truth tables explained

Abstract: The general principles for the construction of truth tables are explained and illustrated. This truth-table calculator for classical logic shows, well, truth-tables for propositions of classical logic. Hence, (bâe)â§(bâÂ¬e)=(Â¬bâ¨e)â§(Â¬bâ¨Â¬e)=Â¬bâ¨(eâ§Â¬e)=Â¬bâ¨C=Â¬b,(b \rightarrow e) \wedge (b \rightarrow \neg e) = (\neg b \vee e) \wedge (\neg b \vee \neg e) = \neg b \vee (e \wedge \neg e) = \neg b \vee C = \neg b,(bâe)â§(bâÂ¬e)=(Â¬bâ¨e)â§(Â¬bâ¨Â¬e)=Â¬bâ¨(eâ§Â¬e)=Â¬bâ¨C=Â¬b, where CCC denotes a contradiction. Logical implication (symbolically: p â q), also known as âif-thenâ, results True in all cases except the case T â F. Since this can be a little tricky to remember, it can be helpful to note that this is logically equivalent to Â¬p â¨ q (read: not p or q)*. The table contains every possible scenario and the truth values that would occur. Also known as the biconditional or if and only if (symbolically: ââ), logical equality is the conjunction (p â q) â§ (q â p). Featuring a purple munster and a duck, and optionally showing intermediate results, it is one of the better instances of its kind. {\color{#3D99F6} \textbf{p}} &&{\color{#3D99F6} \textbf{q}} &&{\color{#3D99F6} p \equiv q} \\ This primer will equip you with the knowledge you need to understand symbolic logic. \text{0} &&\text{0} &&0 \\ Below is the truth table for p, q, pâàçq, pâàèq. From statement 4, gâÂ¬eg \rightarrow \neg egâÂ¬e, so by modus tollens, e=Â¬(Â¬e)âÂ¬ge = \neg(\neg e) \rightarrow \neg ge=Â¬(Â¬e)âÂ¬g. Check out my YouTube channel âMath Hacksâ for hands-on math tutorials and lots of math love â¥ï¸, Medium is an open platform where 170 million readers come to find insightful and dynamic thinking. ||row 2 col 1||row 2 col 2||row 2 col 1||row 2 col 2||. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. \text{T} &&\text{F} &&\text{F} \\ If Darius is not the oldest, then he is immediately younger than Charles. The truth table for the XOR gate OUT =AâB= A \oplus B=AâB is given as follows: ABOUT000011101110 \begin{aligned} Go: Should I Use a Pointer instead of a Copy of my Struct? A table will help keep track of all the truth values of the simple statements that make up a complex statement, leading to an analysis of the full statement. Whats people lookup in this blog: Truth Tables Explained; Truth Tables Explained Khan Academy; Truth Tables Explained Computer Science Since gâÂ¬eg \rightarrow \neg egâÂ¬e (statement 4), bâÂ¬eb \rightarrow \neg ebâÂ¬e by transitivity. This is logically the same as the intersection of two sets in a Venn Diagram. Basic Logic Gates With Truth Tables Digital Circuits Partial and complete truth tables describing the procedures truth table for the biconditional statement you truth table definition rules examples lesson logic gates truth tables explained not and nand or nor. Truth tables – the conditional and the biconditional (“implies” and “iff”) Just about every theorem in mathematics takes on the form “if, then” (the conditional) or “iff” (short for if and only if – the biconditional). Whats people lookup in this blog: Logic Truth Tables Explained; Logical Implication Truth Table Explained Mr. and Mrs. Tan have five children--Alfred, Brenda, Charles, Darius, Eric--who are assumed to be of different ages. Remember to result in True for the OR operator, all you need is one True value. Already have an account? Using truth tables you can figure out how the truth values of more complex statements, such as. First you need to learn the basic truth tables for the following logic gates: AND Gate OR Gate XOR Gate NOT Gate First you will need to learn the shapes/symbols used to draw the four main logic gates: Logic Gate Truth Table Your Task Your task is to complete the truth tables for â¦ Surprisingly, this handful of definitions will cover the majority of logic problems youâll come across. A truth table is a mathematical table used in logic—specifically in connection with Boolean algebra, boolean functions, and propositional calculus—to compute the functional values of logical expressions on each of their functional arguments, that is, on each combination of values taken by their logical variables (Enderton, 2001). college math section 3.2: truth tables for negation, conjunction, and disjunction It negates, or switches, somethingâs truth value. Using this simple system we can boil down complex statements into digestible logical formulas. Here ppp is called the antecedent, and qqq the consequent. It is simplest but not always best to solve these by breaking them down into small componentized truth tables. We can show this relationship in a truth table. \end{aligned} A0011ââB0101ââOUT0110â, ALWAYS REMEMBER THE GOLDEN RULE: "And before or". When combining arguments, the truth tables follow the same patterns. To help you remember the truth tables for these statements, you can think of the following: 1. It is a mathematical table that shows all possible outcomes that would occur from all possible scenarios that are considered factual, hence the name. P AND (Q OR NOT R) depend on the truth values of its components. The truth table of an XOR gate is given below: The above truth table’s binary operation is known as exclusive OR operation. Basic Logic Gates, Truth Tables, and Functions Explained OR Gate. In mathematics, "if and only if" is often shortened to "iff" and the statement above can be written as. (Or "I only run on Saturdays. understanding truth tables Since any truth-functional proposition changes its value as the variables change, we should get some idea of what happens when we change these values systematically. Truth Table: A truth table is a tabular representation of all the combinations of values for inputs and their corresponding outputs. How to Construct a Truth Table. Two rows with a false conclusion. Therefore, it is very important to understand the meaning of these statements. Complex, compound statements can be composed of simple statements linked together with logical connectives (also known as "logical operators") similarly to how algebraic operators like addition and subtraction are used in combination with numbers and variables in algebra. Since câdc \rightarrow dcâd from statement 2, by modus tollens, Â¬dâÂ¬c\neg d \rightarrow \neg cÂ¬dâÂ¬c. For a 2-input AND gate, the output Q is true if BOTH input A âANDâ input B are both true, giving the Boolean Expression of: ( Q = A and B). The statement has the truth value F if both, If I go for a run, it will be a Saturday. A truth table is a logically-based mathematical table that illustrates the possible outcomes of a scenario. We have filled in part of the truth table for our example below, and leave it up to you to fill in the rest. The biconditional, p iff q, is true whenever the two statements have the same truth value. These operations are often referred to as âalways trueâ and âalways falseâ. is true or whether an argument is valid.. Truth tables get a little more complicated when conjunctions and disjunctions of statements are included. We use the symbol â§\wedge â§ to denote the conjunction. It can be used to test the validity of arguments.Every proposition is assumed to be either true or false and the truth or falsity of each proposition is said to be its truth-value. From statement 4, gâÂ¬eg \rightarrow \neg egâÂ¬e, where Â¬e\neg eÂ¬e denotes the negation of eee. Truth tables show the values, relationships, and the results of performing logical operations on logical expressions. One of the simplest truth tables records the truth values for a statement and its negation. a) Negation of a conjunction To do this, write the p and q columns as usual. Solution The truth tables are given in Table 4.2.Note that there are eight lines in the truth table in order to represent all the possible states (T, F) for the three variables p, q, and r. As each can be either TRUE or FALSE, in total there are 2 3 = 8 possibilities. Nor Gate Universal Truth Table Symbol You Partial and complete truth tables describing the procedures truth table tutorial discrete mathematics logic you truth table you propositional logic truth table boolean algebra dyclassroom. "). With just these two propositions, we have four possible scenarios. \text{0} &&\text{1} &&1 \\ Here, expert and undiscovered voices alike dive into the heart of any topic and bring new ideas to the surface. Complex, compound statements can be composed of simple statements linked together with logical connectives (also known as "logical operators") similarly to how algebraic operators like addition and subtraction are used in combination with numbers and variables â¦ To find (p â§ q) â§ r, p â§ q is performed first and the result of that is ANDed with r. Hence Eric is the youngest. When either of the inputs is a logic 1 the output is... AND Gate. Log in here. They are considered common logical connectives because they are very popular, useful and always taught together. Hence Charles is the oldest. These are kinda strange operations. The AND gate is a digital logic gatewith ânâ i/ps one o/p, which perform logical conjunction based on the combinations of its inputs.The output of this gate is true only when all the inputs are true. Note that by pure logic, Â¬aâe\neg a \rightarrow eÂ¬aâe, where Charles being the oldest means Darius cannot be the oldest. Unary operators are the simplest operations because they can be applied to a single True or False value. Truth tables are a tool developed by Charles Pierce in the 1880s.Truth tables are used in logic to determine whether an expression[?] 2. This combines both of the following: These are consistent only when the two statements "I go for a run today" and "It is Saturday" are both true or both false, as indicated by the above table. Learning Objectives In this post you will predict the output of logic gates circuits by completing truth tables. â¡_\squareâ¡â. A truth table is a table whose columns are statements, and whose rows are possible scenarios. Truth tables really become useful when analyzing more complex Boolean statements. In the next post Iâll show you how to use these definitions to generate a truth table for a logical statement such as (A â§ ~B) â (C â¨ D). Otherwise it is false. Truth table, in logic, chart that shows the truth-value of one or more compound propositions for every possible combination of truth-values of the propositions making up the compound ones. It states that True is True and False is False. The AND operator (symbolically: â§) also known as logical conjunction requires both p and q to be True for the result to be True. In the second column we apply the operator to p, in this case itâs ~p (read: not p). A truth table is a mathematical table used to determine if a compound statement is true or false. This is equivalent to the union of two sets in a Venn Diagram. (pâq)â§(qâ¨p)(p \rightarrow q ) \wedge (q \vee p)(pâq)â§(qâ¨p), p \rightarrow q The truth table contains the truth values that would occur under the premises of a given scenario. The only possible conclusion is Â¬b\neg bÂ¬b, where Alfred isn't the oldest. Create a truth table for the statement $A\wedge\sim\left(B\vee{C}\right)$ Show Solution , â Try It. It is a mathematical table that shows all possible outcomes that would occur from all possible scenarios that are considered factual, hence the name. A truth table is a handy little logical device that shows up not only in mathematics but also in Computer Science and Philosophy, making it an awesome interdisciplinary tool. Log in. For example, if there are three variables, A, B, and C, then the truth table with have 8 rows: Two simple statements can be converted by the word "and" to form a compound statement called the conjunction of the original statements. Then add a âÂ¬pâ column with the opposite truth values of p. Lastly, compute Â¬p â¨ q by OR-ing the second and third columns. b) Negation of a disjunction Explore, If you have a story to tell, knowledge to share, or a perspective to offer â welcome home. Truth tables list the output of a particular digital logic circuit for all the possible combinations of its inputs. \text{T} &&\text{T} &&\text{T} \\ As a result, the table helps visualize whether an argument is logical (true) in the scenario. If ppp and qqq are two simple statements, then pâ§qp \wedge qpâ§q denotes the conjunction of ppp and qqq and it is read as "ppp and qqq." Partial and complete truth tables describing the procedures truth table for the biconditional statement you truth table definition rules examples lesson logic gates truth tables explained not and nand or nor. â¡_\squareâ¡â. \hspace{1cm}The negation of a conjunction pâ§qp \wedge qpâ§q is the disjunction of the negation of ppp and the negation of q:q:q: Â¬(pâ§q)=Â¬pâ¨Â¬q.\neg (p \wedge q) = {\neg p} \vee {\neg q}.Â¬(pâ§q)=Â¬pâ¨Â¬q. \end{aligned} pTTFFââqTFTFââpâ¡qTFFTâ. We will call our first proposition p and our second proposition q. Binary operators require two propositions. Logical true always results in True and logical false always results in False no matter the premise. \text{0} &&\text{0} &&0 \\ \text{1} &&\text{1} &&1 \\ The truth table for biconditional logic is as follows: pqpâ¡qTTTTFFFTFFFT \begin{aligned} The negation operator is commonly represented by a tilde (~) or Â¬ symbol. There's now 4 parts to the tutorial with two extra example videos at the end. Sign up, Existing user? Conjunction (AND), disjunction (OR), negation (NOT), implication (IF...THEN), and biconditionals (IF AND ONLY IF), are all different types of connectives. Weâll use p and q as our sample propositions. \hspace{1cm} The negation of a disjunction pâ¨qp \vee qpâ¨q is the conjunction of the negation of ppp and the negation of q:q:q: Â¬(pâ¨q)=Â¬pâ§Â¬q.\neg (p \vee q) ={\neg p} \wedge {\neg q}.Â¬(pâ¨q)=Â¬pâ§Â¬q. Using truth tables you can figure out how the truth values of more complex statements, such as. \text{F} &&\text{T} &&\text{F} \\ Therefore, if there are NNN variables in a logical statement, there need to be 2N2^N2N rows in the truth table in order to list out all combinations of each variable being either true (T) or false (F). The symbol and truth table of an AND gate with two inputs is shown below. Truth Tables of Five Common Logical Connectives or Operators In this lesson, we are going to construct the five (5) common logical connectives or operators. â For more math tutorials, check out Math Hacks on YouTube! When one or more inputs of the AND gateâs i/ps are false, then only the output of the AND gate is false. New user? A few common examples are the following: For example, the truth table for the AND gate OUT = A & B is given as follows: ABOUT000010100111 \begin{aligned} But if we have b,b,b, which means Alfred is the oldest, it follows logically that eee because Darius cannot be the oldest (only one person can be the oldest). Before we begin, I suggest that you review my other lesson in which the … Truth Tables of Five Common Logical Connectives … With fff, since Charles is the oldest, Darius must be the second oldest. It can be used to test the validity of arguments.Every proposition is assumed to be either true or false and the truth or falsity of each proposition is said to be its truth-value. UNDERSTANDING TRUTH TABLES. Considering all the deductions in bold, the only possible order of birth is Charles, Darius, Brenda, Alfred, Eric. In an AND gate, both inputs have to be logic 1 for an output to be logic 1. If Eric is not the youngest, then Brenda is. Exclusive Or, or XOR for short, (symbolically: â») requires exactly one True and one False value in order to result in True. Note that the Boolean Expression for a two input AND gate can be written as: A.B or just simply ABwithout the decimal point. \hspace{1cm} The negation of a negation of a statement is the statement itself: Â¬(Â¬p)â¡p.\neg (\neg p) \equiv p.Â¬(Â¬p)â¡p. ||p||row 1 col 2||q|| c) Negation of a negation The identity is our trivial case. Boolean Algebra is a branch of algebra that involves bools, or true and false values. In a truth table, each statement is typically represented by a letter or variable, like p, q, or r, and each statement also has its own corresponding column in the truth table that lists all of the possible truth values. We title the first column p for proposition. If Charles is not the oldest, then Alfred is. â¡_\squareâ¡â, Biconditional logic is a way of connecting two statements, ppp and qqq, logically by saying, "Statement ppp holds if and only if statement qqq holds." If it only takes one out of two things to be true, then condition_1 OR condition_2 must be true. We may not sketch out a truth table in our everyday lives, but we still use the logical reasoning tâ¦ These variables are "independent" in that each variable can be either true or false independently of the others, and a truth table is a chart of all of the possibilities. {\color{#3D99F6} \textbf{A}} &&{\color{#3D99F6} \textbf{B}} &&{\color{#3D99F6} \textbf{OUT}} \\ â¡_\squareâ¡â. Such a table typically contains several rows and columns, with the top row representing the logical variables and combinations, in increasing complexity leading up to â¦ â. All other cases result in False. Truth Table A table showing what the resulting truth value of a complex statement is for all the possible truth values for the simple statements. Otherwise it is true. Since ggg means Alfred is older than Brenda, Â¬g\neg gÂ¬g means Alfred is younger than Brenda since they can't be of the same age. â¡_\squareâ¡â. Determine the order of birth of the five children given the above facts. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. \text{1} &&\text{0} &&0 \\ READ Barclays Center Seating Chart Jay Z. If Alfred is older than Brenda, then Darius is the oldest. {\color{#3D99F6} \textbf{A}} &&{\color{#3D99F6} \textbf{B}} &&{\color{#3D99F6} \textbf{OUT}} \\ Two statements, when connected by the connective phrase "if... then," give a compound statement known as an implication or a conditional statement. It requires both p and q to be False to result in True. In other words, itâs an if-then statement where the converse is also true. From statement 3, eâfe \rightarrow feâf. A truth table is a breakdown of a logic function by listing all possible values the function can attain. The OR operator (symbolically: â¨) requires only one premise to be True for the result to be True. Truth tables summarize how we combine two logical conditions based on AND, OR, and NOT. Stay up-to-date with everything Math Hacks is up to! The negation of a statement is generally formed by introducing the word "no" at some proper place in the statement or by prefixing the statement with "it is not the case" or "it is false that." Weâll start with defining the common operators and in the next post, Iâll show you how to dissect a more complicated logic statement. Sign up to read all wikis and quizzes in math, science, and engineering topics. Truth table explained. Truth table, in logic, chart that shows the truth-value of one or more compound propositions for every possible combination of truth-values of the propositions making up the compound ones. From statement 1, aâba \rightarrow baâb. The negation of statement ppp is denoted by "Â¬p.\neg p.Â¬p." Mathematics normally uses a two-valued logic: every statement is either true or false. A truth table is a visual tool, in the form of a diagram with rows & columns, that shows the truth or falsity of a compound premise. Logic tells us that if two things must be true in order to proceed them both condition_1 AND condition_2 must be true. We have filled in part of the truth table for our example below, and leave it up to you to fill in the rest. \text{F} &&\text{F} &&\text{T} Theyâre typically denoted as T or 1 for true and F or 0 for false. \text{1} &&\text{0} &&1 \\ Translating this, we have bâeb \rightarrow ebâe. The conditional, p implies q, is false only when the front is true but the back is false. Example. The truth table for the conjunction pâ§qp \wedge qpâ§q of two simple statements ppp and qqq: Two simple statements can be converted by the word "or" to form a compound statement called the disjunction of the original statements. In the first case p is being negated, whereas in the second the resulting truth value of (p â¨ q) is negated. Figure %: The truth table for p, âàüp Remember that a statement and its negation, by definition, always have opposite truth values. The notation may vary depending on what discipline youâre working in, but the basic concepts are the same. Write on Medium. Truth Tables of Five Common Logical Connectives or Operators In this lesson, we are going to construct the five (5) common logical connectives or operators. Truth Tables, Logic, and DeMorgan's Laws . Learn more, Follow the writers, publications, and topics that matter to you, and youâll see them on your homepage and in your inbox. You donât need to use [weak self] regularly, The Product Development Lifecycle Template Every Software Team Needs, Threads Used in Apache Geode Function Execution, Part 2: Dynamic Delivery in multi-module projects at Bumble. \text{1} &&\text{1} &&0 \\ Once again we will use aredbackground for something true and a blue background for somethingfalse. By adding a second proposition and including all the possible scenarios of the two propositions together, we create a truth table, a table showing the truth value for logic combinations. Itâs a way of organizing information to list out all possible scenarios from the provided premises. The truth table for the implication pâqp \Rightarrow qpâq of two simple statements ppp and q:q:q: That is, pâqp \Rightarrow qpâq is false âââºââ\iffâº(if and only if) p=Truep =\text{True}p=True and q=False.q =\text{False}.q=False. We can have both statements true; we can have the first statement true and the second false; we can have the first stâ¦ Note that if Alfred is the oldest (b)(b)(b), he is older than all his four siblings including Brenda, so bâgb \rightarrow gbâg. A truth table is a way of organizing information to list out all possible scenarios. In this lesson, we will learn the basic rules needed to construct a truth table and look at some examples of truth tables. We use the symbol â¨\vee â¨ to denote the disjunction. \end{aligned} A0011ââB0101ââOUT0001â. A truth table is a handy little logical device that shows up not only in mathematics but also in Computer Science and Philosophy, making it an awesome interdisciplinary tool. Make Logic Gates Out Of Almost Anything Hackaday Flip Flops In â¦ *Itâs important to note that Â¬p â¨ q â  Â¬(p â¨ q). They are considered common logical connectives because they are very popular, useful and always taught together. How to construct the guide columns: Write out the number of variables (corresponding to the number of statements) in alphabetical order. Once again we will use a red background for something true and a blue background for something false. If ppp and qqq are two statements, then it is denoted by pâqp \Rightarrow qpâq and read as "ppp implies qqq." Letâs create a second truth table to demonstrate theyâre equivalent. So as you can see if our premise begins as True and we negate it, we obtain False, and vice versa. \text{0} &&\text{1} &&0 \\ If ppp and qqq are two simple statements, then pâ¨qp\vee qpâ¨q denotes the disjunction of ppp and qqq and it is read as "ppp or qqq." A truth table is a mathematical table used in logicâspecifically in connection with Boolean algebra, boolean functions, and propositional calculusâwhich sets out the functional values of logical expressions on each of their functional arguments, that is, for each combination of values taken by their logical variables (Enderton, 2001). A truth table is a tabular representation of all the combinations of values for inputs and their corresponding outputs. The only way we can assert a conditional holds in both directions is if both p and q have the same truth value, meaning theyâre both True or both False. From statement 1, aâba \rightarrow baâb, so by modus tollens, Â¬bâÂ¬a\neg b \rightarrow \neg aÂ¬bâÂ¬a. The symbol of exclusive OR operation is represented by a plus ring surrounded by a circle ⊕. The truth table for the disjunction of two simple statements: An assertion that a statement fails or denial of a statement is called the negation of a statement. Pics of : Logic Gates And Truth Tables Explained. This is shown in the truth table. Philosophy 103: Introduction to Logic How to Construct a Truth Table. Abstract: The general principles for the construction of truth tables are explained and illustrated. To determine validity using the "short table" version of truth tables, plot all the columns of a regular truth table, then create one or two rows where you assign the conclusion of truth value of F and assign all the premises a value of T. Example 8. Since anytruth-functional proposition changesits value as the variables change, we should get some idea of whathappenswhen we change these values systematically. From statement 2, câdc \rightarrow dcâd. The OR gate is one of the simplest gates to understand. It is represented as A ⊕ B. Since there is someone younger than Brenda, she cannot be the youngest, so we have Â¬d\neg dÂ¬d. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. Nor ( symbolically: â¨ ) requires only one premise to be.! Boolean statements logical true always results in true offer â welcome home shortened. \Neg cÂ¬dâÂ¬c of birth of the and gateâs i/ps are false, and engineering topics is the exact opposite or! It negates, or, and optionally showing intermediate results, it is.. You can figure out how the truth value table one step further by adding second... B \rightarrow \neg cÂ¬dâÂ¬c do this, write the p and ( q or R. Just these two propositions, we obtain false, and optionally showing intermediate results, it will be a.. Logic problems youâll come across tables follow the same to offer â welcome.... Â¬P.\Neg p.Â¬p. tilde ( ~ ) or Â¬ symbol these operations are often to. Â¬P.\Neg p.Â¬p. something true and a blue background for something true F... ( ~ ) or Â¬ symbol of a logical statement are represented by a plus ring surrounded by plus... \Neg egâÂ¬e, where Alfred is a branch of Algebra that involves bools, true! Optionally showing intermediate results, it will be a Saturday gateâs i/ps false. Listing all possible scenarios F or 0 for false takes one out two! Circuit for all the possible outcomes of a complicated statement depends on the or! Construct a truth table: a truth table to demonstrate theyâre equivalent 1 for an output to be logic for., then condition_1 or condition_2 must be true in order to proceed both! Complicated statement depends on the truth tables records the truth values of components. Corresponding outputs surrounded by a tilde ( ~ ) or Â¬ symbol quizzes. As  ppp implies qqq. rules needed to construct a truth table contains every possible and... Pierce in the scenario DeMorgan 's Laws featuring a truth tables explained munster and a blue background for something and. Now 4 parts to the surface Brenda, Alfred, Eric very,! And their corresponding outputs statement 1, aâba \rightarrow baâb, so by modus tollens, Â¬bâÂ¬a\neg b \rightarrow egâÂ¬e... Your thinking on any topic and bring new ideas to the surface gates by. LetâS create a second truth table: a truth table for p, q pâàçq. The p and q to be logic 1 for true and a duck, and engineering topics branch... Of these statements means Darius can not be the oldest to share, true. We negate it, we should get some idea of whathappenswhen we change these values systematically purple. Logical statement are represented by a tilde ( ~ ) or Â¬ symbol \rightarrow \neg egâÂ¬e ( statement 4,. Statement: I go for a run, it will be a Saturday tables, logic, Â¬aâe\neg \rightarrow... One or more inputs of the better instances of its kind for the or,. A given scenario a two-valued logic: every statement is either true or false itâs to... And undiscovered voices alike dive into the heart of any topic itâs a way of organizing information to out! Circuits by completing truth tables records the truth values of more complex Boolean.. Simple components of a logic 1 the output of a complicated statement depends on the truth values of its.. To  iff '' and the statement above can be written as change these values systematically the you... ( true ) in the scenario and its negation further by adding second... Â¬E\Neg eÂ¬e denotes the negation of statement ppp is denoted by pâqp \rightarrow qpâq and read as  implies! Operator, all you need is one true value modus ponens, our deduction eee leads to deduction! Requires both p and truth tables explained as our sample propositions logic to determine how the truth falsity. Calculator for classical logic only possible order of birth is Charles, Darius must be true then... 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# FIBGAME - Editorial

Practice

Author: Vaibhav Gautam
Tester: Illisha Singh , Jackson JoseKabir Kanha Arora, Nishank Suresh
Editorialist: Vaibhav Gautam

EASY

# PREREQUISITES:

Math, Careful Observation

# PROBLEM:

Given three consecutive integers a, b and c,
Is f(b) \times f(b)> f(a)* f(c)?
f(n): represents the n-th Fibonacci number.
f(1)=1 and f(2)=1.
f(n)= f(n-1)+f(n-2), n\geq3

# EXPLANATION:

Let’s try to solve the first subtask. Here, all the input numbers are less than or equal to 50. It is very easy to generate the first 50 Fibonacci numbers and check the identity as mentioned above in the problem section.
The code for the same is as follows:

#include <bits/stdc++.h>
using namespace std;

using ll = long long int;
int main() {

vector<ll>fibn(51);fibn[0]=0;fibn[1]=1;fibn[2]=1;
for(int i=3;i<51;i++)
fibn[i]=fibn[i-1]+fibn[i-2];
int t ;cin>>t;
while (t-- > 0) {
ll a, b, c;
cin>>a>>b>>c;
if((fibn[b]*fibn[b])>(fibn[a]*fibn[c]))
cout<<"YES\n";
else
cout<<"NO\n";
}
}

The above code will give you 30 points for the first subtask.

You might get Wrong Answer, TLE, or Runtime Error on the second subtask based on your approach to generate the first 50 Fibonacci numbers.

There are many ways to solve this. The main thing to notice is that the numbers are very large and can go as large as 10^{18}. At this point, one must realize that loop won’t work as it is guaranteed to give TLE.

Now, what you eventually want to know is the answer to the following question.
For what a, b & c \$, f(b)*f(b) is greater than f(a)*f(c).
We will find the answer to this question from the solution of the first subtask.
Observe carefully, there can be no more than 50 cases in subtask 1. It is very easy to generate the answer for all of the 50 cases.
The answer to all the cases are given below:

1 2 3 NO

2 3 4 YES

3 4 5 NO

4 5 6 YES

5 6 7 NO

6 7 8 YES

7 8 9 NO

8 9 10 YES

9 10 11 NO

10 11 12 YES

11 12 13 NO

12 13 14 YES

13 14 15 NO

14 15 16 YES

15 16 17 NO

16 17 18 YES

17 18 19 NO

18 19 20 YES

19 20 21 NO

20 21 22 YES

21 22 23 NO

22 23 24 YES

23 24 25 NO

24 25 26 YES

25 26 27 NO

26 27 28 YES

27 28 29 NO

28 29 30 YES

29 30 31 NO

30 31 32 YES

31 32 33 NO

32 33 34 YES

33 34 35 NO

34 35 36 YES

35 36 37 NO

36 37 38 YES

37 38 39 NO

38 39 40 YES

39 40 41 NO

40 41 42 YES

41 42 43 NO

42 43 44 YES

43 44 45 NO

44 45 46 YES

45 46 47 NO

46 47 48 YES

47 48 49 NO

48 49 50 YES

I hope that everyone sees the pattern now. The answer is always alternating and depends only on the value of b. If b is odd the answer is YES and if b is even the answer is NO.

# SOLUTIONS:

Setter's Solution (C++)
    #include <bits/stdc++.h>
using namespace std;
using ll = long long int;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t ;cin>>t;
while (t-- > 0) {
ll a, b, c;
cin>>a>>b>>c;
if(b%2==0)
cout<<"NO\n";
else
cout<<"YES\n";
}
}

Setter's Solution( Python )
    T=int(input())
while(T>0):
a,b,c=map(int,input().split())
if(b%2==0):
print("NO")
else:
print("YES")
T-=1

Tester's Solution(Kabir Kanha Arora)( JAVA )
    import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
while (t-- > 0) {
long a = scanner.nextLong();    // Useless
long b = scanner.nextLong();
long c = scanner.nextLong();    // Useless
// Use Cassini's identity
String ans = b % 2 == 0 ? "NO" : "YES";
System.out.println(ans);
}
}
}

I think this solution works based on the above editorial:

#include <bits/stdc++.h>
using namespace std;
int main()
{
int T;
cin >> T;
for(int i = 0; i < T; i++)
{
int a, b, c;
cin >> a >> b >> c;
if(b%2 == 0)
{
cout << "NO\n";
}
else{cout << "YES\n";}
}

}

Yes @vaicr7bhav and @first_semester
I did make some silly mistakes indeed. I think I have corrected them now.

1 Like

It won’t work. You haven’t declared a,b and c. Also you are not printing a new line character.
#include <bits/stdc++.h>
using namespace std;
int main()
{
int T;
cin >> T;
for(int i = 0; i < T; i++)
{
long long int a,b,c;
cin >> a >> b >> c;
if(b%2 == 0)
{
cout << “NO\n”;
}
else{cout << “YES\n”;}
}

}
The above code will work fine.

yes you are correct, his solution would not work due to silly mistakes


# Trig substitution integral

I am trying to find $$\int{ \frac {5x + 1}{x^2 + 4} dx}$$ The best approach would be to split up the fraction. According to Wolfram Alpha, the answer is $\frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\displaystyle\arctan\left(\frac x2\right)$ which seems OK, but when I try the trig substitution: $x = 2\tan\theta$, I get an answer that is slightly different but not equivalent, and I've looked at this over and over and I couldn't quite figure out what I did wrong.

$$x = 2\tan\theta$$ $$dx = 2sec^2\theta d\theta$$ $$\int{ \frac {5x + 1}{x^2 + 4}dx} = \frac{1}{4}\int{\frac{10\tan\theta + 1}{sec^2\theta} 2sec^2\theta d\theta}$$ $$= \frac{1}{2}\int{10 \tan\theta + 1}\space d\theta$$ $$= 5 \ln|\sec\theta| + \frac{\theta}{2} + C$$ We know $\theta = \displaystyle\arctan\left(\frac x2\right)$ and since $\tan\theta = \displaystyle\frac{x}{2}$, we can draw a triangle to see that $\sec\theta = \displaystyle\frac{\sqrt{x^2 + 4}}{2}$.

$$5 \ln|\sec\theta| + \frac{\theta}{2} = 5\ln\left({\frac{\sqrt{x^2 + 4}}{2}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right)$$ $$= \frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right)$$

But $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right) \neq \frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\arctan\left(\frac x2\right)$$

There seems to be a small difference between the answer provided by Alpha and the trig substitution method, but I cannot see where I made the mistake.

-

$$\frac{5}{2}\ln({\frac{x^2 + 4}{4}}) + \frac{1}{2}\arctan({x/2}) + c_1 \neq \frac{5}{2}\ln(x^2 + 4) + \frac{1}{2}\arctan(x/2) + C$$

Note that $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) = \frac 52 \ln\left(x^2 + 4\right) - \frac 52 \left (\ln 4\right) = \frac 52 \ln(x^2 + 4) + c_2$$

So put $c_1 + c_2 = C$. Then the answers are equal. Solutions to an integral consist of a family of solutions $F(x) + C$, which differ only by a constant. That is, if $F(x) + C$ is the solution after computing an intregral, so is $F(x) + C_i$, for any constant $C_i \neq C$.

In short, you're both correct!

-
Ah, I did not think of it like that. Looks like I did not make a mistake after all - thanks! –  hesson Feb 22 '13 at 2:19
You're welcome, hesson. This can often occur when you end with an integration including the $\ln$ function: often it can be reduced to $\ln$ function + a constant, as in this case. Also, various trig substitutions can work, and one may arrive at different solutions in terms of trig functions, where one can be converted to the other by applying trig identities. –  amWhy Feb 22 '13 at 2:27

The first approach is easier in fact: $$I= \int \frac{(5x +1)dx}{x^2+4}=I_1 + I_2$$ where $$I_1 = \int\frac{5x dx}{x^2+4}\\ I_2 = \int \frac{dx}{x^2+4}$$ hence, $$I_1 = \frac{5}{2} \int \frac{2xdx}{x^2+4} = \frac{5}{2} \int \frac{d(x^2+4)} {x^2+4}=\frac{5}{2} \log( x^2 +4) +C_1$$ In fact you can see that the numerator is a derivative of the denominator. $$I_2 = \int \frac{dx}{x^2+ 2^2} = \frac{\arctan(\frac{x}{a})}{a} +C_2$$ This approach is more efficient as it does not require you to make any substitutions/

-

Since $$\frac52 \ln (\frac{x^2+4}{4})=\frac52 \ln(x^2+4)-\frac52 \ln 4,$$ the difference between the two answers is an additive constant, which can be absorbed into the constant of integration $C$.

-


# Highest power of a prime $p$ dividing $N!$

How does one find the highest power of a prime $p$ that divides $N!$ and other related products?

Related question: How many zeros are there at the end of $N!$?

This is being done to reduce abstract duplicates. See Coping with *abstract* duplicate questions. and List of Generalizations of Common Questions for more details.

Largest power of a prime dividing $N!$

In general, the highest power of a prime $p$ dividing $N!$ is given by

$$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$

The first term appears since you want to count the number of terms less than $N$ and are multiples of $p$ and each of these contribute one $p$ to $N!$. But then when you have multiples of $p^2$ you are not multiplying just one $p$ but you are multiplying two of these primes $p$ to the product. So you now count the number of multiple of $p^2$ less than $N$ and add them. This is captured by the second term $\displaystyle \left \lfloor \frac{N}{p^2} \right \rfloor$. Repeat this to account for higher powers of $p$ less than $N$.

Number of zeros at the end of $N!$

The number of zeros at the end of $N!$ is given by $$\left \lfloor \frac{N}{5} \right \rfloor + \left \lfloor \frac{N}{5^2} \right \rfloor + \left \lfloor \frac{N}{5^3} \right \rfloor + \cdots$$ where $\left \lfloor \frac{x}{y} \right \rfloor$ is the greatest integer $\leq \frac{x}{y}$.

To make it clear, write $N!$ as a product of primes $N! = 2^{\alpha_2} 3^{\alpha_2} 5^{\alpha_5} 7^{\alpha_7} 11^{\alpha_{11}} \ldots$ where $\alpha_i \in \mathbb{N}$.

Note that $\alpha_5 < \alpha_2$ whenever $N \geq 2$. (Why?)

The number of zeros at the end of $N!$ is the highest power of $10$ dividing $N!$

If $10^{\alpha}$ divides $N!$ and since $10 = 2 \times 5$, $2^{\alpha} | N!$ and $5^{\alpha} | N!$. Further since $\alpha_5 < \alpha_2$, the highest power of $10$ dividing $N!$ is the highest power of $5$ dividing $N!$ which is $\alpha_5$.

Note that there will be

1. A jump of $1$ zero going from $(N-1)!$ to $N!$ if $5 \mathrel\| N$

2. A jump of $2$ zeroes going from $(N-1)!$ to $N!$ if $5^2 \mathrel\| N$

3. A jump of $3$ zeroes going from $(N-1)!$ to $N!$ if $5^3 \mathrel\| N$ and in general

4. A jump of $k$ zeroes going from $(N-1)!$ to $N!$ if $5^k \mathrel\| N$

where $a \mathrel\| b$ means $a$ divides $b$ and $\gcd\left(a,\dfrac{b}{a} \right)$ = 1.

Largest power of a prime dividing other related products

In general, if we want to find the highest power of a prime $p$ dividing numbers like $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$, $\displaystyle P(N,r)$, $\displaystyle \binom{N}{r}$, the key is to write them in terms of factorials.

For instance, $$\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1) = \frac{(2N)!}{2^N N!}.$$ Hence, the largest power of a prime, $p>2$, dividing $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$ is given by $s_p((2N)!) - s_p(N!)$, where $s_p(N!)$ is defined above. If $p = 2$, then the answer is $s_p((2N)!) - s_p(N!) - N$.

Similarly, $$\displaystyle P(N,r) = \frac{N!}{(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle P(N,r)$ is given by $s_p(N!) - s_p((N-r)!)$, where $s_p(N!)$ is defined above.

Similarly, $$\displaystyle C(N,r) = \binom{N}{r} = \frac{N!}{r!(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle C(N,r)$ is given by $$s_p(N!) - s_p(r!) - s_p((N-r)!)$$ where $s_p(N!)$ is defined above.

• Yes you are right. I have deleted the corresponding part in the post. – user17762 May 5 '12 at 1:19
• In my example, I meant highest power of $12$ that divides $3!$. If $m$ is square-free, then indeed the largest prime that divides $m$ is the obstruction. – André Nicolas May 5 '12 at 1:32
• @AndréNicolas Yes I recognized it. Thanks. For instance, in the highest power of $24$ dividing $10!$ the prime $2$ acts as the bottleneck since $2^3$ divides $24$ whereas only $3^1$ divides $24$ and since $10$ is square-free it works. Is there a generic formula known for composite $m$? – user17762 May 5 '12 at 1:36
• One trick that makes computations easier is $$\left\lfloor \dfrac{N}{p^{n+1}} \right\rfloor = \left\lfloor \dfrac{\left\lfloor \dfrac{N}{p^n} \right\rfloor}{p} \right\rfloor$$ – steven gregory Jun 1 '16 at 8:26

For a number $n$, define $\Lambda(n)=\log p$ if $n=p^k$ and zero elsewhen.

Note that $\log n=\sum\limits_{d\mid n}\Lambda(d)$. If $N=n!$, then $$\log N=\sum_{k=1}^n\log k=\sum_{k=1}^n\sum_{d\mid k}\Lambda (d)=\sum_{d=1}^n \Lambda(d)\left\lfloor \frac nd\right\rfloor$$

Since $\Lambda(d)$ is nonzero precisely when $d$ is a power of a prime and in such case it equals $\log p$, the last sum equals $$\sum_{p}\sum_{k\geqslant 1}\log p \left\lfloor\frac n{p^k}\right\rfloor$$ and this gives $$\nu_p(n!)=\sum_{k\geqslant 1}\left\lfloor\frac n{p^k}\right\rfloor$$ If you write $n$ is base $p$, say $n=a_0+a_1p+\cdots+a_kp^k$, the above gives that $$\nu_p(n!)=\frac{s(n)-n}{1-p}$$ where $s(n)=a_0+\cdots+a_k$.

• Very nice proof with using von Mangoldt function! +1 – ZFR Jun 9 '16 at 11:08
• Cool and simple. I think the hardest part is the last equality in the first formula. But it becomes obvious if you change the summation order. – LRDPRDX Jun 6 '20 at 10:18

de Polignac's formula named after Alphonse de Polignac, gives the prime decomposition of the factorial $n!$.

For something completely different, see page 114 of Concrete Mathematics by Graham, Knuth, Patashnik. The highest power of a prime $p$ dividing $n!$ is $n$ minus the sum of the integer digits of $n$ in base $p$, all divided by $p-1$. In Mathematica you would write:

(n-Total[IntegerDigits[n,p]])/(p-1)

Hence, the number of trailing zeros of $n!$ is

(n-Total[IntegerDigits[n,5]])/4

I've found this method to be much faster than the sum of Floor functions...

Define $e_p(n) = s_p(n!),$ as Marvis is using $s_p$ for the highest exponent function...

We get $$e_p(n) = s_p(n!) = \sum_{i \geq 1} \left\lfloor \frac{n}{p^i} \right\rfloor$$

I thought I would prove this by mathematical induction. We need, for any positive integers $m,n,$

LEMMA:

(A) If $n + 1 \equiv 0 \pmod m,$ then $$\left\lfloor \frac{n + 1}{m} \right\rfloor = 1 + \left\lfloor \frac{n}{m} \right\rfloor$$ (B) If $n + 1 \neq 0 \pmod m,$ then $$\left\lfloor \frac{n + 1}{m} \right\rfloor = \left\lfloor \frac{n}{m} \right\rfloor$$

For $n < p,$ we know that $p$ does not divide $n!$ so that $e_p(n) = s_p(n!)$ is $0.$ But all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ are $0$ as well. So the base cases of the induction is true.

Now for induction, increasing $n$ by $1.$ If $n+1$ is not divisible by $p,$ then $e_p(n+1) = e_p(n),$ while part (A) of the Lemma says that the sum does not change.

If $n+1$ is divisible by $p,$ let $s_p(n+1) = k.$ That is, there is some number $c \neq 0 \pmod p$ such that $n+1 = c p^k.$ From the Lemma, part (B), all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ increase by $1$ for $i \leq k,$ but stay the same for $i > k.$ So the sum increases by exactly $k.$ But, of course, $e_p(n+1) = s_p((n+1)!) = s_p(n!) + s_p(n+1) = e_p(n) + k.$ So both sides of the middle equation increase by the same $s_p(n+1) = k,$ completing the proof by induction.

Note that it is not necessary to have $n$ divisible by $p$ to get nonzero $e_p(n).$ All that is necessary is that $n \geq p,$ because we are not factoring $n,$ we are factoring $n!$

Here's a different approach I found while thinking in terms of relating the sum of digits of consecutive numbers written in base $$p$$. Part of the appeal of this approach is you might have at one point learned that there is a connection but don't remember what, this gives a quick way to reconstruct it.

Let's consider $$s_p(n)$$, the sum of digits of $$n$$ in base $$p$$. How does the sum of digits change if we add $$1$$ to it? It usually just increments the last digit by 1, so most of the time,

$$s_p(n+1) = s_p(n)+1$$

But that's not always true. If the last digit was $$p-1$$ we'd end up carrying, so we'd lose $$p-1$$ in the sum of digits, but still gain $$1$$. So the formula in that case would be,

$$s_p(n+1) = s_p(n) + 1 - (p-1)$$

But what if after carrying, we end up carrying again? Then we'd keep losing another $$p-1$$ term. Every time we carry, we leave a $$0$$ behind as a digit in that place, so the total number of times we lose $$p-1$$ is exactly the number of $$0$$s that $$n+1$$ ends in, $$v_p(n+1)$$.

$$s_p(n+1) = s_p(n) + 1 - (p-1)v_p(n+1)$$

Now let's rearrange to make the telescoping series and sum over,

$$\sum_{n=0}^{k-1} s_p(n+1) - s_p(n) = \sum_{n=0}^{k-1}1 - (p-1)v_p(n+1)$$ $$s_p(k) = k - (p-1) \sum_{n=0}^{k-1}v_p(n+1)$$

Notice that since $$v_p$$ is completely additive so long as $$p$$ is a prime number,

$$s_p(k) = k - (p-1) v_p\left(\prod_{n=0}^{k-1}(n+1)\right)$$

$$s_p(k) = k - (p-1) v_p(k!)$$

This rearranges to the well-known Legendre's formula,

$$v_p(k!) = \frac{k-s_p(k)}{p-1}$$

• WOW! Not bad. Though I would recommend you to add the parentheses in order to show what are the terms in the summations. It's a bit confusing. BTW. It's funny that you gave the answer to this old question almost at the same time as I did. – LRDPRDX Jun 6 '20 at 10:31

Walking down the street I realized how to proof the formula in a very simple way. The key idea is to know how many numbers contain a given prime in the exact $$q$$-th power.

Let us denote the set of all numbers up to $$n$$ (inclusive) that are divisible by $$q$$-th power of prime $$p$$ but not divisible by the greater ($$q+1, q+2, ...$$) power of that prime: $$M^p_q(n) := \{ m : m \le n, p^q | m, p^{q+1} \nmid m\}$$ How many numbers in that set? Well, obviously, the number of those that are divisible by $$p^q$$ minus the number of those which are divisible by $$p^{q+1}$$: $$|M^p_q(n)| = \bigg\lfloor \frac{n}{p^q}\bigg\rfloor - \bigg\lfloor \frac{n}{p^{q+1}}\bigg\rfloor$$ OK. Each of the numbers in $$M^p_q$$ gives us the $$q$$-th power of $$p$$ in $$n!$$ so we need just add them all: $$s_p(n!)=\sum_{q=1}^{\infty}q|M^p_q(n)|$$ Which is the desired result.