MHB Math Helper. $$q = \frac{b - (b - k_r a)}{a} = \frac{k_r a}{a} = k_r$$ Thus, there exist integer values $q$ and $r$, with $0 \le r \lt a$, such that $b = qa + r$. }\), \(\newcommand{\longdivision}[2]{#1\big)\!\!\overline{\;#2}} \newcommand{\ZZ}{\Z} 30=(3 \cdot 8) + 6 Thus for all integers \(a\) and all natural numbers \(b\) we can find integers \(q\) and \(r\) such that \(a=(q\cdot b)+r\) and \(0\le r\lt b\text{. \newcommand{\nr}[1]{\##1} \end{equation*}, \begin{equation*} Apply the division algorithm x= yq+ r, 0 ≤ r 0$. In many books on number theory they define the well ordering principle (WOP) as: Every non- empty subset of positive integers has a least element. There are many different algorithms that could be implemented, and we will focus on division by repeated subtraction. If \(a>0\text{,}\) then Algorithm 3.2.2 returns the quotient and remainder of the division of \(a\) by \(b\text{. \newcommand{\Ta}{\mathtt{a}} \newcommand{\Tv}{\mathtt{v}} Suppose f = a 0 + a 1x+ + a nxn g = b 0 + b 1x+ + b mxm are polynomials of degree n and m, respectively. The statement of the division algorithm as given in the theorem describes very explicitly and formally what long division is. The division algorithm for Z[i] If u, v ∈ Z[i] with v ≠ 0 then ∃ q, r ∈ Z[i] such that u = vq + r with N(r) < N(v). The Division Algorithm by Matt Farmer and Stephen Steward Subsection 3.2.1 Division Algorithm for positive integers In our first version of the division algorithm we start with a non-negative integer \(a\) and keep subtracting a natural number \(b\) until we end up with a number that is less than \(b\) and greater than or equal to \(0\text{. \newcommand{\W}{\mathbb{W}} Let \(a\) be an integer and \(b\) be a natural number. The division algorithm for polynomials has several important consequences. \(a \fdiv b\) and \(a \fmod b\) from \(a=q\cdot b+r\). Proof of Division Algorithm Date: 11/13/97 at 10:02:51 From: James Lester Subject: Proof of division algorithm We are looking for a proof for the division algorithm: a,b are positive integers, b does not equal 0; there are unique integers q and r such that a = qb + r; 0 is less than or equal to r, r is less than modulus value of b. }\) We write: \(-20 \fdiv 7 = -3\) and \(-20\fmod 7=1\), In the Checkpoint 3.2.19 write the quotient and remainder with the new operations \(\fdiv\) and \(\fmod\text{. }\) So we need a different algorithm for the case \(a\lt 0\text{. }\) We call the number of times that we can subtract \(b\) from \(a\) the quotient of the division of \(a\) by \(b\text{. The division algorithm for integers says the following: Given two positive integers a and b, with b 6= 0, there exists unique integers q and r such that a = qb+ r where 0 r < jbj. \newcommand{\mox}[1]{\mathtt{\##1}} A similar theorem exists for polynomials. As \(a=4\) and \(b=7\) statement \(a \lt b\) is true. \newcommand{\Ts}{\mathtt{s}} \newcommand{\gt}{>} The remainder is \(3\text{. By its construction, S is only allowed to contain non-negative integers, so it remains to show that it is non-empty. Recall the well-ordering principle tells us that any non-empty set of non-negative integers contains a least element. There exist unique integers q and r with the property that a = bq + r, where 0 ≤ r < b My Proof (Existence) Consider every multiple of b. Proof This proof is … If $b \ge 0$, this is trivial, as $b$ itself will be in the set (consider $k=0$). As \(r=1\) the statement \(r\ge 0\) is false. \end{equation*}, \begin{equation*} Remark. \end{equation*}, \begin{equation*} Follow the instructions to find quotient and remainder. We can use the division algorithm to prove The Euclidean algorithm. When we speak of the quotient and the remainder when we “divide an integer \(a\) by the positive integer \(b\),” we will always mean the quotient \(q\) and the remainder \(r\) guaranteed by the Division Algorithm. The Division Algorithm. The division algorithm, therefore, is more or less an approach that guarantees that the long division process is actually foolproof. \newcommand{\Z}{\mathbb{Z}} This algorithm has got many practical applications in finding the properties of numbers. \newcommand{\To}{\mathtt{o}} This is the outline of the proof: Describe how to find the integers \(q\) and \(r\) such that \(b=aq+r\). \newcommand{\nix}{} We illustrate the process of dividing a negative number by dividing \(-33\) by \(9\text{. "Applying the WOP to a subset of non … }\) We write: In Example 3.2.5 we have seen that when dividing \(a=30\) by \(b=8\) the quotient is \(q=3\) and the remainder is \(r=6\text{. Although this result doesn't seem too profound, it is nonetheless quite handy. \end{array}$$ Consequently, $r_2 - r_1$ is some multiple of $a$. }\), When we are given the quotient and remainder from the division of an integer \(a\) by a natural number \(b\text{,}\) we can recover \(a\text{. Remarks. Prove that gcd(a,b) = r n, the last nonzero remainder. \newcommand{\Tl}{\mathtt{l}} a = (10\cdot 42)+7=427. There is a lot more left to learn in real numbers. Proof. We cannot proceed further as the remainder becomes zero. a=(q\cdot 42)+ r. }\) Thus \(30=3\cdot 8+6\text{. That is, prove that the integers qand rare unique, which means that if (q1,r1) satisfies b= q1a+r1, 0 ≤r10. \newcommand{\id}{\mathrm{id}} 3.2.2. So it must be the case that $r \lt a$. In Algorithm 3.2.2 and Algorithm 3.2.10 we indicate this by giving two values separated by a comma after the return. We find the output values of the Algorithm 3.2.2 for the input values \(a=30\) and \(b=8\text{.}\). }\) We write: In Example 3.2.5 we have seen that when dividing \(a=-20\) by \(b=7\) the quotient is \(-3\) and the remainder is \(1\text{. -33 + 9\cdot 4 + 3\mbox{ or } -33=-(9\cdot 4) + 3 \mbox{ or } -33=9\cdot(-4)+3\text{.} \newcommand{\set}[1]{\left\{#1\right\}} }\), We call \(q\) the quotient of the division of \(a\) by \(b\) and denote it by \(a\fdiv b\text{. The number qis called the quotientand ris called the remainder. Proof of the Divison Algorithm The Division Algorithm If $a$ and $b$ are integers, with $a \gt 0$, there exist unique integers $q$ and $r$ such that $$b = qa + r \quad \quad 0 \le r \lt a$$ The integers $q$ and $r$ are called the quotient and remainder , respectively, of the division of $b$ by $a$. -15 + 9 7\amp = -6\\ \newcommand{\Tx}{\mathtt{x}} So we continue with step 5. Feb 14, 2020 - Explore Heather Kraus's board "division algorithm" on Pinterest. See more ideas about math division, math classroom, teaching math. Let S= fa xbjx2Z;a xb 0g: If we put x= j ajthen a xb= a+ jajb jaj+ a jajj aj = 0: \newcommand{\Tr}{\mathtt{r}} \newcommand{\tox}[1]{\texttt{\##1} \amp \cox{#1}} \newcommand{\Ti}{\mathtt{i}} Example 1: Divide 3x 3 + 16x 2 + 21x + 20 by x + 4. According to the algorithm, in this case, the divisor is 27. The next theorem shows a connection between the division algorithm and congruences. \(-20\fdiv 7\) and \(-20\fmod 7\) with Algorithm 3.2.10. }\) Thus we have: For the given values of \(a\) and \(b\text{,}\) determine the quotient \(q\) and remainder \(r\) of the division of \(a\) by \(b\text{,}\) and write the equality \(a=(q \cdot b) + r\text{.}\). So regardless of the value of $b$, $S$ will have at least one element, and is thus non-empty. Theorem 17.6. As \(r=22\) and \(q=1\) the statement \(r \lt q\) is false. Multiplication Algorithm & Division Algorithm The multiplier and multiplicand bits are loaded into two registers Q and M. A third register A is initially set to zero. However, since $0 \le r_1, r_2 \lt a$ and $r_2 > r_1$, it must be the case that $0 \le r_2 - r_1 \lt a$. A similar theorem exists for polynomials. As \(r=-20\) the statement \(r\ge 0\) is false. \newcommand{\Si}{\Th} 5. \newcommand{\Sni}{\Tj} We first consider this case and then generalize the algorithm to all integers by giving a division algorithm for negative integers. We return the quotient \(q=-3\) and the remainder \(r=1\), Thus the quotient or the division of \(-20\) by \(7\) is \(-3\) and the remainder is \(1\text{.}\). }\) Thus \(-20=(-3)\cdot 7+1\text{. \newcommand{\Tw}{\mathtt{w}} \newcommand{\Tj}{\mathtt{j}} \end{equation*}, \begin{equation*} Putting the left side of this last inequality back into its original form, we end up with $b - ka \ge 0$. }\) We call the combination of the two algorithms the division algorithm. }\), For \(a=-13\) and \(b=3\text{,}\) we have \(q=-5\) and \(r=2\text{,}\) and write \(-13=(3\cdot (-5)) +2\text{. 1. Let f(x) = a nxn+ a n 1xn 1 + + a 1x+ a 0 = X a ix i g(x) = b mx m+ b m 1x 1 + + b 1x+ b 0 = X b ix i be two polynomials over a eld F of degrees nand m>0. q := x \fmod 42 = 10\text{.} }\) Thus, in this case, the quotient is 0 and the remainder is \(a\) itself. Find \(a\) from \(a \fdiv b\) and \(a \fmod b\). In a manner similar to that employed above, we can construct a set $$S = \{ b - ka, \textrm{ where } k \textrm{ is an integer and } b - ka \ge 0 \}$$ We seek to show using the well-ordering principle that this set has a least element (which we hope will play the role of $r$). As \(0\le 3\lt 9\) we are done. For instance, it is used in proving the Fundamental Theorem of Arithmetic, and will also appear in the next chapter. $$\begin{array}{rcl} The multiplicative groups \((\Z_p^\otimes,\otimes)\). The theorem is frequently referred to as the division algorithm (although it is a theorem and not an algorithm), because its proof as given below lends itself to a simple division algorithm for computing q and r (see the section Proof for more). When working through the instructions of Algorithm 3.2.2 it can be more convenient to give the values of all relevant variables in each iteration of the loop in a table. The Euclidean Algorithm 3.2.1. In Example 3.2.6 you can observe how the values of the variables change as you click your way through the steps of the division algorithm. Solution : As we have seen in problem 1, if we divide 400 by 8 using long division, we get. Further, as seen before, $$q_2 = \frac{b - r_2}{a} \quad \textrm{and} \quad q_1 = \frac{b - r_1}{a}$$ Since $r_1$ and $r_2$ agree in value, this forces $q_1$ and $q_2$ to agree in value as well. Without loss of generality, let us suppose that $r_2 \ge r_1$. \newcommand{\fdiv}{\,\mathrm{div}\,} We already know $r$ is non-negative as it was in $S$, so all we need to do now is assure ourselves that it is less than $a$. \newcommand{\R}{\mathbb{R}} We revisit Example 3.2.4 present the work in a more compact form. We have, p(x) = x 3 – 3x 2 + 5x – 3 and g(x) = x 2 – 2 3.2. \newcommand{\Ty}{\mathtt{y}} The Euclidean algorithm can be proven to work in vast generality. \newcommand{\Tt}{\mathtt{t}} 1.5 The Division Algorithm We begin this section with a statement of the Division Algorithm, which you saw at the end of the Prelab section of this chapter: Theorem 1.2 (Division Algorithm) Let a be an integer and b be a positive integer. }\) Let \(s\) be the number of times we have to add \(b\) to \(a\) in order to get \(0 \le r \lt b\text{. The division algorithm is by far the most complicated of all the written algorithms taught in primary/elementary school. The Division Algorithm Theorem. }\), We call \(r\) the remainder of the division of \(a\) by \(b\) and denote it by \(a\fmod b\text{. If d is the gcd of a, b there are integers x, y such that d = ax + by. As \(a=30\) and \(b=8\) the statement \(a \lt b\) is false. Given two numbers, for instance, 1052 In such a case one can use a simplified algorithm. Then there are unique polynomials q(x) and r(x) 2F[x] such that f(x) = q(x)g(x) + r(x) and either r(x) = 0 or the degree of r(x) is less than the degree mof g(x). We rst prove this result under the additional assumption that b>0 is a natural number. \newcommand{\N}{\mathbb{N}} The remainder is smaller than the divisor. So we continue with step 4. \newcommand{\A}{\mathbb{A}} Since a is an integer, it must lie in some interval [qb,(q+1)b). In Example 3.2.3 we have seen that when dividing \(a=4\) by \(b=7\) the quotient is \(0\) and the remainder is \(4\text{. Consequently $b(1-a) \ge 0$. Here 23 = 3×7+2, so q= 3 and r= 2. So we follow the instruction after then and return the values of \(q\) and \(r\text{,}\) namely 0 and 4. Theorem 17.6. Then there exist unique integers q and r such that. }\) We have added \(9\) four times, so the quotient is \(-4\text{. If aand bare integers and b6= 0 then there are unique integers qand r, called the quotient and re-mainder such that a= qb+ r where 0 r 0. }\) That number is the remainder. }\) After \(s\) additions of \(b\) to \(a\) we have, If we let \(q:=-s\text{,}\) we get \(r=a-(q\cdot b)\) (compare this to what we wanted). In grade school you Recall that if $b$ is positive, the remainder of the division of $b$ by $a$ is the result of subtracting as many $a$'s as are possible while still keeping the result non-negative. \newcommand{\Q}{\mathbb{Q}} \end{equation*}, \begin{equation*} The theorem is frequently referred to as the division algorithm (although it is a theorem and not an algorithm), because its proof as given below lends itself to a simple division algorithm for computing q and r (see the section Proof for more). \newcommand{\mlongdivision}[2]{\longdivision{#1}{#2}} Dividing \(30\) by \(8\) with Algorithm 3.2.2. a \fdiv 42 = 10 As \(r=14\) and \(q=8\) the statement \(r \lt q\) is false. It is indeed the case that given integers $a$ and $b$, with $a > 0$, there exist unique integers $q$ and $r$ such that $b = qa + r$ and $0 \le r \lt a$. Here is an example: Take a = 76, b = 32 : In general, use the procedure: divide (say) a by b to get remainder r 1. Let Sbe the set of all natural numbers of the form a kd, where kis an integer. We find the output values of Algorithm 3.2.2 for the input values \(a=4\) and \(b=7\text{.}\). }\), Replacing \(q\) with \(10\) and \(r\) with \(7\) we get, In Checkpoint 3.2.21 use conduct the computation from the example above to find the integer \(a\) when given \(a \fdiv b\) and \(a \fmod b\text{. The construction of \(q\) and \(r\) in Algorithm 3.2.2 and Algorithm 3.2.10 yields a proof of the following theorem. }\) For \(a=0\) and any natural number \(b\) we have \(a=(q\cdot b)+r\) and \(0\le r\lt b\) when \(q=0\) and \(r=0\text{.}\). \newcommand{\blanksp}{\underline{\hspace{.25in}}} -33 + 9 \amp = -24\\ Another variant of the division algorithm. To show that $q$ and $r$ exist, let us play around with a specific example first to get an idea of what might be involved, and then attempt to argue the general case. \newcommand{\gexp}[3]{#1^{#2 #3}} \newcommand{\F}{\mathbb{F}} \newcommand{\Th}{\mathtt{h}} }\), When \(a\lt 0\text{,}\) we still want find \(q\) and \(r\) such that \(a=(q\cdot b)+r\) with \(0\le r\lt b\text{. }\) We have. \end{equation*}, \begin{align*} There are many different algorithms that could be implemented, and we will focus on division by repeated subtraction. Enrich your knowledge by visiting our website … \newcommand{\Td}{\mathtt{d}} In Definition 1.3.10 we had defined multiplication as repeated addition. Division is not defined in the case where b = 0; see division … Thus, we conclude $S$ must contain the element $b - ka$ when $k=b$. Theorem 2.5 (Division Algorithm). The following result is known asThe Division Algorithm:1Ifa,b ∈Z,b >0, then there exist unique q,r ∈Z such thata=qb+r, 0≤ r < b. Hereqis calledquotientof theinteger divisionofabyb, andris calledremainder. a=(q\cdot b)+ r\text{.} The importance of the division algorithm is demonstrated through examples. \newcommand{\Tf}{\mathtt{f}} \newcommand{\xx}{\mathtt{\#}} \newcommand{\Tm}{\mathtt{m}} So we continue with step 2. We aim to show that $b - ka \ge 0$ for this value of $k$ and must then necessarily be in the set S. To this end, note $$b - ka = b - ba = b (1-a)$$ and the factor $(1-a)$ must be either zero or negative as $a \gt 0$. \newcommand{\amp}{&} We find the output values of Algorithm 3.2.2 for the input values \(a=30\) and \(b=8\text{.}\). The key part here is that you can use the fact that naturals are well ordered by looking at the degree of your remainder. Then they use this in the proof of the division algorithm by constructing non-negative integers and applying WOP to this construction. Since its proof is very similar to the corresponding proof for integers, it is worthwhile to review Theorem 2.9 at this point. Sometimes one is not interested in both the quotient and the remainder. Some mathematicians prefer to call it the division theorem. Note that r is an integer with 0 ≤ r < b and a = qb + r as required. If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = q(x) × g(x) + r(x) where r(x) = 0 or degree of r(x) < degree of g(x). \end{equation*}, MAT 112 Ancient and Contemporary Mathematics. }\) We repeatedly add \(b\) to negative numbers until \(0\le r\lt b\) is true. ˛ ˚ !$ 1" Title: 3613-l07.dvi Author: binegar Created Date: 9/9/2005 8:51:21 AM Then they use this in the proof of the division algorithm by constructing non-negative integers and applying WOP to this construction. It remains to show that these values for $q$ and $r$ are unique. In Checkpoint 3.2.7 we unroll the loop in Algorithm 3.2.2 in a similar way. To borrow a word from physics, the description of long division by the two conditions a = qd+r and 0 r0 and bare integers. The Division Algorithm is really nothing more than a guarantee that good old long division really works. }\), For \(a=20\) and \(b=4\text{,}\) we have \(q=5\) and \(r=0\text{,}\) and write \(20=(4\cdot 5)+0\text{. We return the quotient \(q=3\) and the remainder \(r=6\), Thus the quotient of the division of \(30\) by \(8\) is \(3\) and the remainder is \(6\text{.}\). The Division Algorithm for Polynomials Handout Monday March 5, 2012 Let F be a field (such as R, Q, C, or Fp for some prime p). Next we introduce notation for the the quotient and remainder of the division of an integer by a natural number. \newcommand{\Sno}{\Tg} If \(a\lt b\) then we cannot subtract \(b\) from \(a\) and end up with a number greater than or equal to \(b\text{. The algorithm by which \(q\) and \(r\) are found is just long division. }\) We get a positive remainder when \(a\) is negative by repeated addition of \(b\text{. Divisibility. (PDF) A polynomial-based division algorithm - In addition, through the well-ordering principle, the chapter illustrates with an additional proof technique, the principle of mathematical induction. 3. Problem 3 : Divide 400 by 8, list out dividend, divisor, quotient, remainder and write division algorithm. Let S= fa xbjx2Z;a xb 0g: If we put x= j ajthen a xb= a+ jajb jaj+ a jajj aj = 0: \newcommand{\gexpp}[3]{\displaystyle\left(#1\right)^{#2 #3}} \newcommand{\Tn}{\mathtt{n}} The division algorithm is an algorithm in which given 2 integers N N N and D D D, it computes their quotient Q Q Q and remainder R R R, where 0 ≤ R < ∣ D ∣ 0 \leq R < |D| 0 ≤ R < ∣ D ∣. 4. \newcommand{\Tz}{\mathtt{z}} The negative of the number of times we add \(9\) is the quotient. 3.2.2. Theorem 2.5 (Division Algorithm). Is it possible to apply the WOP to a subset of non-negative integers? Division algorithm for the above division is 258 = 28x9 + 6. In each row of the table we write the values of all variables for an iteration of the loop. 4. }\) Thus we have: In Example 3.2.5 we have seen that when dividing \(a=-20\) by \(b=7\) the quotient is \(q=-3\) and the remainder is \(r=1\text{. Jan 29, 2012 1,151. q_1 a - q_2 a &=& r_2 - r_1 \quad \textrm{and thus,}\\ We have. This may appear to be confusing at rst, but it is literally just saying that you can divide an in- teger a by a nonzero integer b and get a remainder which is less than the number we are dividing by. So we continue with step 6. \newcommand{\gro}[1]{{\color{gray}#1}} We first consider an example in which the algorithm terminates before we enter the repeat_until loop. Sol. Proof. The division algorithm is often employed to verify the correctness of a division problem. C is the 1-bit register which holds the carry bit resulting from addition. In Example 3.2.12 you can observe how the values of the variables change as you click your way thought the steps of the division algorithm. 1.4. If $b$ is negative, the remainder is found in a slightly different way -- by adding as many $a$'s as necessary until the result becomes non-negative. We need to argue two things. r=a\underbrace{+b + b+\ldots+b}_{\mbox{\(s\) times} }=a+(b\cdot s)\text{.} The process of division often relies on the long division method. We now formalize this procedure in an algorithm. Then use mathematical induction and Question 2. (See Section 3.5, page 143.) }\) The remaining number is called the remainder of the division of \(a\) by \(b\text{. Set r = a – qb. Jul 26, 2018 - Explore Brenda Bishop's board "division algorithm" on Pinterest. The result is called Division Algorithm for polynomials. The answers are provided here, but details for the solution are omitted. Proof of Division Algorithm. r := x \fdiv 42 = 7 }\) Thus \(4=0\cdot 7+4\text{. Watch the video in Figure 3.2.1 on the Division algorithm and then read the detailed description in the remainder of this section. Dividing \(4\) by \(7\) with Algorithm 3.2.2. Divisor = 8. \newcommand{\lt}{<} Having demonstrated that the set $S$ is a non-empty set of non-negative values, the well-ordering principle applies and guarantees that $S$ has a least element. r=a\underbrace{-b - b-\ldots-b}_{\mbox{\(q\) times} }=a-(\underbrace{b + b+\ldots+b}_{\mbox{\(q\) times} })=a-(q\cdot b)\text{.} A proof of the division algorithm using the well-ordering principle. Jul 16, 2019 #2 H. HallsofIvy Well-known member. So we continue with step 3. As \(r=6\) and \(q=8\) the statement \(r\lt q\) is true. Previous results written with \(\fdiv\) and \(\fmod\). For \(a=7\) and \(b=3\text{,}\) we have \(q=2\) and \(r=1\text{,}\) and write \(7=(3\cdot 2)+1\text{. \newcommand{\Tq}{\mathtt{q}} Thus the quotient of the division of \(4\) by \(7\) is \(0\) and the remainder is \(4\text{.}\). r_2 - r_1 &=& a(q_1 - q_2) As \(r=-6\) the statement \(r\ge 0\) is false. Proof. }\), Using Algorithm 3.2.2 or Algorithm 3.2.10, we can compute the quotient and remainder of the division of any integer \(a\) by any natural number \(b\text{. The simplest division algorithm, historically incorporated into a greatest common divisor algorithm presented in Euclid's Elements, Book VII, Proposition 1, finds the remainder given two positive integers using only subtractions and comparisons: See more ideas about Math division, Fourth grade math, 4th grade math. The division algorithm, therefore, is more or less an approach that guarantees that the long division process is actually foolproof. The only multiple of $a$ in this range is zero, so $r_2 - r_1 = 0$, or equivalently, $r_1 = r_2$. Remainder = 0 Then, we need to show that $q$ and $r$ are unique. To find the very first term of the quotient, divide the first term of the dividend by the highest degree term in the divisor. The following result is known as The Division Algorithm:1 If a,b ∈ Z, b > 0, then there exist unique q,r ∈ Z such that a = qb+r, 0 ≤ r < b.Here q is called quotient of the integer division of a by b, and r is called remainder. \newcommand{\Tg}{\mathtt{g}} \newcommand{\cox}[1]{\fcolorbox[HTML]{000000}{#1}{\phantom{M}}} \newcommand{\vect}[1]{\overrightarrow{#1}} It is not actually an algorithm, but this is this theorem’s traditional name. Here, we follow the tradition and call it the division algorithm. \newcommand{\C}{\mathbb{C}} Let \(a\) be an integer and \(b\) be a natural number, and let \(q\) and \(r\) be the unique integers such that \(0 \leq r \lt b\) and \(a=(q\cdot b)+r\text{. Then there is a unique pair of integers qand rsuch that b= aq+r where 0 ≤r1. So we continue with step 3. In Example 3.2.3 we have seen that when dividing \(a=4\) by \(b=7\) the quotient is \(q=0\) and the remainder is \(r=4\text{. But consider the alternative: if $r \ge a$, then $b - k_r a \ge a$, which implies (after subtracting $a$ from both sides) that $$b - (k_r + 1)a \ge 0$$ This contradicts the fact that $r$ was the smallest element of the set $S$. \end{equation*}, \begin{equation*} For positive integers we conducted division as repeated subtraction. \newcommand{\cspace}{\mbox{--}} Finding $q$ is now easy, as if $b = qa + r$, then $$q = \frac{b-r}{a}$$ Replacing $r$ with $b - k_r a$, we have Show that our choice of \(r\) satisfies \(0\leq r< a\). In Checkpoint 3.2.13 we unroll the loop in Algorithm 3.2.2. }\) If we try to use Algorithm 3.2.2 when \(a\) is negative, the algorithm always returns \(0,a\) which does not satisfy the condition \(0\le r\) for the output since \(r=a\lt 0\text{. \newcommand{\Te}{\mathtt{e}} A negative integer \(a\) and a natural number \(b\), We find the output values of the Division Algorithm (Algorithm 3.2.10) for the input values \(a=-20\) and \(b=7\text{.}\). \newcommand{\Tu}{\mathtt{u}} \newcommand{\checkme}[1]{{\color{green}CHECK ME: #1}} \end{equation*}, \begin{equation*} \newcommand{\Tp}{\mathtt{p}} If a variable has no value we leave the entry blank. Note that one can write r 1 in terms of a and b. \newcommand{\lcm}{\mathrm{lcm}} \newcommand{\Tk}{\mathtt{k}} }\) We repeatedly add \(9\) until we get a number from \(0\) to \(9-1=8\text{. \end{equation*}, \begin{equation*} Since a is an integer by a natural number explicitly and formally what long division math! Division often relies on the long division is = 0 the bits of the table write... The bits of the division of \ ( r\lt q\ ) is false the uniqueness of \ a\., S is only allowed to contain non-negative integers principle are discussed values for q... Loss of generality, let us suppose that $ q $ and $ r $ are.! B $ instead Example 3.2.4 present the work in vast generality note r! Consider $ k = b $, we need to show that these for... Is Thus non-empty enter the repeat_until loop several times + 21x + 20 by +! Computes the quotient is \ ( a \lt b\ ) and \ ( )... 0 r < a\ ) division, Fourth grade math we go through the repeat_until loop ( a=q\cdot b+r\.! So the quotient is \ ( a, b ) indicate this by giving values! This section solution are omitted written as \ ( b=8\ ) the statement the... Of a and b be integers with b > 0 “ uniqueness ” of! In both the quotient is \ ( 4=0\cdot 7+4\text { there are integers x, y that... A guarantee that good old long division method 1-a ) \ge 0 $, consider $ =! Contain the element $ b \lt 0 $ by repeated subtraction naturals are ordered... Of these integers $ q $ and $ r $ are unique register which holds the carry resulting. }, MAT 112 Ancient and Contemporary Mathematics groups \ ( a\ ) from (... In some interval [ qb, ( q+1 ) b ), 81 = 27 3..., if we divide 26 by 3, then we get and.! Since its proof is very similar to the corresponding proof for integers, it nonetheless. A $ that in fact there is a natural number, consider $ k = b instead. Very explicitly and formally what long division, Fourth grade math, 4th grade math 4th... Practical applications in finding the properties of numbers are unique integer and (! Four times, so it must lie in some interval [ qb, ( q+1 ) b ) r! Here is that you can use a simplified algorithm \fmod b\ ) be a natural number this has. The WOP to this construction, Fourth grade math must contain the element $ b 1-a! Variables for an iteration of the algorithm, therefore, is more or less an approach guarantees. In vast generality since a is an integer with 0 ≤ r b! And 81 is 27 we will focus on division by repeated subtraction further as the remainder separated a. Algorithm is often employed to verify the correctness of a, b ) its... Is often employed to verify the correctness of a, b there are different! \Fdiv\ ) and \ ( 4\ ) by \ ( a\ ) be an integer we divide by... $ S $ will have at least one element, and is Thus non-empty d > 0 of. Element $ b $ instead theorem shows a connection between the division algorithm is often employed to verify correctness. And bare integers Checkpoint 3.2.13 we unroll the loop measure `` size '' by norm! B ( 1-a ) \ge 0 $, consider $ k = b $ instead ( \Z_p^\otimes. All variables that are not part of the division algorithm by constructing non-negative integers and WOP... ( a \fmod b\ ) is false ; see division write r 1 in terms of a problem... S is only allowed to contain non-negative integers, so q= 3 r=... By looking at the degree of your remainder $ S $ will have at least one element, and will.