Freshmen Algebra “Mistakes”?

History behind this problem: In our Abstract Algebra class, a problem was posed concerning what (x+1)⁴ would look like in Z_n[x]. In context, we were discussing situations for which (a+b)ⁿ might equal aⁿ + bⁿ. Of course, if we are working with coefficients over the real numbers, these expressions are equivalent only in very special cases (such as when a=b=0 – boring!). So we decided to look at a few rings (or fields) for which such expressions might indeed be equivalent for all values in the ring (or field).

We have seen what happens to a polynomial when we are working in the field of real numbers. However, what happens when we work under a more “specialized” type of ring or field of coefficients? Or, put another way, how does the binomial theorem look different when polynomials are reduced mod (n), for some natural number n?

For this discussion, let the notation R[x] denote the set of all polynomials with real number coefficients, Q[x] with rational coefficients, Z[x] with integer coefficients, and Z_n[x] with coefficients in Z mod (n).

Z₂ = {0, 1}

Z₃ = {0, 1, 2}

Z₅ = {0, 1, 2, 3, 4}

Z₇ = {0, 1, 2, 3, 4, 5, 6}

Z₁₁ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

First example: f(x) = (x+1)² = x² + 2x +1 in R[x]

Expansion in Z₂[x]: f(x) = (x+1)² = x² + 2x + 1 = x² + 1. The moral of this story is the following: when high school algebra students make this “mistake,” clearly they must have been working in Z_p[x]!

Let’s first look at various representations of Pascal’s Triangle for different Z_n.

First, for Z we have (noted later as “Z-triangle”):

																1
															1		1
														1		2		1
													1		3		3		1
												1		4		6		4		1
											1		5		10		10		5		1
										1		6		15		20		15		6		1
									1		7		21		35		35		21		7		1
								1		8		28		56		70		56		28		8		1
							1		9		36		84		126		126		84		36		9		1
						1		10		45		120		210		252		210		120		45		10		1
					1		11		55		165		330		462		462		330		165		55		11		1
				1		12		66		220		495		792		924		792		495		220		66		12		1
			1		13		78		286		715		1287		1716		1716		1287		715		286		78		13		1
		1		14		91		364		1001		2002		3003		3432		3003		2002		1001		364		91		14		1
	1		15		105		455		1365		3003		5005		6435		6435		5005		3003		1365		455		105		15		1
1		16		120		560		1820		4368		8008		11440		12870		11440		8008		4368		1820		560		120		16		1

For Z₂:

																1
															1		1
														1		0		1
													1		1		1		1
												1		0		0		0		1
											1		1		0		0		1		1
										1		0		1		0		1		0		1
									1		1		1		1		1		1		1		1
								1		0		0		0		0		0		0		0		1
							1		1		0		0		0		0		0		0		1		1
						1		0		1		0		0		0		0		0		1		0		1
					1		1		1		1		0		0		0		0		1		1		1		1
				1		0		0		0		1		0		0		0		1		0		0		0		1
			1		1		0		0		1		1		0		0		1		1		0		0		1		1
		1		0		1		0		1		0		1		0		1		0		1		0		1		0		1
	1		1		1		1		1		1		1		1		1		1		1		1		1		1		1		1
1		0		0		0		0		0		0		0		0		0		0		0		0		0		0		0		1

I have color-coded the 1’s in red and 0’s in gold to help us see better. One thing this immediately shows us is the odd and even numbers for each row of the Pascal’s Triangle in Z. Note that rows 1, 3, 7, and 15 of the Z-triangle have no even numbers in them. Can you predict the next row of the triangle that will contain all odd numbers?

For the Z₃ triangle, I was surprised at what I observed.

0																				1
1																			1		1
2																		1		2		1
3																	1		0		0		1
4																1		1		0		1		1
5															1		2		1		1		2		1
6														1		0		0		2		0		0		1
7													1		1		0		2		2		0		1		1
8												1		2		1		2		1		2		1		2		1
9											1		0		0		0		0		0		0		0		0		1
10										1		1		0		0		0		0		0		0		0		1		1
11									1		2		1		0		0		0		0		0		0		1		2		1
12								1		0		0		1		0		0		0		0		0		1		0		0		1
13							1		1		0		1		1		0		0		0		0		1		1		0		1		1
14						1		2		1		1		2		1		0		0		0		1		2		1		1		2		1
15					1		0		0		2		0		0		1		0		0		1		0		0		2		0		0		1
16				1		1		0		2		2		0		1		1		0		1		1		0		2		2		0		1		1
			1		2		1		2		1		2		1		2		1		1		2		1		2		1		2		1		2		1
		1		0		0		0		0		0		0		0		0		2		0		0		0		0		0		0		0		0		1
	1		1		0		0		0		0		0		0		0		2		2		0		0		0		0		0		0		0		1		1
1		2		1		0		0		0		0		0		0		2		1		2		0		0		0		0		0		0		1		2		1

Observe the parts colored in gray. It appears that this triangle is repeating itself both on the left and on the right. Rows 3 and 9 of this triangle contain coefficients that are all divisible by 3 (except for the first and last coefficients of the polynomial).

Let’s take a look at another binomial expansion. Let f(x) = (x+y)⁴ .

Z₂[x]: f(x) = (x+y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴ = x⁴+ y⁴ .

Z₃[x]: f(x) = (x+y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴ = x⁴ + x³y + xy³ + y⁴

(note the squared term ran away because Z₃ hurt its feelings by reducing the 6 coefficient to 0).

Furthermore, note that this expression factors into f(x) = (x³ + y³) (x + y) = (x + y)(x² – xy + y²) (x + y) = (x + y)² (x² – xy + y²),

and since -1 = 2 in Z₃, we may finally write f(x) = (x + y)² (x² + 2xy + y²). However, the second factor now has the form of a perfect square trinomial, so we may factor that one down even further to obtain: f(x) = (x + y)² (x + y)² .

In summary, there are multiple factorizations for f(x) in this case. I will write them all at once:

f(x) = (x+y)⁴ = (x³ + y³) (x + y) = (x + y) (x + y) (x² – xy + y²) = (x + y)² (x² – xy + y²) = (x + y)² (x + y)² .

Z₄[x]: f(x) = (x+y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴ = x⁴ + 2x²y² + y⁴ = (x² + y²)² --- how strange!

Z₆[x]: f(x) = (x+y)⁴ = x⁴ + 4x³y + 4xy³ + y⁴ = x⁴ + 4x³y + 4xy³ + y⁴

So, this begs a question: “Which other rows of Pascal’s Triangle will (in Z_n) will have coefficients (other than the first and last) that are zero?” Is this true for all Z_n or just special cases? (I’m suspicious that this may be the case for all Z_p where p is prime.)

This is in fact what we’re saying in this case. Assume that p is prime and q = pⁿ, where n is a natural number. Then . (*)

This gets to the essence of our “Freshman Mistake.” We’ve seen the common “error” (among the Reals) whereby new and unsuspecting algebra students claim (x + y)ⁿ = xⁿ + yⁿ. I am certainly not advocating that we teach our high school students how to do binomial expansions among other domains (nor am I necessarily saying we should not do so). However, I think it is helpful for mathematics educators to know of situations for which this “calculation” is actually correct. There are indeed worlds for which this calculation is correct. In essence, for any Z_p, we find that for all p = q^x (for all natural numbers, x), this calculation is no longer “in error.”

Proof for (*):

Conclusion:

Our students may make this mistake even if they aren’t first year students (even in college). Even though we cringe when we see these kinds of mistakes, it may be helpful for teachers to know when certain “mistakes” are actually not mistakes, but valid mathematics. Hopefully this investigation essay will help to illustrate this important issue.

Questions? Comments? Email me.