Writing Math Solutions

The Art of Problem Solving Bookstore now offers two titles from the creator of Math Olympiads in the Elementary and Middle Schools. Click here and here to check them out.

How to Write a Solution
U s e S p a c e
by Richard Rusczyk

imagineyoutrytoreadaparagraphoftextthathasnopuncuationnocapitalsandjustenough spaceinittobreakuplinessoitdoesntmessupbrowsersitsreallytoughtoreadandpretty soonyoulldecidethatitsnotworthreadingandyoullgoandreadsomethingelseyouwont realizehowterriblyharditistotypelikethisitshardbecausewhenyoureusedtowritingcle arlyandusingspaceandpunctuationandsentencestructureitgetsreallyhardtowritewithou utitsimilarlyonceyougetusedtoproperlyusingspaceinwritingyoursolutionsitwillbesec ondnatureandyoullactuallyfindithardtowriteanindecipherableproof

When you write your solution, you should:

Give each important definition or equation its own line.
Don't bury too much algebra in a paragraph. You can write line after line of algebra, but put each step on its own line. Don't cram the algebra in a paragraph.
Label equations or formulas or lemmas or cases you will use later very clearly.
Remember that there's always more paper.

Here's a sample problem:

Problem: Let p(x) be a polynomial with degree 98 such that p(n) = 1/n for n= 1, 2, 3, 4, . . . , 99. Determine p(100).

Have fun reading this solution:

How Not to Write the Solution: Let r(x) = x (p(x) - 1/x) = x p(x) - 1. Since p(x) is a polynomial with degree 98, r(x) is a polynomial with degree 99. Since r(x) = x (p(x) - 1/x), and we are given that (p(x) - 1/x) = 0 for x = 1, 2, 3, . . . , 99, r(x) has roots 1, 2, . . . , 99. Since r(x) has degree 99, these are the only roots of r(x), which must thus have the form r(x) = c(x - 1)(x - 2)(x - 3) . . . (x - 99) for some constant c. To find c, we first let x = 0 in equation r(x) = x p(x) - 1, yielding r(0) = -1. Letting x = 0 in r(x) = c(x - 1)(x - 2)(x - 3) . . . (x - 99) yields r(0) = -c(99!); hence, c = 1/99!. Thus, we have r(x) = (x - 1)(x - 2)(x - 3) . . . (x - 99)/99!. We can combine the equations r(x) = x p(x) - 1 and r(x) = (x - 1)(x - 2)(x - 3) . . . (x - 99)/99! and let x = 100 to find 100p(100) - 1 = (100 - 1)(100 - 2)(100 - 3) . . . (100 - 99)/99!, so 100p(100) - 1 = 99!/99! = 1, so p(100) = 1/50.

Here's the same solution, with nearly the same wording.

How to Write the Solution: Let

r(x) = x (p(x) - 1/x) = x p(x) - 1.

(1)

Since p(x) is a polynomial with degree 98, r(x) is a polynomial with degree 99. Since r(x) = x (p(x) - 1/x), and we are given that (p(x) - 1/x) = 0 for x = 1, 2, 3, . . . , 99,

r(x) has roots 1, 2, . . . , 99.

Since r(x) has degree 99, these are the only roots of r(x), which must thus have the form

r(x) = c(x - 1)(x - 2)(x - 3) . . . (x - 99)

(2)

for some constant c. To find c, we first let x = 0 in equation (1), yielding r(0) = -1. Letting x = 0 in (2) yields r(0) = -c(99!); hence, c = 1/99!. Thus, we have

r(x) = (x - 1)(x - 2)(x - 3) . . . (x - 99)/99!

(3)

We can combine equations (1) and (3) and let x = 100 to find

100p(100) - 1 = (100 - 1)(100 - 2)(100 - 3) . . . (100 - 99)/99!
100p(100) - 1 = 99!/99! = 1
p(100) = 1/50.

Which would you rather read?

Previous Section: Readers Are Not Interpreters

Next Section: sdrawkcaB knihT, Write Forwards

NEW! NEW! NEW!
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's NEW Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!