Fun Fractions? You've Got to Be Kidding!

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  • Fun Fractions? You've Got to Be Kidding!Author(s): Richard KreminskiSource: The Mathematics Teacher, Vol. 91, No. 7 (October 1998), pp. 572-575Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27970688 .Accessed: 17/05/2014 16:03

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  • Fun Fractions? You've Got to Be Kidding!

    w

    The decimal

    expansion

    eventually involves

    digits that

    "bump into one another"

    ell, no, I am not kidding. Consider the following fractions; their decimal expansions, given in paren theses, may surprise you. I urge you to check these

    expansions on a calculator:

    (1) = fg = 1.020408163265306... 49

    (2) = = 1.0305070911131517...

    (3) s = WS = 1.04091625364964820122457...

    (4) f= 3g = 1.01020305081321345590463683...

    The letters denoting the constants were chosen as abbreviations for powers of two, odds, squares, and Fibonacci numbers to remind us of the patterns in the nonzero digits of the decimal expansions. Are these fractions just a fluke? A computer and such software as Derive, Maple, or Mathematica might be better than a calculator for checking the following:

    (10 p' = 500 499 = 1.002004008016032064128256513024...

    (20 o' = 1001000 998001 = 1.003005007009011013015017...

    (30 s' = 1001000000 997002999 = 1.004009016025036049...

    (40 f = 1000000 998999 = 1.001002003005008013021034...

    Just for fun, if you have found a calculator or com

    puter by now, try 10000/9801, too. In this article, I explain how my class obtained

    these fractions, as well as how you and your stu dents can find many more. Before doing so, notice one difference between the fractions shown in

    expressions (30 and (40 and their unprimed counter

    parts in expressions (3) and (4), respectively. The

    digits in the decimal expansions for the unprimed fractions, that is, those with the smaller denomina

    tors, eventually "bump into one another," a phenom enon that we shall call digit overlap. For instance, in expression (3), the squares 1,4,9,16,25,36, 49, and 64 are all apparent in the decimal expansion 1.04091625364964... ; but the digits "... 820122...

    "

    seem to break the pattern. However, if we allocate

    only two digits for any one square, the correct val ues of the squares really are present. We shall elab orate later, but for now, consider the sum of the

    squares that seem incorrectly displayed:

    81 100

    121 144

    + 169 82012245.

    Compare that result with the final digits in

    expression (3).

    Digit overlap is apparent in expression (1), and

    expression ( ) has some digit overlap at the end of its display, as well. We shall see why this overlap occurs, but we first consider a final example that is familiar to many of us from elementary school. It is a case of digit overlap that many people have

    undoubtedly seen before. Recall that 1/7 = .142857142857142857 ..., with

    the pattern of "14? double to 28?roughly double

    again, to 57; then repeat." How many people noticed, as children, that this expansion of 1/7 con

    ceals genuine doubling? Perhaps many of us no

    longer remember having discovered that if we dou ble 14 to 28, to 56, to 112, to 224, and so on, we get

    precisely the correct decimal expansion?if the dig its are aligned properly:

    Rick Kreminski, kremin@boisdarc.tamu-commerce.edu, teaches college algebra through graduate-level mathe

    matics courses at Texas A&M University?Commerce,

    Commerce, TX 75429. His three children, ages 1.5 to 15.0 when this article was written, keep him interested in pre K-12 education.

    572 THE MATHEMATICS TEACHER

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  • 1/7 = .14 + .0028 + .000056 + .00000112 +

    (5) i .0000000224 + .000000000448 + .00000000000896 + .0000000000001792 +???

    = .14285714285714...

    Perhaps expressions (1) through (5) have made us wonder what is going on.

    Some algebra can help us understand the frac tions and their decimal expansions. For explana tions of the expansions shown in expressions (1), (2), and (3) and their primed versions, a knowledge of geometric series is needed, but only as far as

    knowing that

    1 + r + r2 + r3 + r4 + ? 1

    1-r

    if Irl < 1. One can develop the geometric-series for mula from scratch using the methods of this article; however, important convergence issues should be addressed. See the appendix.

    Two words can be used to explain why the frac tions p, o, s, and f and their primed counterparts, behave as they do: generating function. If a0, ah a2,... are the terms in a sequence, then its gener ating function is defined as the formal power series A(x) = a0 + a\x + a2x2 + a3x3 +

    ? ? ?. We illustrate the idea with examples, beginning with and p' first. Consider the sequence 1,2,4,8,16,... of powers of 2; its generating function, Py is given by

    (6) P(x) = 1 + 2x + 4x2 + 8x3 + 16x4 + ? ? ?.

    Note that in this particular case, is just a geo metric series, with ratio r = 2x:

    P(x) = 1 + 2x + (2jc)2 + (2x)3 + (2x)4 +

    Since it is a geometric series, we know that if 12?t I < 1, the series sums to

    1 1-2*.

    For such sufficiently small x, we can conclude that

    (7) P(jc) = ?^=1

    + 2jc + 4x2 + 8x3 +

    16jc4 + 32x5 + 64x6 + 128x7 + 256x8 + ? ? ?.

    After we have conceded equation (7), we simply insert - 1/100, obtaining

    2 + 100 1002 1003 1 100

    + 16? ttL+32-^ + 64-ttL+128- 1

    1004 1005 1006 1007

    + 256? 1008

    = 1 +

    .02 +

    .0004 +

    .000008 +

    .00000016 +

    .0000000032 +

    .000000000064 +

    .00000000000128 +

    .0000000000000256 + ...

    Hence 100/98 = 1.020408163265306 ..., which

    explains our first fractional curiosity in expression (I) , including the digit-overlap phenomenon. Using = 1/1000 instead of = 1/100 in equation (7), we

    see how p' was obtained to yield the result of

    expression ( ). We next explain the values of /and / 'in expres

    sions (4) and (40. Recall that the Fibonacci num bers can be defined recursively as follows: fa = 1, fa =

    1, and fn+2 = fm\ + fn f?r n - 1> yielding the sequence 1,1,2,3, 5,8,13,21,34, 55,89,144,... ; the next term in the sequence is the sum of the previous two. Consider the generating function, F, for the Fibonacci numbers,

    (8) Fix) =fi+f2x+fsx2 + fa* + h 4 + *5 + ? ? * -

    Then

    (9) xF(x) = frx+ f2x2 + fzxz + f4xA + frx5 + - - -

    and

    (10) x2F(x) = fx2 + f2xz + fax4 + fax5 + - -..

    Next, add equations (9) and (10), and simplify using the recursion relations f$ = f2 + /l, U = f% + f2, and so forth:

    (II) xF(x) + x2F(x) = fx+fax2 + fax3+fax4+fax5 + . ? ?.

    By comparing expression (8) with expression (11) and by recalling that fa = 1 = fa, we get

    xF(x) + x2F(x) = F(x)-l.

    We can solve this equation for F(x) and so conclude that

    (12) Fix) = fa + fax + fax2 + fax* + fax4 + fax5 + - ? -

    1 1-x-x2.

    Plugging in = 1/100 and = 1/1000 explains the outcomes in expressions (4) and (4'), respectively.

    To understand o and o' in expressions (2) and (20, we should consider

    (13) Oix)= 1 + 3jc + 5x2 + 7jc3 + 9x4+11jc5 + ....

    Twice the geometric series 1 + + 2 + 3 + ? ? ? yields

    the equation

    (14) T^ = 2 + 2x + 2*2 + 2x3 + 2x4 + 2x5 + .... 1-x

    How many

    people noticed, as

    children, that the expansion of 1/7 conceals

    genuine

    doubling?

    Vol. 91, No. 7 ? October 1998 573

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  • Does a

    paradox exist?

    Adding equations (13) and (14) gives

    0(x) + = 3 + 5x + 7x2 + 9x3 + llx4 + 13x5 + , L-X

    which looks a lot like the original 0(x) itself; a moment's thought reveals that we really have

    ?,, 2 -

    ^ = ? -

    Next we solve for 0{x); after some rearranging, we

    conclude that

    (15) 0(x) = 1 + Sx + 5x2 + 7x3 + 9x4 + Ik5 + ? ? ?

    2x 1-x

    + 1

    1-jc

    Inserting = 1/100 and =1/1000 in equation (15)

    yields the results in expression (2) and expression (2'), respectively.

    I leave expressions (3) and (3') as exercises for the reader, with some hints given in the following exercises. Finally, to verify equation (5), note that if we let U denote the right-hand side of equation (5), namely, U= .14 + .0028 + .000056 + .00000112 + .0000000224 + ??? ,then

    C7=-14'1 + 2'?M + 4'

    1002 + 8?

    1 1003

    which, except for the factor of .14, we recognize as the geometric series in equation (6) evaluated at

    1/100; hence from expression (7), we see that U is

    il 100

    1

    1-2 100

    or

    or

    14 98'

    1 7'

    which explains equation (5). Note that 1/7 is exactly .14 times p, which may be surprising. Perhaps this discussion settles an open question that some of us

    had forgotten long ago. To summarize the technique, when seeking a

    fraction whose decimal expansion displays terms from some sequence, first write down the corre

    sponding generating function, F. If F can be

    expressed as a rational function of x, simply evalu ate the rational function with a suitable x, such as

    = 1/100 or 1/1000.

    EXERCISES 1. UseMx) = 1 + 2x + Sx2 + 4x3 + 5x4 + ? ? ? to find a

    fraction whose decimal expansion begins 1.002003004005 .... Hints: Analogous to the

    derivation for 0(x) that resulted in equation (15) _at/..\ . .9 3 . consider N(x) + 1+ +

    2 + 3 + ..., that i, is,

    N(x) + l .

    how does it compare with (N(x) - l)lx ... ? For students who have taken some calculus, we sup ply a different hint:

    N(x) = -^(x

    + x2 + xs+xU--.\

    which means

    dix

    mx)=Tx\i^r 2. Find a fraction whose decimal expansion begins

    1.02040608101214....

    3. Explain the phenomenon in expressions (3) and (30. Hints: consider S(x) = l + 4x + 9x2 + I6x3 + 25x4 + ? ? ? ; how does S(x) + 3 + 5x + 7x2 + 9jc3 + llx4 + ? ? ? compare with S(x), that is, how does

    S(x) + 0(x)-l

    compare with S(x)... ? For those with calculus

    experience, how can one use the derivative of

    xiV(x)tofindS(x)?

    4. Plug different values for x, such as = 1/10000, into the formulas for P, O, S, F, and iV to obtain different fractions. Predict the decimal expan sions for these new fractions then test your pre dictions on a computer.

    5. Evaluate

    and 1000000/98999901

    1000000000/998999999001, and generalize. Our favorite fraction is

    1000999001000000000/999997000002999999.

    6. Many students think that decimal expansions of fractions must eventually repeat. Others think that this statement implies the existence of a

    paradox, since a fraction as shown in expression (2') does seem to contain all the odd numbers, in

    order, in its decimal expansion, and they clearly do not repeat. Does a paradox exist?

    APPENDIX Comments on Convergence Issues

    The manipulation of series, especially those similar to the one shown in expression (8), requires care. As an

    example, the derivation of expression (12), as we have

    presented it, is incomplete without a discussion of con

    vergence issues?that is, for which the equality is valid. For instance, expression (12) is not valid for >

    574 THE MATHEMATICS TEACHER

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  • 0.7. Our manipulations were all legitimate because we

    performed valid operations on series in their domain of convergence. For more information on convergence issues in series, see virtually any calculus textbook, for example, Finney and Thomas (1994).

    We present the following derivation of the geometric series formula, which may help demonstrate why cau tion is needed. We wish to show that

    1 + X + X2 + XS + X4 + ? ? ? = rs-i- ? 1-x

    We let

    GW = l + x + x2 + 3 + %4+"-.

    Then xG(x) surely is + 2 + 3 + 4 + ? ? ?, which looks

    a lot like G{x); in fact, it is G(x) -1. So xG(x) = G(x) -1. Then we simply solve for G{x) by rearranging, and we indeed have verified that

    G(x) = 1 + + 2 + 3 + 4 + ? ? ? =

    1- X

    Or have we? The foregoing derivation may seem to work regardless of the value of x. Not so! The function G is not even defined as a real number if > 1 or if < -1. Explaining why the derivation is valid for suffi ciently small -values is left as an exercise for those familiar with series.

    Finally, I comment on digit overlap and its relation to divergent series. I previously asked readers to eval uate 10000/9801; the result is 1.020304050607 ....

    Many students wonder whether the pattern will con tinue forever. In this case, the answer is no; digit over

    lap creeps in, and after almost two hundred digits, the decimal expansion is... 91929394959697990001.... But this outcome just begs the question: Does a frac tion exist whose decimal expansion is

    1.23456789101112131415..., that is, a fraction whose decimal expansion is all the integers, in exact ascending order? It is not hard to show that the answer is no, but the result, somewhat surprisingly, depends on the fact that 1 + 1/2 + 1/3 + 1/4 + ? ? ? is a divergent series. For details, see, for instance, Mercer (1994).

    For more information on generating functions, see almost any book on combinatorics; two with discus sions that should be accessible to teachers and some students are Brualdi (1992) and Straight (1993).

    REFERENCES Brualdi, Richard A. Introductory Combinatorics. Englewood Cliffs, N.J.: Prentice Hall, 1992.

    Finney, Ross L., and George B. Thomas Jr. Calculus. 2nd ed. Reading, Mass.: Addison-Wesley Publishing Co., 1994.

    Mercer, A. McD. "A Note on Some Irrational Decimal Fractions." American Mathematical Monthly 101 (June-July 1994): 567-68.

    Straight, H. Joseph. Combinatorics: An Invitation. Pacific Grove, Calif.: Brooks/Cole Publishing Co., 1993.?

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