Detached mathematics Amount representations

 25 June 05:19   You are already accustomed with autograph a amount down, and accepting it beggarly a assertive aggregate of tens, hundreds, and so on. This is one anatomy of amount representation, but there are others. We will attending at amount bases and connected fractions.

    You are already accustomed with base-10 amount representation. For example, the amount 2818 is the aforementioned as

    : 2×10³+8×10²+1×10¹+8×10⁰

    We can alter the amount 10 actuality with any amount and we access a altered number. In general, we can represent an accumulation n in a abject b by the following:

    : a_kb^k+a_k-1b^k-1+...+a₀b⁰

    where the a_i are all beneath than b.

    We address a amount abject b as (a_ka_k-1...a₀)_b.

    For example, if we accept the character 243 in abject 6, we address it (243)₆. If we are in abject 10 we artlessly address the number: for archetype the character 155 in abject 10 is artlessly accounting 155.

    However, accustomed a amount in a abject b, how can we catechumen it to our accustomed abject 10 system? Likewise, how can we catechumen a amount from our abject 10 arrangement to abject b?

    The first is almost easy, the additional added difficult.

    We artlessly use the analogue of a base-b amount to catechumen a base-b amount to abject 10 - that is we artlessly accumulate out.

    For example

    :(919)₁₂=9×12²+12¹+9=1317.

    This assignment about is hardly added difficult, and there are several means of assuming this task.

    One adjustment is to address anniversary move using the analysis algorithm from the section. For example, if we ambition to catechumen 1317 to abject 12:

    : 1317 = 12 × 109 + 9

    : 109 = 12 × 9 + 1

    : 9 = 12 × 0 + 9

    So in abject 12, (919)₁₂=1317.

    Weve just apparent how we can catechumen integers from abject to base, but how do we plan with converting absolute numbers?

    Recall in abject 10 we address a amount such as 11.341 as

    : 1×10¹+1×10⁰+3×10^-1+4×10^-2+1×10^-3

    and so we can extend our analogue aloft of a base-b amount to be

    :

    a_kb^k+a_k-1b^k-1+...+a₀b⁰+a_-1b^-1+...

    where the a_i are all beneath than b, and the sum could abolish or go on indefinately.

    Again, how are we to catechumen these numbers from abject to base? We can catechumen the basic part, but what about the apportioned allotment (the allotment beneath than 1)?

    Say we ambition to catechumen .341 in abject 10 to abject 6.

    We address a table, using the afterward rules

    :$c\_i\; =\; lfloor\}$
floor

    :$gamma\_i\; =\; 6gamma\_-c\_i$

     i c_i γ_i 6γ_i

     0 0 .341 2.046

     1 2 .046 0.276

     2 0 .276 1.656

     3 1 .656 3.936

     4 3 .936 5.616

     5 5 .616 3.696

     6 3 .696 4.176

     7 4 .176 1.056

     8 1 .056 0.336

     9 0 .336 2.016

    It looks like this amplification will go on forever! We charge to account for added ethics of i to see whether we hit a echo amount of γ_i to see whether we get a repetition.

    So we accept the approximation that .341 is abreast to (.20135341)₆. (Calculate this using the definition. How abutting is our approximation?)

    

    If we access a base-b representation for example, that looks something like (.18191819181918191819...)_b we alarm the representation periodic. We generally address this as

    : $(.overline)\_b$

    We use this aforementioned action to catechumen a apportioned amount to base-b by replacing the 6 aloft with b.

    We accept a chic ambush we can use to catechumen a apportioned n in base-b to abject 10 provided the representation repeats. Let us attending at an example:

    Consider $(.overline)\_7=alpha$.

    Now then

    : $(3145.overline)\_7\; =\; 7^4alpha$

    And now

    : $(3145)\_7+(.overline)\_7=7^4alpha$

    which is

    : $(3145)\_7+alpha=7^4alpha$

    Then

    : $(3145)\_7=7^4alpha\; -\; alpha$

    : $(3145)\_7=(7^4-1)alpha$

    And finally

    : $=alpha$

    On converting (3145)₇ to abject 10, we access the following

    : $alpha=1111/2400$

    In a sense, the base-b representation is nice, but it has a few shortcomings in account to accuracy. We cannot anxiously represent the amount $sqrt$ using base-b representation. This is area the connected atom representation comes in handy, which has some nice backdrop apropos boxlike irrationals.

    A connected atom is a amount in the form

    :$q\_0\; +\; \}\}$

    Since this characters is acutely rather cumbersome, we abridge the aloft to

    :$$

    Again we ask ourselves how can we catechumen from and to this amount representation? Afresh converting from is simpler than converting to.

    We artlessly use our analogue of the connected atom to catechumen from a connected fraction. This may attending difficult, but in actuality is absolutely simple depending on which end one starts with. Lets attending at an example

    Consider

    :$alpha=$

    Now, we plan from appropriate to left. We first accept the atom

    :$$

    The next chiffre 2 tells us to perform

    :$2+=11/5$

    and then yield the alternate to get

    :$$

    Now the next chiffre 1 tells us to perform

    :$1+=$

    and then to yield the alternate to get

    :$$

    Now we haveto add q₀ which is consistently greater than 1 and we get the result:

    :$alpha=$

    Again, we draw up a table.

    Consider the atom 12/22, and use the afterward rules in the table

    : θ_i=θ_i-1-q_i-1

    : $q\_i\; =\; lfloor\; heta\_i^$
floor

     i θ_i θ_i^-1 q_i

     0 12/22 . 0

     1 12/22 22/12 1

     2 5/6 6/5 1

     3 1/5 5/1 5

    (stop back next row will be abounding of 0s)

    So now the connected atom for 12/22 is .

    Firstly, we create agenda of a nice acreage of alternate connected fractions (where the arrangement of q_i repeat), that

    : every alternate connected atom is a amount in the anatomy

    :: $over\; c\}\; mathrm\; a,\; b,\; c\; inmathbb$

    For example, accede the connected atom

    :

    alpha & = & & = & end

    Now

    :$=$

    which we can carbon as

    :$alpha=2+\}\}\}\}$

    Now we can abridge to obtain

    :$alpha=2+\}\}\}$

    :$alpha=2+\}\}$

    :$alpha=2+\}$

    :$alpha=2+$

    :$alpha=$

    :$alpha=$

    :$alpha-5=7\; alpha$

    :$3alpha^2+2alpha-5=7\; alpha$

    :$3alpha^2+2alpha-7\; alpha-5=0$

    :$3alpha^2-5\; alpha-5=0$

    which is a boxlike blueprint and can be apparent to obtain

    :$alpha=\}\}\}$.

    Note that we can consistently cycle up the connected atom into the anatomy (aα+b)/(cα+d)=α, which demonstrates the point that every boxlike aberrant has a repeating connected atom representation

    Say we accept a connected atom which represents a amount n. Let us appraise the afterward alternation of fractions , , and so on. Anniversary aspect of the alternation is accepted as a convergent. It turns out that the alternation of convergents accommodate the best rational approximations to n.

    These can be affected from the connected atom representation, but aswell from the adding table. Let us yield $sqrt$.

    Continue as before, but abode an added -1 row, and set u_-1=1, v_-1=0. Iterate with the rules

    :$u\_=q\_u\_i+v\_$

    :$v\_=q\_v\_i+v\_$

    :

    i & heta_i & heta_i^ & q_i & u_i & v_i-1 & & & & 1 & 0 \

    0 & sqrt & & 2 & 2 & 1 1 & 2-sqrt & 1+sqrt & 2 & 5 & 2 2 & -1+sqrt & 2+sqrt & 4 & 22 & 9 3 & -2+sqrt & 1+sqrt & 2 & 49 & 20 \end

    

    and the arrangement repeats and the connected atom is . We can abide the action to accomplish added convergents - the convergents are 2, 5/2, 22/9, 49/20, ...