums and products of the roots.

If you don't understand this you have no hope of understanding higher degree equations.

On the other hand if you do understand this you can also solve cubics.

E.g. noting that the roots of a quadratic are solved for as sums of other numbers, let's try that for a cubic, and let's put X = u+v and then expand

X^3 = (u+v)^3 = u^3 + v^3 + 3 u^2v + 3uv^2

= u^3 + v^3 + 3(u+v)(uv) = u^3+v^3 + 3Xuv.

Thus to solve X^3 = 3uvX + (u^3+v^3) we can take X = u+v.

Hence to solve X^3 = bX + c, we need to find numbers u,v, such that

3uv = b and u^3+v^3 = c.

Equivalently we need 27u^3v^3 = b^3 and u^3 +v^3 = c.

Thus we are given the product b^3/27 and the sum c, of the two numbers u^3 and v^3.

But by my solution of the quadratic, that means we know how to find u^3 and v^3,

namely by solving the quadratic equation t^2 - ct + b^3/27 = 0. I.e. then the two solutions are t = u^3 and v^3.

Then after finding u^3 and v^3 this way, we get the three solutions of X^3 = bX+c,

as X = u+v, where we let u be anyone of the three cube roots of u^3,

and we let v = 27u/b^3, for each value of u.

e.g. to solve X^3 = 3X +2, take u=v=1, and then X = u+v = 2.

to solve X^3 = 6X + 9, take u=1, v=2, so that 6 = 3uv and 9 = 1^3 + 2^3, and X = 1+2 = 3.

Try solving cubic equations, on the other hand, by "completing the cube".

 

[symbolipoint]

The high school elementary and intermediate Algebra of the mid 1970's, at least where I was, did not really show much with solving cubic equations. Instead, those of us in Elementary Functions/PreCalculus dealt with cubics and higher order polynomials with Remainder and Rational Roots theorems, and the use of synthetic division, and Descartes Law of Signs. PreCalc also examined limits of functions. Much of what you showed in #35 about quadratic equations and functions (with your two roots, r and s) was also part of elementary & intermediate Algebra - both in high school and in college, but we were not shown that X=u+v substitution. That was clever.

About two or three years ago, I tried on my own to cube a binomial and look for corresponding coefficients, but I did not get the kind of results I hoped for. I did not know about that X=u+v substitution. I was making things more complicated too quickly in trying to use something like (x+k)^3. So far, having used three different College Algebra & PreCalculus textbooks during the last several years, I have not seen any kind of general solution to a cubic equation nor any derivation like you showed in post #35.

Interesting - last night I tried looking at the result for r+s, and (r)(s). Neat, simple expressions using a, b, and c from the general form ax^2+bx+c=0. Not certain if those resulting simple expressions have practical value, but interesting.

 

[symbolipoint]

If roots of quadratic equation are r and s, then r+s = -(b/a); rs = c/a,
For equation of the form ax^2 + bx + c = 0

(The forum's typeset formatting tool seems to be missing otherwise I hoped to typeset those.)

 

[mathwonk]

the brilliant insights i have shared here on quadratic and cubic equations are due to lagrange and euler. i recommend reading their algebra books! after more than 40 years of teaching undergraduate and graduate algebra i realized knew nothing compared to these giants.