From 4f670b750af5c11e1eac16d9cd8556455f89f46a Mon Sep 17 00:00:00 2001 From: Juan Marín Noguera Date: Fri, 16 May 2025 22:18:44 +0200 Subject: Changed layout for more manageable volumes --- 4.1.lyx | 1897 --------------------------------------------------------------- 1 file changed, 1897 deletions(-) delete mode 100644 4.1.lyx (limited to '4.1.lyx') diff --git a/4.1.lyx b/4.1.lyx deleted file mode 100644 index aba30d7..0000000 --- a/4.1.lyx +++ /dev/null @@ -1,1897 +0,0 @@ -#LyX 2.4 created this file. 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-\begin_body - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc2[24] -\end_layout - -\end_inset - -Consider the following four number systems: -\end_layout - -\begin_layout Enumerate -binary (signed magnitude); -\end_layout - -\begin_layout Enumerate -negabinary (radix -\begin_inset Formula $-2$ -\end_inset - -); -\end_layout - -\begin_layout Enumerate -balanced ternary; - and -\end_layout - -\begin_layout Enumerate -radix -\begin_inset Formula $b=\frac{1}{10}$ -\end_inset - -. -\end_layout - -\begin_layout Standard -Use each of these four number systems to express each of the following three numbers: -\end_layout - -\begin_layout Enumerate -\begin_inset Formula $-49$ -\end_inset - -; -\end_layout - -\begin_layout Enumerate -\begin_inset Formula $-3\frac{1}{7}$ -\end_inset - - (show the repeating cycle); -\end_layout - -\begin_layout Enumerate -\begin_inset Formula $\pi$ -\end_inset - - (to a few significant figures). -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - - -\end_layout - -\begin_layout Enumerate -\begin_inset Formula $-49\to-110001$ -\end_inset - -. - Multiplying by 2, - -\begin_inset Formula $\frac{1}{7}\to\frac{2}{7}\to\frac{4}{7}\to1+\frac{1}{7}$ -\end_inset - -, - so -\begin_inset Formula $-3\frac{1}{7}\to-11.\overbrace{001}$ -\end_inset - -. - Finally, - -\begin_inset Formula $\pi\to11.00100100001111...$ -\end_inset - - -\end_layout - -\begin_layout Enumerate -The trick is to express the number in binary and, - for the odd positions (the ones where -\begin_inset Formula $(-2)^{k}$ -\end_inset - - is negative), - add 1 to the part of the number on the right, - because we are subtracting -\begin_inset Formula $2^{k}$ -\end_inset - - instead of adding so we compensate for that. - For negative numbers, - we represent them in two's complement before this change, - effectively adding -\begin_inset Formula $2^{M}$ -\end_inset - - for some large -\begin_inset Formula $M$ -\end_inset - -, - and after it we drop that upper bit to subtract -\begin_inset Formula $2^{M}$ -\end_inset - -. - Thus -\begin_inset Formula $-49\to...111001111\to11010011$ -\end_inset - -, - -\begin_inset Formula $-3\frac{1}{7}\to...1100.\overbrace{110110}\to1101.\overbrace{001011}$ -\end_inset - -, - -\begin_inset Formula $\pi\to111.01100100010000...$ -\end_inset - -. -\end_layout - -\begin_layout Enumerate -We proceed as in the text: - -\begin_inset Formula -\begin{align*} --49 & \to-1211\to...11102200\to\overline{1}11\overline{1}\overline{1};\\ --3\frac{1}{7} & \to-10.\overbrace{010212}\to\dots1101.\overbrace{100122}\to\overline{1}0.\overbrace{0\overline{1}\overline{1}011};\\ -\pi & \to10.010211012\dots\to...1121.122022201\dots\to10.011\overline{1}111\overline{1}0\dots. -\end{align*} - -\end_inset - - -\end_layout - -\begin_layout Enumerate -This is like the decimal system but backwards, - so -\begin_inset Formula $-49\to-9.4$ -\end_inset - -, - -\begin_inset Formula $-3\frac{1}{7}\to-\overbrace{758241}3$ -\end_inset - -, - -\begin_inset Formula $\pi\to...51413$ -\end_inset - -. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -exerc5[00] -\end_layout - -\end_inset - -Explain why a negative integer in nines' complement notation has a representation in ten's complement notation that is always one greater, - if the representations are regarded as positive. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -Ten's complement of a negative number -\begin_inset Formula $s$ -\end_inset - - is -\begin_inset Formula $10^{M}+s$ -\end_inset - -, - for a suitably large integer -\begin_inset Formula $M$ -\end_inset - -, - while nines' complement is -\begin_inset Formula $(10^{M}-1)+s$ -\end_inset - -, - so the ten's complement is one greater. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -exerc8[M10] -\end_layout - -\end_inset - -Prove Eq. - (5). -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -The part at the left is -\begin_inset Formula $\sum_{j}a_{j}b^{j}$ -\end_inset - -, - while the part at the right is -\begin_inset Formula $\sum_{j}A_{j}b^{kj}$ -\end_inset - -, - but -\begin_inset Formula $A_{j}=\sum_{i=0}^{k-1}a_{kj+i}b^{i}$ -\end_inset - -, - so this is -\begin_inset Formula -\[ -\sum_{j}A_{j}b^{kj}=\sum_{j}\sum_{i=0}^{k-1}(a_{kj+i}b^{i})b^{kj}=\sum_{j}\sum_{i=0}^{k-1}(a_{kj+i}b^{kj+i})=\sum_{j}a_{j}b^{j}. -\] - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc9[15] -\end_layout - -\end_inset - -Change the following -\emph on -octal -\emph default - numbers to -\emph on -hexadecimal -\emph default - notation, - using the hexadecimal digits -\family typewriter -0 -\family default -, - -\family typewriter -1 -\family default -, - ..., - -\family typewriter -9 -\family default -, - -\family typewriter -A -\family default -, - -\family typewriter -B -\family default -, - -\family typewriter -C -\family default -, - -\family typewriter -D -\family default -, - -\family typewriter -E -\family default -, - -\family typewriter -F -\family default -: - 12; - 5655; - 2550276; - 76545336; - 3726755. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -We use Eq. - (5) with base 2. -\end_layout - -\begin_layout Standard -\align center -\begin_inset Tabular - - - - - - - -\begin_inset Text - -\begin_layout Plain Layout -Octal -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout -Binary -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout -Hexadecimal -\end_layout - -\end_inset - - - - -\begin_inset Text - -\begin_layout Plain Layout -12 -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout -1 010 -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout - -\family typewriter -A -\end_layout - -\end_inset - - - - -\begin_inset Text - -\begin_layout Plain Layout -5655 -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout -101 110 101 101 -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout - -\family typewriter -BAD -\end_layout - -\end_inset - - - - -\begin_inset Text - -\begin_layout Plain Layout -2550276 -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout -10 101 101 000 010 111 110 -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout - -\family typewriter -AD0BE -\end_layout - -\end_inset - - - - -\begin_inset Text - -\begin_layout Plain Layout -76545336 -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout -111 110 101 100 101 011 011 110 -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout - -\family typewriter -FACADE -\end_layout - -\end_inset - - - - -\begin_inset Text - -\begin_layout Plain Layout -3726755 -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout -11 111 010 110 111 101 101 -\end_layout - -\end_inset - - -\begin_inset Text - -\begin_layout Plain Layout - -\family typewriter -FADED -\end_layout - -\end_inset - - - - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset Note Comment -status open - -\begin_layout Plain Layout -Is this what academics do when they get bored? -\end_layout - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc13[M21] -\end_layout - -\end_inset - -In the decimal system there are some numbers with two infinite decimal expansions; - for example, - -\begin_inset Formula $2.3599999...=2.3600000...$ -\end_inset - -. - Does the -\emph on -negadecimal -\emph default - (base -\begin_inset Formula $-10$ -\end_inset - -) system have unique expansions, - or are there real numbers with two different infinite expansions in this base also? -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -There are real numbers with two different infinite expansions: - -\begin_inset Formula -\[ -\tfrac{1}{11}=0.090909...=(0.090909...)_{-10}=(1.909090...)_{-10}. -\] - -\end_inset - -Effectively, - -\begin_inset Formula $(0.090909...)_{-10}=\sum_{k\geq1}9\cdot(-10)^{-2k}=9\sum_{k\geq1}100^{-k}=9\frac{\frac{1}{100}}{\frac{99}{100}}=\frac{1}{11}$ -\end_inset - -, - but -\begin_inset Formula $(1.909090...)_{-10}=1+\sum_{k\geq1}9\cdot(-10)^{-2k+1}=1-90\sum_{k\geq1}100^{-k}=1-90\frac{1}{99}=1-\frac{10}{11}=\frac{1}{11}$ -\end_inset - -. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc19[23] -\end_layout - -\end_inset - -(David W. - Matula.) Let -\begin_inset Formula $D$ -\end_inset - - be a set of -\begin_inset Formula $b$ -\end_inset - - integers, - containing exactly one solution to the congruence -\begin_inset Formula $x\equiv j\pmod b$ -\end_inset - - for -\begin_inset Formula $0\leq j0$ -\end_inset - -, - then all numbers in -\begin_inset Formula $D$ -\end_inset - - are negative, - but the integers between -\begin_inset Formula $l$ -\end_inset - - and -\begin_inset Formula $u$ -\end_inset - - are positive so they couldn't be represented in the given form, -\begin_inset Formula $\#$ -\end_inset - - and a similar thing would happen if -\begin_inset Formula $u<0$ -\end_inset - -. - Now, - in the algorithm above, - -\begin_inset Formula $q-k=\lfloor\frac{m}{b}\rfloor-k=\frac{m-r}{b}-\frac{x-r}{b}=\frac{m-x}{b}$ -\end_inset - -. - With this, - if -\begin_inset Formula $m>u$ -\end_inset - -, - then -\begin_inset Formula $m(b-1)>-\min D$ -\end_inset - -, - but and therefore -\begin_inset Formula -\[ -m-(q-k)=m-\frac{m-x}{b}=\frac{m(b-1)+x}{b}>\frac{-\min D+\min D}{b}=0, -\] - -\end_inset - -so -\begin_inset Formula $q-km-\frac{\max D}{b}\geq m+l, -\] - -\end_inset - -so -\begin_inset Formula $q-k\geq l$ -\end_inset - -. - This means that, - on each step, - -\begin_inset Formula $m$ -\end_inset - - is smaller by at least 1, - but it doesn't become smaller than -\begin_inset Formula $l$ -\end_inset - -, - so it eventually reaches the interval -\begin_inset Formula $[l,u]$ -\end_inset - - in at most -\begin_inset Formula $|m-u|$ -\end_inset - - steps. - And if -\begin_inset Formula $mm$ -\end_inset - -, - and -\begin_inset Formula -\[ -m+(q-k)=\frac{m(b+1)-x}{b}0$ -\end_inset - -, - we take -\begin_inset Formula $k\in\mathbb{N}$ -\end_inset - - such that -\begin_inset Formula $|2^{k}-n|$ -\end_inset - - is minimum, - write a 1 at position -\begin_inset Formula $k$ -\end_inset - -, - and then write -\begin_inset Formula $n-2^{k}$ -\end_inset - - on the -\begin_inset Formula $k$ -\end_inset - - positions to the right. - If -\begin_inset Formula $n<0$ -\end_inset - -, - we do the same except that first we write -\begin_inset Formula $\overline{1}$ -\end_inset - - instead of 1. -\end_layout - -\begin_layout Standard -Now we prove that this in fact results in the representation fewest number of nonzero digits. - We only need to prove it for positive -\begin_inset Formula $n$ -\end_inset - -. -\end_layout - -\begin_layout Standard -Now, - if -\begin_inset Formula $n=2^{k}$ -\end_inset - - or -\begin_inset Formula $3\cdot2^{k}$ -\end_inset - - for some -\begin_inset Formula $k\in\mathbb{N}$ -\end_inset - -, - the statement is trivial. - Otherwise, - if -\begin_inset Formula $2^{k}2^{k-1}$ -\end_inset - -, - -\begin_inset Formula $t$ -\end_inset - - would need to have its most significant digit in position -\begin_inset Formula $k$ -\end_inset - - or -\begin_inset Formula $k-1$ -\end_inset - -, - but any representation of -\begin_inset Formula $t$ -\end_inset - - that starts at position -\begin_inset Formula $k$ -\end_inset - - can be converted into one of -\begin_inset Formula $2^{k}-n$ -\end_inset - - by removing the 1 at that position, - and any one starting at position -\begin_inset Formula $k-1$ -\end_inset - - can be converted into one of -\begin_inset Formula $2^{k}-n$ -\end_inset - - by swapping the sign in that position, - so in general -\begin_inset Formula $\sigma(t)=\sigma(2^{k+1}-n)\geq\sigma(2^{k}-n)=\sigma(n-2^{k})$ -\end_inset - -. -\end_layout - -\begin_layout Standard -A similar argument shows that, - if -\begin_inset Formula $n>3\cdot2^{k-1}$ -\end_inset - -, - then -\begin_inset Formula $\sigma(n-2^{k})\geq\sigma(2^{k+1}-n)$ -\end_inset - -. -\end_layout - -\end_body -\end_document -- cgit v1.2.3