Freshmen Algebra ÒMistakesÓ?
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 a^{n}
+ b^{n}. 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 ÒspecializedÓ 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).
Z_{2} = {0, 1}
Z_{3} = {0, 1, 2}
Z_{5} = {0, 1, 2, 3, 4}
Z_{7} = {0, 1, 2, 3, 4, 5,
6}
Z_{11} = {0, 1, 2, 3, 4, 5,
6, 7, 8, 9, 10}
First
example: f(x) = (x+1)^{2}
= x^{2} + 2x +1 in R[x]
Expansion
in Z_{2}[x]: f(x) = (x+1)^{2} = x^{2}
+ 2x + 1 = x^{2} + 1. The moral of this story is the following: when
high school algebra students make this Òmistake,Ó clearly they must have been
working in Z_{p}[x]!
LetÕs
first look at various representations of PascalÕs Triangle for different Z_{n}.
First, for
Z we have (noted later as ÒZtriangleÓ):
















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 Z_{2}:
















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
colorcoded the 1Õs in red and 0Õs in gold to help us see better. One thing
this immediately shows us is the odd and even numbers for each row of the
PascalÕs Triangle in Z. Note that rows 1, 3, 7, and 15 of the Ztriangle have
no even numbers in them. Can you
predict the next row of the triangle that will contain all odd numbers?
For the Z_{3 }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).
LetÕs take
a look at another binomial expansion. Let f(x) = (x+y)^{4} .
Z_{2}[x]: f(x) = (x+y)^{4} = x^{4}
+ 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4}
= x^{4}+ y^{4} .
Z_{3}[x]: f(x) = (x+y)^{4} = x^{4}
+ 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4}
= x^{4} + x^{3}y + xy^{3} + y^{4}
(note the
squared term ran away because Z_{3} hurt its feelings by reducing the 6
coefficient to 0).
Furthermore,
note that this expression factors into f(x) = (x^{3}
+ y^{3}) (x + y) = (x + y)(x^{2}
– xy + y^{2}) (x + y) = (x + y)^{2} (x^{2}
– xy + y^{2}),
and since
1 = 2 in Z_{3}, we may finally write f(x) = (x + y)^{2} (x^{2}
+ 2xy + y^{2}). 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} = (x^{3} + y^{3}) (x + y) = (x + y) (x +
y) (x^{2} – xy + y^{2})
= (x + y)^{2} (x^{2} – xy + y^{2}) = (x + y)^{2}
(x + y)^{2} .
Z_{4}[x]: f(x) = (x+y)^{4} = x^{4}
+ 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4}
= x^{4} + 2x^{2}y^{2} + y^{4} = (x^{2}
+ y^{2})^{2}  how strange!
Z_{6}[x]: f(x) = (x+y)^{4} = x^{4}
+ 4x^{3}y + 4xy^{3}
+ y^{4} = x^{4} + 4x^{3}y + 4xy^{3} + y^{4}
So, this
begs a question: ÒWhich other rows of PascalÕs Triangle will (in Z_{n}) will have
coefficients (other than the first and last) that are zero?Ó Is this true for all Z_{n} or just special
cases? (IÕm suspicious that this
may be the case for all Z_{p} where p is prime.)
This is in
fact what weÕre saying in this case.
Assume that p is prime and q = p^{n}, where n is a natural
number. Then . (*)
This gets to
the essence of our ÒFreshman Mistake.Ó WeÕve seen the common ÒerrorÓ (among the
Reals) whereby new and unsuspecting algebra students claim (x + y)^{n}
= x^{n} + y^{n}. 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 ÒcalculationÓ 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 Òin error.Ó
Proof
for (*):
Conclusion:
Our
students may make this mistake even if they arenÕt 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 ÒmistakesÓ are actually not mistakes,
but valid mathematics. Hopefully this investigation essay will help to illustrate
this important issue.
Questions?
Comments? Email
me.