I love it when math works as it should. Such was the case last night when I was looking for a problem to explore. I began by checking my favourite blogs for a quick puzzle before bed. Nothing really stuck, so I tried some searches on twitter. (#math #mathchat #puzzle usually get the job done). I found a puzzle that intrigued me and began to work.
Before I mention the puzzle, I want to say a quick word about school maths. I have a great personal interest in mathematics, and am able to move on past problems that don’t pique my curiosity; we do not afford students the same opportunity. I wonder, if students were given more freedom, if they too would find puzzles and topics that interest them. This is a not the topic for this post, but is an interesting debate in mathematics education. I digress…
Back to the puzzle. It is as follows:
Prove the following subtraction cannot be true:
E L E V E N
– T H R E E
E I G H T
The intriguing part of this problem is how I initially read it. Because of the spacing and my mathematical mind, I did not see the words initially. When I presented the problem to others, they immediately read “11-3=8”, but all I saw was a collection of variables. The second, and more fulfilling, outcome of this problem could not be predicted from the start. I ended up learning far more than the designer could have ever intended. Again, this personal driven learning is lacking in school mathematics. The more I experiment with maths, the more apparent this injustice becomes.
This is where I began. I assumed that each letter stood for a unique number. I figured if I showed that two variables must represent the same integer, then it was impossible. I counted 9 letters (E, L, V, N, T, H, R, I, G) which meant that there was not much room for ambiguous cases. If there are 10 digits in play (0-9) then there would only be 1 left out of the problem. Again, this is an if, because no base was indicated. Already I was on a trail that was my own, but had no idea where it would take me.
I began on the far left-hand side. I knew that with the subtraction, we lost a digit. Regardless of base, this meant that it had to be “carried”. If it disappeared when carried, then we know that E=1. We also know that L<T to enable the carry to make sense. Even if it was borrowed from previously (to help the E to its right), L would still have to be less than T because if L-1 is less than T but L wasn’t, then L would have to equal T. This is impossible by my assumptions. Such is the logic work the puzzle intended; I was going a much different direction.
Now I knew L<T and then the subtraction gave me this:
L+10 – T = E
but E=1 so…
L + 10 – T = 1
(assuming base 10 on the carry) more generally…
let b = base
L + b – T = 1
L-T = 1-b
T = L + (b-1)
Now this last conclusion fits perfectly with L<T because we must add a positive number to get to T. In fact, because T must be a digit [0, 9] (more generally [0, (b-1)]) we know that L must be 0 and T must equal b-1. It was here where my thoughts veered off track.
These two statements only work together if b>0. If ‘b’ was a negative number, then adding (b-1) to L would not yield a larger number. Therefore L<T. There seemed to be an incongruence. I had never encountered negative bases; my mind had never gone there. I had seen binary in grade school, and experimented with alternate (positive) bases at leisure, but never negative. Was this possible? What did base -10 look like?
I abandoned the problem and began to play with its precipitate. Could all numbers be written in Base -10? How could we count in this base? It turns out, it is possible, and creates some very revealing patterns. Instead of a 1s, 10s, 100s place etc., the sign alternates as the negative is exponentiated. We are left with a 1s, -10s, 100s, -1000s place. etc. For example:
472 in base ten is expanded as
4x(10)^2 + 7x(10)^1 + 2x(10)^0
4×100 + 7×10 + 2×1
but 472 in base negative ten is expanded as
4x(-10)^2 + 7x(-10)^1 + 2x(-10)^0
4×100 + 7x(-10) + 2×1
400 -70 + 2
The result is very different. In fact, the difference between the two numbers always seemed to be twice the absolute value of every negative component. Try counting to 20 in base negative 10. What pattern occurs? Take it further. Try an addition. Any quirks? Try anything you like and attempt to make sense of it. I discovered patterns with exploration, but I can imagine how school math would approach this. More to come on that point later.
After playing for a little while, I was developing a list of rules that allowed me to become more comfortable with the base. Along with the absolute value pattern, I noticed that every number with an even number of digits was negative, and all positive numbers had an odd number of digits. What made the rules so powerful? They had meaning because they were developed! My odd/even rule made sense because I had dissected the facts and devised the rule from a logical pattern. After I had grasped the new number system, I sat back and mused on how schools would probably transfer this knowledge to students. It would probably look something like this:
Rule #1: Every odd digited number is positive
Rule #2: Every even digited number is negative
Rule #3: To convert to base-10, the ones digit remains constant, every pair of subsequent digits (ab), becomes 10a-b in the base-10.
Rule#4: If there is no pair, the answer is negative. Subtract what you have from the place value of the remaining digit.
Crystal clear. Now do problems 1-50 odd. Hopefully these rules sound like garbage to you; they would to me except I wrote them. Actually, I did one better–I created them.
Now, in my curriculum, there is no space for negative bases. In fact, alternate bases only make an appearance in the form of binary. (If you don’t count the bases in logarithms). I would love to take the idea of an alternate base arithmetic system into a math club and explore. Such environments would greatly benefit from discussions and tinkerings with binary, hexadecimal, and neg-ecimal… (to coin a phrase).
It is tough to incorporate true inquiry in high school these days. Guided inquiry with carefully constructed tasks keeps students on the pathway to the outcomes. The more I experience the freedom I never had, the more I realize that being “good” at math, is much deeper than I ever thought possible.
PS. I solved the problem the next day. The notation makes it tough to blog, but maybe I will try someday. I encourage you to solve it as well. Also, I used my odd/even rule to show that the base in this problem cannot be negative. ELEVEN has 6 digits, so it is negative. THREE has 5 so it is positive. The result (EIGHT) has 5 digits, so it is positive. A positive subtracted from a negative can never be positive; this explains why my L and T notation didn’t work out if b<0. Don’t you love how the math comes full circle!