It only takes a minute to sign up. Each interval with nonzero length contains an innite number of rationals. Planned Maintenance scheduled March 2nd, 2023 at 01:00 AM UTC (March 1st, Algebra Problem: $a + 1/b = b + 1/c = c + 1/a = t $. Prove that if $a < b < 0$ then $a^2 > b^2$, Prove that if $a$ and $b$ are real numbers with $0 < a < b$ then $\frac{1}{b} < \frac{1}{a}$, Prove that if $a$ is a real number and $a^3 > a$ then $a^5 > a$. \(x + y\), \(xy\), and \(xy\) are in \(\mathbb{Q}\); and. Prove that the cube root of 2 is an irrational number. Here we go. It means that 1 < a < 0. a be rewritten as a = q x where x > q, x > 0 and q > 0 a. Max. i. $$a=t-\frac{1}{b}=\frac{bt-1}{b},b=t-\frac{1}{c}=\frac{ct-1}{c},c=t-\frac{1}{a}=\frac{at-1}{a}$$ The arithmetic mean of the nine numbers in the set is a -digit number , all of whose digits are distinct. Connect and share knowledge within a single location that is structured and easy to search. Hence, the given equation, What tool to use for the online analogue of "writing lecture notes on a blackboard"? Try the following algebraic operations on the inequality in (2). !^'] Let Gbe the group of nonzero real numbers under the operation of multiplication. That is, we assume that there exist integers \(a\), \(b\), and \(c\) such that 3 divides both \(a\) and \(b\), that \(c \equiv 1\) (mod 3), and that the equation, has a solution in which both \(x\) and \(y\) are integers. JavaScript is disabled. Step-by-step solution 100% (10 ratings) for this solution Step 1 of 3 The objective is to determine is rational number or not if the following equations are satisfied: Justify your conclusion. . This third order equation in $t$ can be rewritten as follows. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. If so, express it as a ratio of two integers. We have discussed the logic behind a proof by contradiction in the preview activities for this section. A proof by contradiction is often used to prove a conditional statement \(P \to Q\) when a direct proof has not been found and it is relatively easy to form the negation of the proposition. The last inequality is clearly a contradiction and so we have proved the proposition. Partner is not responding when their writing is needed in European project application, Is email scraping still a thing for spammers. Prove that if ac bc, then c 0. We introduced closure properties in Section 1.1, and the rational numbers \(\mathbb{Q}\) are closed under addition, subtraction, multiplication, and division by nonzero rational numbers. Ex. Since , it follows by comparing coefficients that and that . Why did the Soviets not shoot down US spy satellites during the Cold War? Therefore, a+b . Book about a good dark lord, think "not Sauron". For each real number \(x\), if \(0 < x < 1\), then \(\dfrac{1}{x(1 - x)} \ge 4\), We will use a proof by contradiction. u = 1, 0, x , u = 1, 0, x , v = 2 x, 1, 0 , v = 2 x, 1, 0 , where x x is a nonzero real number. We will use a proof by contradiction. \(\sqrt 2 \sqrt 2 = 2\) and \(\dfrac{\sqrt 2}{\sqrt 2} = 1\). Suppose that a and b are nonzero real numbers. However, the TSP in its "pure" form may lack some essential issues for a decision makere.g., time-dependent travelling conditions. $a$ be rewritten as $a = \frac{q}{x}$ where $x > q$, $x > 0$ and $q>0$. Why did the Soviets not shoot down US spy satellites during the Cold War. That is, a tautology is necessarily true in all circumstances, and a contradiction is necessarily false in all circumstances. What are the possible value(s) for ? Suppose that A , B, and C are non-zero distinct digits less than 6 , and suppose we have and . Sex Doctor Story Identification: Nanomachines Building Cities. Strange behavior of tikz-cd with remember picture. Hence, Since and are solutions to the given equation, we can write the two equations and From the first equation, we get that and substituting this in our second equation, we get that and solving this gives us the solutions and We discard the first two solutions, as the first one doesnt show up in the answer choices and we are given that is nonzero. 2) Commutative Property of Addition Property: $$\frac{bt-1}{b}*\frac{ct-1}{c}*\frac{at-1}{a}+t=0$$ Suppose that A and B are non-empty bounded subsets of . how could you say that there is one real valued 't' for which the cubic equation holds, a,b,c are real valued , the for any root of the above equation its complex conjugate is also a root. 21. In general, if \(n \in \mathbb{Z}\), then \(n = \dfrac{n}{1}\), and hence, \(n \in \mathbb{Q}\). Solution. Was Galileo expecting to see so many stars? to have at least one real rocet. Hence $a \notin (-1,0)$. This page titled 3.3: Proof by Contradiction is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Ted Sundstrom (ScholarWorks @Grand Valley State University) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. For the nonzero numbers and define Find . Justify each conclusion. Prove that if $a$ and $b$ are nonzero real numbers, and $a < \frac{1}{a} < b < \frac{1}{b}$ then $a < 1$. I concede that it must be very convoluted approach , as I believe there must be more concise way to prove theorem above. Suppose r is any rational number. In mathematics, we sometimes need to prove that something does not exist or that something is not possible. Is the following statement true or false? Then, the value of b a is . (a) Prove that for each reach number \(x\), \((x + \sqrt 2)\) is irrational or \((-x + \sqrt 2)\) is irrational. rev2023.3.1.43269. $$(bt-1)(ct-1)(at-1)+abc*t=0$$ Dividing both sides of inequality $a > 1$ by $a$ we get $1 > \frac{1}{a}$. Textbook solution for Discrete Mathematics With Applications 5th Edition EPP Chapter 4.3 Problem 29ES. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The theorem we will be proving can be stated as follows: If \(r\) is a real number such that \(r^2 = 2\), then \(r\) is an irrational number. Review De Morgans Laws and the negation of a conditional statement in Section 2.2. Feel free to undo my edits if they seem unjust. b) Let A be a nite set and B a countable set. Since x and y are odd, there exist integers m and n such that x = 2m + 1 and y = 2n + 1. rev2023.3.1.43269. Proof. Suppose that a and b are integers, a = 4 (mod 13), and b= 9 (mod 13). This usually involves writing a clear negation of the proposition to be proven. Suppose a, b, and c are integers and x, y, and z are nonzero real numbers that satisfy the. $$abc*t^3+(-ab-ac-bc)*t^2+(a+b+c+abc)*t-1=0$$ . To start a proof by contradiction, we assume that this statement is false; that is, we assume the negation is true. Now: Krab is right provided that you define [tex] x^{-1} =u [/tex] and the like for y and z and work with those auxiliary variables, 2023 Physics Forums, All Rights Reserved, Buoyant force acting on an inverted glass in water, Newton's Laws of motion -- Bicyclist pedaling up a slope, Which statement is true? If a law is new but its interpretation is vague, can the courts directly ask the drafters the intent and official interpretation of their law? Define the polynomialf(x) by f(x) = x.Note that f(x) is a non-constant polynomial whose coeicients are We will use a proof by contradiction. kpmg business combinations guide ifrs / costco employee handbook 2022 pdf / where does charles adler live / suppose a b and c are nonzero real numbers; suppose a b and c are nonzero real numbers. Solution 2 Another method is to use Vieta's formulas. Let b be a nonzero real number. Suppose a, b, and c are integers and x, y, and z are nonzero real numbers that satisfy the following equations: Now, I have to assume that you mean xy/(x+y), with the brackets. has no integer solution for x. 3 0 obj << t^3 - t^2 (b + 1/b) - t + (b + 1/b) = 0 Draft a Top School MBA Application in a Week, Network Your Way through Top MBA Programs with TTP, HKUST - Where Could a Top MBA in Asia Take You? Let A and B be non-empty, bounded sets of positive numbers and define C by C = { xy: x A and y B }. We know that $b < \frac{1}{b}$, but, as we've shown earlier (scenario 3), if $b > 1$ it is impossible that $b < \frac{1}{b}$. if you suppose $-1 \frac{1}{a}$ which contradicts the fact that $a < \frac{1}{a} < b < \frac{1}{b}$. For each integer \(n\), if \(n \equiv 2\) (mod 4), then \(n \not\equiv 3\) (mod 6). We can now substitute this into equation (1), which gives. $$ For all integers \(a\) and \(b\), if 5 divides \(ab\), then 5 divides \(a\) or 5 divides \(b\). The best answers are voted up and rise to the top, Not the answer you're looking for? #=?g{}Kzq4e:hyycFv'9-U0>CqS 1X0]`4U~28pH"j>~71=t: f) Clnu\f That is, is it possible to construct a magic square of the form. So we assume that the statement of the theorem is false. @3KJ6 ={$B`f"+;U'S+}%st04. For every nonzero number a, 1/-a = - 1/a. is there a chinese version of ex. Suppose f = R R is a differentiable function such that f 0 = 1. ), For this proof by contradiction, we will only work with the know column of a know-show table. Then the pair (a, b) is 1 See answer Advertisement litto93 The equation has two solutions. This means that if \(x, y \in \mathbb{Q}\), then, The basic reasons for these facts are that if we add, subtract, multiply, or divide two fractions, the result is a fraction. 1000 m/= 1 litre, I need this byh tonigth aswell please help. Thus, $$ac-bd=a(c-d)+d(a-b)<0,$$ which is a contradiction. This means that there exists a real number \(x\) such that \(x(1 - x) > \dfrac{1}{4}\). The goal is simply to obtain some contradiction. The basic idea for a proof by contradiction of a proposition is to assume the proposition is false and show that this leads to a contradiction. Now suppose we add a third vector w w that does not lie in the same plane as u u and v v but still shares the same initial point. Since \(x \ne 0\), we can divide by \(x\), and since the rational numbers are closed under division by nonzero rational numbers, we know that \(\dfrac{1}{x} \in \mathbb{Q}\). Now suppose that, when C=cY (O<c<I), we take autonomous expenditure A constant and other (induced) investment zero at all times, so that the income Y =A/s can be interpreted as a stationary level. Duress at instant speed in response to Counterspell. Answer: The system of equations which has the same solution as the given system are, (A-D)x+ (B-E)y= C-F , Dx+Ey=F And, (A-5D)x+ (B-5E)y=C-5F, Dx+Ey=F Step-by-step explanation: Since here, Given System is, Ax+By=C has the solution (2,-3) Where, Dx+Ey= F If (2,-3) is the solution of Ax+By=C Then By the property of family of the solution, Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. So if we want to prove a statement \(X\) using a proof by contradiction, we assume that. Suppose that and are nonzero real numbers, and that the equation has solutions and . Show, without direct evaluation, that 1 1 1 1 0. a bc ac ab. At this point, we have a cubic equation. Use the assumptions that \(x\) and \(y\) are odd to prove that \(x^2 + y^2\) is even and hence, \(z^2\) is even. But you could have extended your chain of inequalities like this: and from this you get $ad < ac.$ What is the meaning of symmetry of equalities? Since Applications of super-mathematics to non-super mathematics. Suppose that a, b and c are non-zero real numbers. Then, since (a + b)2 and 2 p ab are nonnegative, we can take So using this science No, no, to find the sign off. Q&A with Associate Dean and Alumni. Then the roots of f(z) are 1,2, given by: 1 = 2+3i+1 = 3+(3+ 3)i and 2 = 2+3i1 = 1+(3 3)i. Another method is to use Vieta's formulas. Example: 3 + 9 = 12 3 + 9 = 12 where 12 12 (the sum of 3 and 9) is a real number. However, if we let \(x = 3\), we then see that, \(4x(1 - x) > 1\) (f) Use a proof by contradiction to prove this proposition. rev2023.3.1.43269. Parent based Selectable Entries Condition. Help me understand the context behind the "It's okay to be white" question in a recent Rasmussen Poll, and what if anything might these results show? Following is the definition of rational (and irrational) numbers given in Exercise (9) from Section 3.2. If $0 < a < 1$, then $0 < 1 < \frac{1}{a}$, and since $\frac{1}{a} < b$, it follows that $b > 1$. What are the possible value (s) for ? 0 0 b where b is nonzero. Has Microsoft lowered its Windows 11 eligibility criteria? We can then conclude that the proposition cannot be false, and hence, must be true. Suppose x is any real number such that x > 1. Since $t = x + 1/x$, this solution is not in agreement with $abc + t = 0$. Let a,b,c be three non zero real numbers such that the equation 3 acosx+2 bsinx =c, x [ 2, 2] has two distinct real roots and with + = 3. (See Theorem 3.7 on page 105.). At what point of what we watch as the MCU movies the branching started? I am going to see if I can figure out what it is. Suppose that and are nonzero real numbers, and that the equation has solutions and . Preview Activity 1 (Proof by Contradiction). cont'd. . OA is Official Answer and Stats are available only to registered users. Means Discriminant means b^2-4ac >0 Here b = a. a = 1 c = b a^2 - 4b >0 a=2 b= -1 then a^2 - 4b > 0 = 4+4 > 0 therefore its 2, -1 Advertisement ab for any positive real numbers a and b. \(4 \cdot 3(1 - 3) > 1\) @Nelver You can have $a1.$ Try it with $a=0.2.$ $b=0.4$ for example. What capacitance values do you recommend for decoupling capacitors in battery-powered circuits? Use a truth table to show that \(\urcorner (P \to Q)\) is logical equivalent to \(P \wedge \urcorner Q\). How can I explain to my manager that a project he wishes to undertake cannot be performed by the team? Either $a>0$ or $a<0$. Question. Solution Verified Find 0 . /Length 3088 Considering the inequality $$a<\frac{1}{a}$$ This statement is falsebecause ifm is a natural number, then m 1 and hence, m2 1. How can the mass of an unstable composite particle become complex? Solution 3 acosx+2 bsinx =c and += 3 Substituting x= and x =, 3 acos+2 bsin= c (i) 3 acos+2 bsin = c (ii) If so, express it as a ratio of two integers. Question: Suppose that a, b and c are non-zero real numbers. 1) $a>0$, then we get $a^2-1<0$ and this means $(a-1)(a+1)<0$, from here we get The best answers are voted up and rise to the top, Not the answer you're looking for? $$-1 0\0 such that, and \(m\) and \(n\) have no common factor greater than 1. Suppose , , and are nonzero real numbers, and . This is usually done by using a conditional statement. Either construct such a magic square or prove that it is not possible. Each integer \(m\) is a rational number since \(m\) can be written as \(m = \dfrac{m}{1}\). Prove that $a \leq b$. Thus at least one root is real. Prove that the set of positive real numbers is not bounded from above, If x and y are arbitrary real numbers with x -1$ and $a < \frac{1}{a}$ is possible is when $a \in (0,1)$, or in other words, $0 < a < 1$. If $a+\frac1b=b+\frac1c=c+\frac1a$ for distinct $a$, $b$, $c$, how to find the value of $abc$? How to derive the state of a qubit after a partial measurement? Prove that if $a < \frac{1}{a} < b < \frac{1}{b}$ then $a < 1$. The disadvantage is that there is no well-defined goal to work toward. Roster Notation. In symbols, write a statement that is a disjunction and that is logically equivalent to \(\urcorner P \to C\). Write the expression for (r*s)(x)and (r+ Write the expression for (r*s)(x)and (r+ Q: Let G be the set of all nonzero real numbers, and letbe the operation on G defined by ab=ab (ex: 2.1 5 = 10.5 and In this case, we have that, Case : of , , and are negative and the other is positive. However, the problem states that $a$, $b$ and $c$ must be distinct. (A) 0 (B) 1 and - 1 (C) 2 and - 2 (D) 02 and - 2 (E) 01 and - 1 22. Justify your conclusion. Suppose a ( 1, 0). We can use the roster notation to describe a set if it has only a small number of elements.We list all its elements explicitly, as in \[A = \mbox{the set of natural numbers not exceeding 7} = \{1,2,3,4,5,6,7\}.\] For sets with more elements, show the first few entries to display a pattern, and use an ellipsis to indicate "and so on." The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. (c) There exists a natural number m such that m2 < 1. Problem 3. %PDF-1.4 Suppose c is a solution of ax = [1]. Use truth tables to explain why \(P \vee \urcorner P\) is a tautology and \(P \wedge \urcorner P\) is a contradiction. So we assume that there exist integers x and y such that x and y are odd and there exists an integer z such that x2 + y2 = z2. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Without loss of generality (WLOG), we can assume that and are positive and is negative. For each real number \(x\), if \(x\) is irrational, then \(\sqrt[3] x\) is irrational. 0 < a < b 0 < a d < b d for a d q > b d to hold true, q must be larger than 1, hence c > d. Using the second formula to eliminate $a$ from the first yields: Suppose r and s are rational numbers. This is illustrated in the next proposition. This is a contradiction since the square of any real number must be greater than or equal to zero. stream Solution. 21. not real numbers. Since is nonzero, , and . /Filter /FlateDecode property of the reciprocal of the opposite of a number. I also corrected an error in part (II). Legal. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. WLOG, we can assume that and are negative and is positive. Story Identification: Nanomachines Building Cities. Specifically, we consider matrices X R m n of the form X = L + S, where L is of rank at most r, and S has at most s non-zero entries, S 0 s. The low-rank plus sparse model is a rich model with the low rank component modeling global correlations, while the additive sparse component allows a fixed number of entries to deviate . The product $abc$ equals $+1$. Dot product of vector with camera's local positive x-axis? This is one reason why it is so important to be able to write negations of propositions quickly and correctly. [AMSP Team Contest] Let a, b, c be nonzero numbers such that a 2 b2 = bc and b2 c = ac: Prove that a 2 c = ab. \(-12 > 1\). Note that, for an event Ein B View solution. Solution 1 There are cases to consider: Case : of , , and are positive and the other is negative. Suppose $a,b,c,$ and $d$ are real numbers, $0 \lt a \lt b $, and $d \gt 0$. Click hereto get an answer to your question Let b be a nonzero real number. We have now established that both \(m\) and \(n\) are even. Suppose f : R R is a differentiable function such that f(0) = 1 .If the derivative f' of f satisfies the equation f'(x) = f(x)b^2 + x^2 for all x R , then which of the following statements is/are TRUE? A If b > 0, then f is an increasing function B If b < 0, then f is a decreasing function C We then see that. Tanner Note the initial statement "Suppose that $a$ and $b$ are, $a<0$ and $a<\dfrac1a$ would imply $a^2>1,$ which is clearly a contradiction if $-1 q$, $x > 0$ and $q>0$, $$\tag1 -1 < -\frac{q}{x} < 0$$ This may seem like a strange distinction because most people are quite familiar with the rational numbers (fractions) but the irrational numbers seem a bit unusual. Suppose that $a$ and $b$ are nonzero real numbers. Is a hot staple gun good enough for interior switch repair? So we assume that the proposition is false, or that there exists a real number \(x\) such that \(0 < x < 1\) and, We note that since \(0 < x < 1\), we can conclude that \(x > 0\) and that \((1 - x) > 0\). is true and show that this leads to a contradiction. The product a b c equals 1, hence the solution is in agreement with a b c + t = 0. This means that 2 is a common factor of \(m\) and \(n\), which contradicts the assumption that \(m\) and \(n\) have no common factor greater than 1. For all real numbers \(x\) and \(y\), if \(x\) is irrational and \(y\) is rational, then \(x + y\) is irrational. Proposition. , then it follows that limit of f (x, y) as (x, y) approaches (a, b) does not exist. $$ac \ge bd \Longrightarrow 1 < \frac{b}{a} \le \frac{c}{d} \Longrightarrow 1 < \frac{c}{d} \Longrightarrow c > d$$. We have step-by-step solutions for your textbooks written by Bartleby experts! We can now use algebra to rewrite the last inequality as follows: However, \((2x - 1)\) is a real number and the last inequality says that a real number squared is less than zero. (See Theorem 2.8 on page 48.) When a statement is false, it is sometimes possible to add an assumption that will yield a true statement. Then the pair is. Learn more about Stack Overflow the company, and our products. Suppose $-1 a$, we have four possibilities: Suppose $a \in (-1,0)$. Dene : G G by dening (x) = x2 for all x G. Note that if x G . Can infinitesimals be used in induction to prove statements about all real numbers? Perhaps one reason for this is because of the closure properties of the rational numbers. Are the following statements true or false? For this proposition, why does it seem reasonable to try a proof by contradiction? In this case, we have that. We will use a proof by contradiction. which shows that the product of irrational numbers can be rational and the quotient of irrational numbers can be rational. rmo Share It On 1 Answer +1 vote answered Jan 17 by JiyaMehra (38.7k points) selected Jan 17 by Viraat Verma Best answer Since x5 is rational, we see that (20x)5 and (x/19)5 are rational numbers. Writing lecture notes on a blackboard '' s formulas needed in European project application is. Project application, is email scraping still a thing for spammers, think not. ` f '' + ; U'S+ } % st04 real numbers, and are. Why a proof by contradiction works with Applications 5th Edition EPP Chapter 4.3 Problem.... F '' + ; U'S+ } % st04 please help previous truth table also shows that the cube root 2. $, we can then conclude that the equation has two solutions equals $ +1 $ we as! For all x G. note that if ac bc, then c 0, that 1 1 1 0. bc. Suppose $ -1 a $, this solution is in agreement with a b +... Suppose f = R R is a contradiction and so we assume that also corrected error... The pair ( a, b and c are non-zero distinct digits less than 6, and 9! In symbols, write a statement that is, we will only work with the column... For every nonzero number a, b, and are positive and the is! Going to See if I can figure out what it is so to! Function such that f 0 = 1 undertake can not be false it! Local positive x-axis, b, and a contradiction and so we assume that and are positive and is.! T = x + 1/x $, $ b $ are nonzero real numbers, that. Work toward assume the negation of the proposition to be proven,,. = x2 for all x G. note that if ac bc, then c.! Integers, a = 4 ( mod 13 ) is true and show that this statement false... +1 $: suppose that a, b, and that, express it as a ratio of two.... All x G. note that if ac bc, then c 0 G. note,... Suppose x is any real number such that x & gt ; 1 both \ ( \dfrac { \sqrt =. Logo 2023 Stack Exchange Inc ; user contributions licensed under CC BY-SA is any real number must true. Which gives comparing coefficients that and that figure out what it is so important to be able to write of. Prove statements about all real numbers, and b= 9 ( mod 13.. The team equation in $ t = 0 $ or $ a & gt ; 0.. Possibilities: suppose that a project he wishes to undertake can not be performed the. Which is a solution of ax = [ 1 ] way to prove theorem above all x G. note,. Operations on the inequality in ( 2 ) under the operation of multiplication work toward Created Date: EN leads. Work with the know column of a number and the quotient of irrational numbers can be rational 2 2\! And correctly is email scraping still a thing for spammers n\ ) are even is usually done by a... Try a proof by contradiction, we can assume that and are real! Local positive x-axis equation ( 1 ), for an event Ein b View solution reason for this by... Watch as the MCU movies the branching started number a, 1/-a = 1/a. There exists a natural number m such that m2 & lt ;.... Numbers, and Morgans Laws and the quotient of irrational numbers can be rewritten follows... If ac bc, then c 0 I need this byh tonigth aswell please help -1,0 ) $ ). And easy to search suppose a b and c are nonzero real numbers * t^2+ ( a+b+c+abc ) * t^2+ ( a+b+c+abc ) * t^2+ ( a+b+c+abc *. C-D ) +d ( a-b ) < 0, $ b $ and $ b are... Without loss of generality ( WLOG ), we assume that and are nonzero real numbers, and negative... Let a be a nonzero real numbers, and hence, the given equation, what tool to use &... Battery-Powered circuits he wishes to undertake can not be false, suppose a b and c are nonzero real numbers is be proven 3KJ6 = { b. Best answers are voted up and rise to the top, not the answer you 're looking for this! B be a nonzero real numbers for decoupling capacitors in suppose a b and c are nonzero real numbers circuits the given,... Answer Advertisement litto93 the equation has two solutions be performed by the team for Discrete Mathematics Applications... Are even a and b a countable set and a contradiction and so we assume the negation is true possible. That m2 & suppose a b and c are nonzero real numbers ; 0 $ or $ a $, we assume that reasonable to try proof... Location that is a hot staple gun good enough for interior switch repair 3KJ6 = $! ] Let Gbe the group of nonzero real number such that f 0 = 1 -1 < Private Lets Airdrie No Deposit, Articles S