Algebraic Structures - Mother Teresa University PDF

Machine Translated by Google "MOTHER" UNIVERSITY TERESA,,- SKOPJE ALGEBRAIC STRUCTURES Save Rexhepi A book...

Machine Translated by Google "MOTHER" UNIVERSITY TERESA,,- SKOPJE ALGEBRAIC STRUCTURES Save Rexhepi A booklet 2020-SKOPJE 1 Chapter 1: Mathematical Logic Judgment is the declarative sentence which can be true or false. Usually judgments are marked with small letters, for example: p,q,r,s,t....others. Example: Discuss whether the following sentences are judgments or not? 1. The dog has 6 legs. 2. 2+3=5 3. The rectangle is negative. 4. 4+5=16 5. Close the door Machine Translated by Google 6. What time is it? 7. Apple is the best fruit 8. Tetova is the capital of North Macedonia Solution : 1. False judgment. 2. True judgment 3. It is not a judgment because the sentence does not make sense. 4. False judgment 5. It is not a judgment (imperative sentence). 6. It is not a judgment (interrogative sentence). 7. It is not a judgement, since one cannot speak about the truth of the sentence 8. Judgment not true. 2 A judgment has only two truth possibilities and that is: true or false. Symbolically, the truth is marked as follows: (ÿÿ) = T true. (ÿÿ) = 1 true. (ÿÿ) = ÿ Not true. (ÿÿ) = 0 Not true. For two judgments there are 4 possibilities of truth and that: pq Machine Translated by Google TT Tÿ ÿT ÿÿ So the number of judgments is 2 , The number of possible truths is 24=2 3 For three judgments, the number of possibilities of truth will be 8 = 2 pQR TTT TTÿ TÿT Tÿÿ ÿTT ÿTÿ ÿÿT ÿÿÿ If the number of trials is n , then we will have 2 possibilities 3 truthfulness. Negation (Opposite, inverse of Judgment) Negation is marked with: ¬ÿÿ, ÿÿ, ÿÿ. p¬ Machine Translated by Google ÿ ÿ Tÿ ÿT The following relations apply: ¬(¬ÿÿ) = ÿÿ ¬(¬)(¬ÿÿ) = ¬ÿÿ Example: Find the negation of the following judgments: 1. Today is a sunny day. 2. Tomorrow is Friday. 3. 4+5=9 4. 2 + 3 ÿ 4 Solution: The negations of the above judgments will be: 1. Today is not a sunny day. 2. Tomorrow is not Friday. 3. 4 + 5 ÿ 9 4. 2 + 3 > 4 Compound judgments: A judgment which is formed by two or more other judgments with connecting words is called a compound judgment. 4 Below are some of the most popular composite judgments: Conjunction: The judgment composed of judgments p and q, which has the connecting word "and" in the middle, is called conjunction ip and q and is marked with: ÿ ÿÿ “ ÿ”(AND). (read: judgment p and judgment q) Machine Translated by Google The conjunction is true if both statements are true, otherwise it is false. The conjunction in the truth table is given as follows: P q ÿÿ ÿ ÿÿ TT T Tÿÿ ÿT ÿ ÿÿÿ The relations apply: ÿÿ ÿ ÿÿ = ÿÿ ÿÿ ÿ ÿ = ÿ ÿÿ ÿ ÿÿ = ÿÿ ÿÿ ÿ ¬ÿÿ =ÿ Disjunction : Judgment composed of judgments p and q, which are connected by the conjunction "or" is called disjunction ip and q, and is marked ÿ ÿÿ , “ÿ”.(OR) ( read judgment p or judgment q). The disjunction is not true if the two statements are not true, otherwise it is a true judgment. , 5 The truth table for the disjunction is given as follows: p q ÿÿ ÿ ÿÿ TT T Machine Translated by Google TÿT ÿT T ÿÿÿ The following relations apply: ÿÿ ÿ ÿÿ = ÿÿ ÿÿ ÿ ¬ÿÿ = ÿÿ ÿÿ ÿ ÿÿ = ÿÿ ÿÿ ÿ ¬ÿÿ = ÿ ÿÿ ÿ ÿ = ÿÿ Some properties of Conjunction and Disjunction: 1. Commutative property: ÿÿ ÿ ÿÿ ÿ ÿÿ ÿ ÿÿ ÿÿ ÿ ÿÿ ÿ ÿÿ ÿ ÿÿ 2. Associative Property: ÿÿ ÿ (ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ) ÿ ÿÿ 3. Demogran's law: ¬(ÿÿ ÿ ÿÿ) ÿ ¬ ÿÿ ÿ ¬ ÿÿ ¬(ÿÿ ÿ ÿÿ) ÿ ¬ ÿÿ , ÿ ¬ ÿÿ 4. Distributive Law: ÿÿ ÿ (ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ) ÿÿ ÿ (ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ) 6 Example: a) Use the truth table to show the following statements Machine Translated by Google ÿÿ ÿ (ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ) ÿ ÿÿ p q r ÿÿ ÿÿ ÿ (ÿÿ ÿÿ ÿ (ÿÿ ÿ ÿ ÿ ÿÿ) ÿÿ ÿÿ) ÿ ÿÿ ÿÿ T T T T T T T T T ÿ ÿ ÿ T ÿ T ÿ T ÿ ÿ ÿ ÿ T ÿ ÿ ÿ ÿ ÿ ÿ ÿT T ÿ ÿ ÿ ÿ ÿT ÿ ÿ ÿ ÿ ÿ ÿ ÿ T ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ b) D.SH : ÿÿ ÿ (ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ)? ? ? ÿÿ) [(ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ¬ÿÿ)] ÿ ¬ÿÿ = ÿÿ] i) Verification by means of the truth table? pq¬ ¬ (ÿÿ ÿ (ÿÿ ÿ (ÿÿ ÿ ÿÿ) ÿ [(ÿÿ ÿ ÿÿ) ÿ ÿ ÿ ÿÿ) ¬ÿÿ) (ÿÿ ÿ ¬ÿÿ) (ÿÿ ÿ ÿ ÿ ¬ÿÿ )] ÿ ¬ÿÿ TT ÿ ÿ T T T T TÿÿT T T T T ÿ TT ÿ T ÿ ÿ T ÿÿTT ÿ T ÿ T ii) Proof by means of logical laws: Machine Translated by Google 7 [(ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ¬ÿÿ)] ÿ ¬ÿÿ ÿ ÿ [ÿÿ ÿ (ÿÿ ÿ ¬ÿÿ)] ÿ ¬ÿÿ ÿ ÿ [ÿÿ ÿÿ] ÿ ¬ÿÿ ÿ ÿÿ ÿ ¬ÿÿ ÿ ÿÿ d) To prove: (ÿÿ ÿ ¬ÿÿ) ÿ ((¬ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ¬ÿÿ)) ÿ 0 Verification by truth table: pqÿ ¬ ¬ (¬ÿÿ ÿ (ÿÿ ÿ (ÿÿ ÿ (¬ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿ ÿ ÿ ÿÿ) ¬ÿÿ) ¬ÿÿ) (ÿÿ ÿ ¬ÿÿ) ¬ÿÿ) ÿ ÿ ÿ ((¬ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ¬ÿÿ)) 11100 0 0 0 0 0 11000 0 0 0 0 0 10110 0 0 1 0 0 10010 0 0 1 0 0 01101 1 1 0 1 0 01001 1 0 0 1 0 0011 1 0 1 0 1 0 0001 1 0 0 0 0 0 Proof by means of logical laws: (ÿÿ ÿ ¬ÿÿ) ÿ [(¬ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ¬ÿÿ)] ÿ (ÿÿ ÿ ¬ÿÿ) ÿ [¬ÿÿ ÿ (¬ÿÿ ÿ ÿÿ)] ÿ (ÿÿ ÿ ¬ÿÿ) ÿ ¬ÿÿ ÿ (¬ÿÿ ÿ ÿÿ) ÿ 0 ÿ ¬ÿÿ ÿ (ÿÿ ÿ ÿÿ) ÿ 0 A judgment which is true in all cases is called Tautology , while one Machine Translated by Google judgment which is untrue in all cases Contradiction. Example: Determine the truth of judgments p, q and r if the relation holds: 8 ÿÿ ÿ ¬(¬ÿÿ ÿ ÿÿ) ÿ ÿÿ Solution: So we have: ÿÿ ÿ ÿÿ , ¬(¬ÿÿ ÿ ÿÿ) ÿ ÿÿ ÿÿ ÿ ÿÿ , ¬ÿÿ ÿ ÿÿ ÿÿ ÿÿ ÿ ÿÿ , ¬ÿÿ ÿÿ, ÿÿ ÿÿ ÿÿ ÿ ÿÿ , ÿÿ ÿ ÿÿ , ÿÿ ÿÿ The implication The implication is formed if the judgments p and q are closed with the conjunction "If p then q,, Or,, p implies q,, or ,,from p follows q,,. The implication is false if the first statement is true and the second statement is not real , otherwise it is true in all cases. The truth table for the implication is given as follows: p q ÿÿ => ÿÿ TT T Tÿ ÿ ÿT T ÿÿ T The following statement applies: ÿÿ => ÿÿ ÿ ¬ÿÿ => ¬ÿÿ Machine Translated by Google which we prove by means of the table. pq¬ ¬ ÿÿ => ¬ÿÿ ¬ÿÿ => ÿ ÿ ÿÿ ÿ ÿÿ ¬ÿÿ ÿ ÿ 9 TTÿÿ T T T TÿÿT ÿ ÿ ÿ ÿTTÿ T T T ÿÿTT T T T Example: If the values ÿÿ ÿ ÿÿ = > ÿÿ ÿ 0, ÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿÿÿÿÿÿÿÿÿ ¬[(ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ)] => (ÿÿ => ¬ÿÿ) =? Solution: So we have ÿÿ ÿ ÿÿ ÿ 1 ÿÿÿÿÿ ÿÿ ÿ 0 For pÿ 1 , ÿÿ ÿ 1 ÿÿÿÿÿ ÿÿ ÿ 0 [(1 ÿ 0) ÿ (1 ÿ 0)] => (0 => 0) ÿ [0 ÿ 1] => 1 = ¬1 => 1 = 0 => 1 = 1 Example: Prove that: ÿÿ ÿ {(ÿÿ => ÿÿ) => [¬ÿÿ ÿ (ÿÿ => ÿÿ)]} ÿ 0. i) Verification by truth table: pq¬ ÿÿ => ÿÿ ÿ (ÿÿ (ÿÿ => ÿÿ) => [¬ÿÿ ÿ ÿÿ ÿ {(ÿÿ => ÿÿ) = ÿ ÿÿ => ÿÿ) (ÿÿ => ÿÿ)] > [¬ÿÿ ÿ (ÿÿ ÿ => ÿÿ)]} 110 1 0 0 0 101 0 0 1 0 Machine Translated by Google 010 1 0 0 0 001 1 1 1 0 ii) Proof by logical statements: ÿÿ ÿ {(ÿÿ => ÿÿ) => [¬ÿÿ ÿ (ÿÿ => ÿÿ)]} ÿ 0 10 ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿ ÿÿÿÿÿÿÿÿÿÿÿ ÿÿ => ÿÿ = ¬ÿÿ ÿ ÿÿ ÿ ÿÿ ÿ {¬(¬ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ (¬ÿÿ ÿ ÿÿ))} = ÿ ÿÿ ÿ {(ÿÿ ÿ ¬ÿÿ) ÿ (¬ÿÿ ÿ ¬ÿÿ) ÿ (ÿÿ ÿ ÿÿ)} = ÿ ÿÿ ÿ {(ÿÿ ÿ ¬ÿÿ) ÿ (¬ÿÿ ÿ ¬ÿÿ)} = ÿ ÿÿ ÿ {¬ÿÿ ÿ (ÿÿ ÿ ¬ÿÿ)} = ÿÿ ÿ {¬ÿÿ ÿ 1} ÿ ÿÿ ÿ ¬ÿÿ ÿ 0 Equivalence Equivalence: it is formed by connecting two judgments p,q connected with the conjunction "if". and only if” and is symbolically denoted by ÿÿ ÿÿ. Equivalence is a true judgment if two propositions have the same truth , otherwise it is a false judgment. The truth table for equivalence is given as follows: p q ÿÿ ÿÿ T T T T ÿ ÿ ÿ T ÿ ÿ ÿ T Machine Translated by Google The following relations apply: ÿÿ ÿÿ ÿ (ÿÿ => ÿÿ) ÿ (ÿÿ => ÿÿ) ÿÿ ÿÿ ÿ (¬ÿÿ ÿ ÿÿ) ÿ (¬ÿÿ ÿ ÿÿ) Example: To examine the accuracy (truth) of the judgments below: 11 1. Earth is a planet if and only if coal is black. So it is a correct judgment, since TT=T. 2. A flower is a tree if and only if the snow is red. So it is an incorrect judgment, since Tÿ=ÿ Mathematical Quantifiers We distinguish these mathematical quantifiers: ÿx, (ÿÿëÿÿ ÿÿÿÿÿÿ ÿÿ) ÿx, (ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿ) ÿ!x, (ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿÿÿÿÿÿ ÿÿÿÿÿÿ ÿÿ) "ÿ" , (ÿÿÿÿÿÿ ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ) Example: To find the truth of judgments: 1. Some apples are green. / Correct. 2. All eagles have four legs./ Incorrect. 3. Some prime numbers are odd./ Correct. 4. All months of the year have 30 days./ Incorrect. 5. ÿx, 6. There , so that x-1=0./ Correct because for x=1,1-1=0. 2 exists x such that < ÿÿ/ÿÿÿÿÿÿÿÿë ÿÿÿÿÿÿÿÿÿÿ Machine Translated by Google 1 12 1 1 1 ÿÿÿÿÿÿ ÿÿ = 2,( 2) < 8= 4< 2 7. ÿÿÿ x:x=1./ He saw it correctly because 2 0:0=? (Unspecified form). 8. ÿÿÿ + 4 > 0./ Correct. Example: Find the truth and negation of the following sentences: 1. ÿÿÿ ÿ ÿÿ, 2ÿÿ ÿ 1 < 0./ ÿÿÿÿÿÿÿÿë. 2. ÿÿÿ ÿ ÿÿ, 3ÿÿ + 5 = 0./ ÿÿÿÿ ÿÿÿÿÿÿÿÿë. 2 3. ÿÿÿ ÿ ÿÿ, ÿÿ ÿ 0 ÿ (ÿÿÿÿÿ ÿ ÿÿ, ÿÿ ÿÿ) ÿ ÿÿ ÿ ÿÿ ÿ ÿÿ. 12 2 ÿÿ 4. (ÿÿÿ ÿ ÿÿ, ÿÿ + 1 = 0) ÿ (ÿÿÿ ÿ , 2 ÿÿ = ÿÿ)ÿÿÿ ÿÿ = ÿÿ 2 5. [(ÿÿÿ ÿ ÿÿ, 2ÿÿ + 3 = 0) ÿ (ÿÿÿ ÿ ÿÿ, ÿÿ ÿ ÿÿ )] => (ÿÿÿ ÿ 2 ÿÿ, ÿÿ ÿ ÿÿ = 0) =ÿ ÿÿ= ÿÿ 2 6. ÿÿÿ ÿ ÿÿ, ÿÿÿÿÿ, (ÿÿ + 2ÿÿ ÿ 0) ÿ ÿÿÿÿÿÿÿÿë Negation For homework. Some examples for repetition: 1) (ÿÿ ÿ ÿÿ) ÿ [(ÿÿ ÿ ÿÿ) ÿ ((ÿÿ => ÿÿ))] PQ r ÿÿ => (ÿÿ ÿ ÿÿ) ÿ (ÿÿ (ÿÿ ÿ (ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ) ÿ ÿ ÿÿ => ÿÿ) ÿÿ) (ÿÿ => ÿÿ) ÿÿ 111 1 1 1 1 1 110 1 0 0 0 0 101 1 1 1 1 1 100 1 0 0 0 0 011 1 1 1 0 1 010 1 1 1 0 1 001 0 1 0 0 0 000 0 1 0 0 0 2)(ÿÿ ÿ ÿÿ) (ÿÿ => ÿÿ) PQ r (ÿÿ ÿ (ÿÿ => (ÿÿ ÿ ÿÿ) (ÿÿ => ÿÿ) ÿÿ) ÿÿ) 1 1 1 1 1 1 1 10 0 1 0 Machine Translated by Google 101 1 1 1 100 0 1 0 01 1 0 0 1 010 0 0 1 001 0 1 0 000 0 1 0 3)[(ÿÿ => ÿÿ) ÿ (¬ÿÿ => ÿÿ)] ÿ (ÿÿ ÿÿ) Fill in the table: 13 pqr ¬ (ÿÿ => (¬ÿÿ => [(ÿÿ => ÿÿ) ÿ (¬ÿÿ (ÿÿ [(ÿÿ = ÿ ÿÿ) ÿÿ) => ÿÿ)] ÿÿ) > ÿÿ) ÿ ÿ (¬ÿÿ = > ÿÿ)] ÿ (ÿÿ < => ÿÿ) 111 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4)(ÿÿ => ÿÿ) ÿ (ÿÿ ÿÿ). Fill in the table: Machine Translated by Google PQ r (ÿÿ => (ÿÿ (¬ÿÿ ÿ ÿÿ) ÿ {(¬ÿÿ ÿ (¬ÿÿ ÿ ÿÿ) ÿÿ) 1 1 1 1 1 ÿÿ))} ÿ {(¬ÿÿ ÿ (¬ÿÿ ÿ ÿÿ))} ÿ 1 1 1 (¬ÿÿ ÿ ÿÿ) ÿ [(ÿÿ => ÿÿ) ÿ (ÿÿ 1 1 1 => ÿÿ)] 1 ÿ (¬ÿÿ ÿ ÿÿ) ÿ (¬ÿÿ ÿ ÿÿ) ÿ (¬ÿÿ ÿ ÿÿ) 01 1 1 ÿ (¬ÿÿ ÿ ÿÿ) 01 1 1 0 1 1 36. 0 1 1 (ÿÿ => ¬ÿÿ) => (¬ÿÿ => ÿÿ) 5) We simplify with mathematical laws the expression: (ÿÿ ÿ ÿÿ) ÿ ÿÿ (ÿÿ => ÿÿ) ÿ (ÿÿ ÿÿ) 14 (ÿÿ => ¬ÿÿ) => (¬ÿÿ => ÿÿ) ÿ (¬ÿÿ ÿ ¬ÿÿ) => (ÿÿ ÿ ÿÿ) Machine Translated by Google pqr ¬ ¬ (ÿÿ => (¬ÿÿ => (ÿÿ => ¬ÿÿ) => (¬ÿÿ => ÿ ÿ ¬ÿÿ) ÿÿ) ÿÿ) ÿ ÿ 1110 0 0 1 1 1100 1 0 1 1 101 1 0 1 1 1 1001 1 1 0 0 0110 0 1 1 1 0100 1 1 1 1 001 1 0 1 1 1 0001 1 1 0 0 Electric Circuits 1. Serial combination of electrical circuits: ÿÿ ÿ ÿÿ 2. Combination of electrical circuits in parallel: ÿÿ ÿ ÿÿ Machine Translated by Google 15 1. Draw the electrical circuits ÿÿ ÿ ( ÿÿ ÿ ÿÿ)? 2. ÿÿ ÿ [(ÿÿ ÿ ÿÿ) ÿ ÿÿ] 3. Should the lower circles be expressed in the form of judgments? (ÿÿ ÿ ÿÿ) ÿ [ÿÿ ÿ ÿÿ ÿ Machine Translated by Google ÿÿ ÿ ÿÿ ÿ ÿÿ] 16 a) {ÿÿ ÿ [(ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ)]} b) (ÿÿ ÿ ÿÿ) ÿ {(ÿÿ ÿ ÿÿ) ÿ ÿÿ ÿ (ÿÿ ÿ ÿÿ)} Axioms and Theorems: Definition 1: Statements that are taken to be true without proof are called axioms or postulate. Definition 2: Statements that can be proven using facts auxiliary, such as axioms, are called theorems. Types of proofs of theorems: 17 1. Direct proofs: They start with hypotheses and go through known facts in order to reach the correct conclusion. Example : To prove that the sum of two odd numbers is an even number Obtaining that number x and y are odd numbers x=2k+1 , in ÿÿ ÿ , z=2t+1 , in ÿÿ ÿ Machine Translated by Google then x+z=2k+1+2t+1=2k+2t+2=2(k+t+1)=2s where ÿÿ ÿ ÿÿ. 2. Indirect Way: It is usually expressed through contradiction. Example: Prove that the square of an even number is an even number. Verification: Suppose that there is at least one odd number x such that ÿÿ 2 is an even number, then x=2m+1, ÿÿ ÿ ÿÿ, 2 2 2 2 = 4ÿÿ + 2 · 2ÿÿ + 1 = ÿÿÿ + 4ÿÿ + 1 = 2 ÿÿ = (2ÿÿ + 1) 2 2(2ÿÿ + 2ÿÿ) + 1 = 2ÿÿ + 1, ÿÿ ÿ ÿÿ. we obtained the converse which means that the assertion in the task, the square of an even number is an even number. 18 Chapter II: Communities The way of giving (presenting) the communities: 1. Communities can be presented with a Venn diagram: Machine Translated by Google Example: Present the following communities with a Venn diagram: a) A={1,2,3} b) ÿÿ = {ÿÿ, 1, ÿÿ, ÿ4} 2. In listed form (Written) for example: Example: Write the following communities in written form? 1.{ÿÿÿÿÿÿëÿÿ ÿÿ ÿÿÿÿÿÿëÿÿ} ÿÿ ÿÿ ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ } 2.{ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿ ÿÿÿÿÿÿÿÿëÿÿ ÿÿ 1.{ÿÿ ÿÿëÿÿë , ÿÿ ÿÿÿÿÿÿÿÿë , ÿÿ ÿÿëÿÿÿÿÿÿÿÿë, ÿÿ ÿÿÿÿÿÿÿÿ, ÿÿ ÿÿÿÿÿÿÿÿÿÿÿÿ, ÿÿ ÿÿÿÿÿÿÿÿÿÿÿ, ÿÿ ÿÿÿÿÿÿÿÿÿÿ} 2. ÿÿ = {ÿÿ, ÿÿ, ÿÿ, ÿÿ,ÿÿ,ÿÿ} 3. In descriptive form: Example: Should the sentences be written in descriptive form? ÿÿ = {0,2,4,6,8,10} ÿÿ = {ÿÿ ÿÿÿÿÿÿÿÿÿÿ, ÿÿ ÿÿëÿÿÿÿÿÿÿÿë} ÿÿ = {ÿÿ = 2ÿÿ , 0 ÿ ÿÿ ÿ 6 ÿÿ ÿ ÿÿ} 19 ÿÿ = {ÿÿ ÿÿ ëÿÿÿÿÿë ÿÿÿÿÿÿë ÿÿ ÿÿÿÿÿÿÿÿÿÿ ÿÿë ÿÿÿÿÿÿÿÿÿÿÿÿÿÿë ÿÿÿÿ ÿÿÿÿÿÿÿÿÿÿÿÿÿëÿÿ ÿÿ} Example: Mark the elements of the communities: 2 1.{ÿÿ ÿÿ + 1 = ÿÿ ÿ ÿÿ} 2.{ÿÿ ÿÿ(ÿÿ ÿ 1)(ÿÿ ÿ 2) = 0, ÿÿ ÿ ÿÿ} Machine Translated by Google 3.{ÿÿ ÿÿ ÿÿÿÿÿÿÿÿÿÿ ÿÿ ÿÿÿÿÿÿÿÿÿÿÿÿë ÿÿë ÿÿ ÿÿÿÿÿÿëÿÿ ÿÿÿÿ 50}{ÿÿ ÿÿ ÿ ÿÿ, ÿÿ ÿ ÿÿ} ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿ 1. ÿÿ = ÿ 2 2 ÿÿ + 1 = 0ÿÿ1,2 =ÿÿÿ ± ÿÿÿ ÿ 4 · ÿÿ · ÿÿ 2ÿ4·1·1 2 · ÿÿÿÿ1,2 =ÿ0 ± ÿ0 2 · 1ÿÿ1,2 =ÿ ± ÿÿ4 ±2ÿÿ 2ÿ ÿÿ4 = ÿÿ4 · (ÿ1) = 2ÿÿ1 = 2ÿÿÿÿ1,2 = 2ÿÿ1 = ÿÿÿÿ2 = ÿ1 2. ÿÿ = {0,1,2} 3. ÿÿ = {2,3,5,7,11,13,17,19,23,29,31,37,39,41,43,47} Types of communities: 1. The empty set : The set which does not contain any elements is called the empty set and denoted by { }, ÿÿÿÿÿÿ ÿ. Example: ÿÿ ) ÿÿ = {ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿ ÿÿÿÿÿÿÿÿÿÿÿÿë ÿÿÿÿÿÿÿÿ ÿÿë ÿÿÿÿÿÿÿÿÿÿ 1961} ÿÿÿÿÿÿ ÿ ÿÿ = ÿ. ÿÿ) ÿÿ = {ÿÿÿÿÿÿëÿÿ ÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿë ÿÿÿÿÿÿÿÿëÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿÿÿ ÿÿÿÿÿÿÿÿ} ÿÿÿÿÿÿ ÿÿ = ÿ. ÿÿ) ÿÿ = {ÿ} ÿ ÿÿÿÿÿÿ ëÿÿÿÿÿë ÿÿÿÿÿÿÿÿÿëÿÿÿÿ ÿÿ ÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿÿÿ ÿÿÿÿë ÿÿÿÿÿÿÿÿÿÿÿÿ. worth relation: ÿ ÿ {ÿ} Machine Translated by Google 20 2. Finite set, infinite set: Definition: If the number of elements of a set is finite, then it is called a finite set, otherwise it is called an infinite set. Set A is a subset of B if every element of set A also belongs to set B, written ÿÿ ÿ ÿÿ = { ÿÿ ÿÿ ÿ ÿÿ ÿ ÿÿ ÿ ÿÿ}. Properties: The empty set is a subset of every set ÿ ÿ ÿÿ. Each set represents its own subsets ÿÿ ÿ ÿÿ. Partitive Community (Partial) : A partition set is called the set of all subsets of a given set. Example: Find the partitive communities of all lower communities.? a) ÿÿ = {1} , b) ÿÿ = {1, ÿÿ}, c) ÿÿ = {1, ÿ2, ÿÿ}, d) ÿÿ = {1, ÿÿ, 2 , } Solution : 1 ÿÿ) ÿÿ(ÿÿ) = {ÿ,{1}} ÿ ÿÿ(ÿÿ) = 1 ÿ ÿÿ(ÿÿ(ÿÿ)) = 2 = 2 ÿÿ) ÿÿ(ÿÿ) = {ÿ,{1},{ÿÿ},{1, ÿÿ}} ÿ ÿÿ(ÿÿ) = 2 ÿ 2 ÿÿ(ÿÿ(ÿÿ)) = 4 = 2 ÿÿ) ÿÿ(ÿÿ) = {ÿ,{1},{ÿÿ},{ÿ2},{1, ÿ2},{1, ÿÿ},{ÿ2, ÿÿ},{1, ÿ2, ÿÿ}} ÿ ÿÿ(ÿÿ) = 3 Machine Translated by Google 21 3 ÿ ÿÿ(ÿÿ(ÿÿ)) = 8 = 2 ÿÿ) ÿÿ(ÿÿ) = {ÿ,{1}, {ÿÿ},{2},{ÿÿ},{1, ÿÿ},{1,2},{1, ÿÿ},{ÿÿ, 2},{ÿÿ, ÿÿ},{2, ÿÿ},{1, ÿÿ, 2},{1, ÿÿ, ÿÿ},{ÿÿ, 2, ÿÿ},{1,2, ÿÿ, ÿ ÿÿ(ÿÿ) = 4 ÿ 4 ÿÿ(ÿÿ(ÿÿ)) = 16 = 2 ÿÿ...!!? If n(A)=m, then m(P(n))= 2 Example: If the number of partition sets of the set A has 2 3ÿÿÿ8 elements, where the set A has n- elements, find the value of m. Solution: n(A)=n ÿÿ 3ÿÿÿ8 = 2 = ÿÿ(ÿÿ(ÿÿ)) = 2 = ÿÿ = 3ÿÿ ÿ 8 =ÿÿ ÿ 3ÿÿ = ÿ8 = ÿ2ÿÿ = ÿ8 2 = ÿÿ = 8 = ÿÿ = 4 The number of the subset with r-elements where the set has n-elements is calculated with the formula: ÿÿ ÿÿ! ÿÿ) = ( ÿÿ!(ÿÿÿÿÿ)! Example:: 3 3! ( 2)= 3! 2!(3ÿ2)!= 2!.3 211!= 2!= 3 4 ( 2) =4! Machine Translated by Google 2! (4 ÿ 2)!=4! 2! 2!=2! · 3 · 4 2! 2!= 6 22 4! · 5 · 6 · 7 7 ( 4) =7! 4! 3!=5 · 6 · 7 4! (7 ÿ 4)!=7! 6= 35 4! 3! Remarks: ÿÿ ÿÿ ÿÿ ÿ ( 0) + ÿÿ(ÿÿ) = ÿÿ ÿ ÿÿ(ÿÿ(ÿÿ)) = 2 ( 1) ÿÿ ÿÿ + ( 2) +.... ( ÿÿ ÿÿ 2 ÿÿ ÿÿ = ( 0) + ÿÿ) = 2 ( 1) + ÿÿ ÿÿ ÿÿ ( 2) +... ÿÿ) = 2. ( Example: If the number of subsets with two elements is equal to the number of sets with 5 elements, then we have: a) Number of subsets with 3 elements? b) Number of subsets with at most 2-elements ? c) Number of subsets with at least 3 elements? a) 5 ( 3) =5! 20 3! 2!=3! · 4 2= · 5 3! 2! 10 Machine Translated by Google ÿÿ ÿÿ ÿÿ b) ( 2) = ( 5) , ÿ ( ÿÿ) = (ÿÿ ÿÿ ÿ ÿÿ) , ÿ ÿÿ = 7 ÿÿ = 2 , ÿ ÿÿ ÿ ÿÿ = 5 ÿ ÿÿ ÿ 2, = 5 , ÿ ÿÿ = 7 7 7 7 c) ( 3) + ( 4) + ÿ.. + ( 7) The largest community that contains all the communities is called a community universal and marked with "ÿÿ". 23 Real Number Intervals a) [ÿÿ, ÿÿ] = {ÿÿ ÿÿ ÿ ÿÿ ÿ ÿÿ} b) [ÿÿ, ÿÿ] = {ÿÿ ÿÿ < ÿÿ ÿ ÿÿ} c) [ÿÿ, ÿÿ] = {ÿÿ ÿÿ ÿ ÿÿ < ÿÿ} d) [ÿÿ, ÿÿ] = {ÿÿ ÿÿ < ÿÿ < ÿÿ} Machine Translated by Google 24 Unions ÿÿ ÿ ÿÿ = {ÿÿ ÿÿ ÿ ÿÿ ÿÿÿÿÿÿ ÿÿ ÿ ÿÿ} Example: ÿÿ = {ÿ1, ÿ2,0} ÿÿ = {ÿÿ 2 < ÿÿ ÿ 6, ÿÿ ÿ ÿÿ} ÿÿ = {8,4,5,6} ÿÿ ÿ ÿÿ = {ÿ2,0, ÿ1,8,4,5,6} Prairies ÿÿ ÿ ÿÿ = {ÿÿ ÿÿ ÿ ÿÿ ÿÿÿÿÿ ÿÿ ÿ ÿÿ} Machine Translated by Google 25 Example: ÿÿ ÿ ÿÿ = ÿ ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿÿÿÿÿÿÿ ÿÿ ÿ ÿÿ ÿÿëÿÿÿÿ ÿ ÿÿ = {ÿÿ ÿ2 < ÿÿ < 2, ÿÿ ÿ ÿÿ} 2 ÿÿ = {ÿÿ ÿÿ ÿ 4, ÿÿ ÿ ÿÿ} ÿÿ = {1} ÿÿ = {0,1,2, ÿ2, ÿ1} ÿÿ ÿ ÿÿ = {1} The difference ÿÿ\B = {xx ÿ A and x ÿ B} Example: A = {1} B = {a, 1,2, ÿ1, ÿ2} Machine Translated by Google A\B = ÿ 26 B\A = {a, 2, ÿ2, ÿ1} A\B ÿ It belongs to A and it does not belong to B. B\A ÿ It belongs to B and it does not belong to A ÿ s.. Complement of Community (opposite of community) ÿÿ ÿÿ = {ÿÿ ÿÿ ÿ ÿÿ} Example: ÿÿ = (3,8], ÿÿ = [ÿ4,5), ÿÿ = (3,7) Set : ÿÿ ÿ , ÿÿ ÿ ÿÿ, ÿÿ\ÿÿ, ÿÿ\ÿÿ, ÿÿ\ÿÿ, ÿÿ\ÿÿ, ÿÿ\ÿÿ ÿÿ ÿ ÿÿ = {3,5} , ÿÿ ÿ ÿÿ = {ÿ4,8}, ÿÿ\ÿÿ = {5,8}, ÿÿ\ÿÿ = {ÿ4, ÿ3}, 27 Machine Translated by Google ÿÿ\ÿÿ = {ÿ4,3}, ÿÿ ÿÿ\ÿÿ = {5,7}, ÿÿ = {ÿÿ, ÿ3} ÿ {8, ÿ} Number ÿÿ( ÿÿ ÿ ÿÿ) = ÿÿ(ÿÿ) + ÿÿ(ÿÿ) = ÿÿ( ÿÿ ÿ ÿÿ) Example: For the sets A and B if ÿÿ( ÿÿ ÿ ÿÿ) = 4 n(A) and n(B)= ÿÿ( ÿÿ ÿ ÿÿ) = 14 find the number of elements of the partition set of the set B? Solution: 1. ÿÿ + 4 = ÿÿ + 4 2. ÿÿ + ÿÿ = 10 3. ÿÿ(ÿÿ) = 4 + 5 = 9 ÿÿ 3 = ÿÿ 2ÿÿ = 10 ÿÿ(ÿÿ(ÿÿ)) = 2 ÿÿ + 4 + ÿÿ = 14 ÿÿ = 5 ÿÿ + ÿÿ = 10 Properties of the operation of communities 1. ÿÿ ÿ ÿÿ = ÿÿ ÿ ÿÿ Commutative law ÿÿ ÿ ÿÿ = ÿÿ ÿ 2. ÿÿ ÿ ( ÿÿ ÿ ÿÿ) = ( ÿÿ ÿ ÿÿ) ÿ ÿÿ Associative law ÿÿ ÿ ( ÿÿ ÿ ÿÿ) = ( ÿÿ ÿ ÿÿ) ÿ 3. ÿÿ ÿ ( ÿÿ ÿ ÿÿ) = ( ÿÿ ÿ ÿÿ) ÿ ( ÿÿ ÿ ÿÿ) Distributive law ÿÿ ÿ ( ÿÿ ÿ ÿÿ) = ( ÿÿ ÿ ÿÿ) ÿ ( ÿ ÿÿ) 28 ÿÿ ÿÿ ÿÿ = ÿÿ ÿ ÿÿ 4. (ÿÿ ÿ ÿÿ) Democratic Law ÿÿ ÿÿ ÿÿ = ÿÿ ÿ ÿÿ 5. (ÿÿ ÿ ÿÿ) Machine Translated by Google Example: If ÿÿ ÿ ÿÿ = (ÿ1,2,3 ) ÿ ÿÿ = (3,4,5) Assign : ÿÿ ÿ ( ÿÿ ÿ ÿÿ). ? ÿ ÿÿÿÿÿÿÿÿ ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿ ( ÿÿ ÿ ÿÿ) ÿ ( ÿÿ ÿ ÿÿ ÿÿ { ÿ1,2,3} ÿ {3,4,5} = {ÿ1,2,3,4,5} ) Example: ÿÿ ÿ (ÿÿ ÿ ÿÿ) {ÿÿ\(ÿÿ ÿ ÿÿ)} ÿ {(ÿÿ ÿ ÿÿ)\ÿÿ} (ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ) ( ÿÿ ÿ ÿÿ)\ ÿÿ or ( ÿÿ ÿ ÿÿ)\ 29 Some properties of community operations: ÿÿ\ÿÿ = ÿ, ÿÿ ÿ ÿÿ = ÿÿ Machine Translated by Google ÿÿ ÿ ÿ = ÿ, ÿÿ ÿ ÿ = ÿÿ, ÿÿ ÿÿ= ÿÿ, ÿÿ ÿÿ=ÿ, ÿÿ ÿ= ÿ, ÿ/ ÿÿ = ÿÿ ÿÿ , ÿÿ ÿÿ ÿ ÿÿ = ÿÿ, ÿÿ ÿÿ\ÿÿ = ÿÿ ÿ ÿÿ. Example: To draw the congeners: a). [(ÿÿ ÿ ÿÿ) ÿ ÿÿ] b). [(ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ ÿ ÿÿ)] c). [(ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ)] ÿ ÿÿ ÿÿ d). ÿÿ ÿ (ÿÿÿÿÿÿ) 30 Machine Translated by Google Example: Simplify the following communities: ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ a) [(ÿÿ ÿ ÿÿ ) ÿ (ÿÿ ÿ ÿÿ)] ÿ (ÿÿ ÿ ÿÿ ) b) ÿ [(ÿÿ ÿ ÿÿ) ÿ ÿÿ] Solution: ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ a). [(ÿÿ ÿ ÿÿ ) ÿ (ÿÿ ÿ ÿÿ)] ÿ (ÿÿ ÿ ÿÿ ) = b). ÿÿ ÿ [(ÿÿ ÿ ÿÿ) ÿ ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ] = [(ÿÿ ÿ ÿÿ) ÿ (ÿÿ ÿ ÿÿ)] ÿ (ÿÿ ÿ ÿÿ ) = ÿÿ ÿ [(ÿÿ ÿ ÿÿ )ÿ ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ ] = [ÿÿ ÿ (ÿÿ ÿ ÿÿ)] ÿ (ÿÿ ÿ ÿÿ ) = ÿÿ ÿ [(ÿÿ ÿ ÿÿ ) ÿ (ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ ÿ ÿÿ )] = ÿÿ ÿ (ÿÿ ÿ ÿÿ) = ÿÿ ÿ [ÿÿ ÿ (ÿÿ ÿ ÿÿ )] = ÿÿ ÿ ÿÿ. ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ ÿÿ ÿ (ÿÿ ÿ ÿÿ ) = (ÿÿ ÿ ÿÿ ) ÿ ÿÿ = ÿÿ ÿ ÿÿ = ÿÿ 31 Chapter III. Relations, functions and operations Cartesian product: The pair of the form A, B formed by the two components, the first component is A and e dzta is B is called an ordered pair and is symbolically denoted (A,B) and worth a)( x y ,3 5,. ) = (abcdac dheb d ÿ( ) , , ) = =ÿ = ( ) b) Example: (4 ,3 7,6. xyxy 3 Find x and ÿ=ÿ)() Solution: a) y = ÿ = 5, 3. x ÿ ÿÿ=ÿ Machine Translated by Google 21 b) 4 7 /· 3 ( ) 12 3 x y x y ÿ+ ÿÿ 6 18 x =ÿ ÿÿ=ÿ 63xy ÿÿ=633xy = ÿ33 ÿ= ÿÿ x ÿ =ÿ=ÿ=ÿ=== ÿ yx to xy 4 7 4·3 7 12 7 5 , 3, 5. Let A,B be two sets, where the ordered pair is obtained in such a way that the component the first belongs to A while the second component belongs to B. The union of these couples the sequence is called the Cartesian product of the set A and B, and mathematically marked: AxB aba Adheb B = ÿ ÿ {( ; | ) } example AB ab : 1,3, , , = ÿ = { } { } AxB BxA , AxB ababab = ÿ ÿ {( 1, , 1, , 3, , 3, , 6, , 6, ) ( ) ( ) ( ) ( ) ( )} BxA aaabbb = ÿ ÿ {( , ( ) ( ) ( ) ( )} 1 , ,3 , ,6 , , 1 , ,3 , ,6 ) ( ) 32 If the number of the set n(A)=a and the number of the set n(B)=b then n(AxB)= a ÿ ÿÿ. Example: If 2 2 ÿÿ × ÿÿ = {(ÿÿ, ÿÿ)|(ÿÿ , ÿÿ , ÿÿ) = (9,2)(ÿÿ, ÿÿ) ÿ ÿÿÿÿÿÿ} Solution: Then find A x B , n(A x B) and n(p(A x B))=? 2 ÿÿ 2 = 9 ÿÿ ÿ 2 = 2 A={ -3 , 3} 2 |ÿÿ| = ÿ9 ÿÿ ÿ 2 + 2 B={ -2 , 2} 2 |ÿÿ| = 3 ÿÿ = 4 A x B = { (-3,-2),(-3,2),(3,-2),(3,2) } x = 3 ose x = -3 |ÿÿ| = ÿ4 n ( A x B ) Machine Translated by Google = n ( A ) xn ( B )=2 x 2 = 4 | | = 2 sepse A,B can be the number. y = 2 or y = ÿ2 n (p (A x 4 B) ) = 2 = 16 elements. Cartesian Product on Intervals Example: To present the Cartesian products graphically: a) {1,2} x {-3} b)[1,2] x [2,4] c) [-3,5]X[-2,2) solution: a) {(1,-3),(2,-3)} 33 b)[1,2] x [2,4] c) [-3,5]X[-2,2) Machine Translated by Google 34 Relationships In everyday life, we constantly encounter relationships between objects, between people, in mathematics, in computing, etc. In some examples of relations we encountered in the previous chapters, such as, for example, a logical expression is equivalent to another logical expression. Similarly, the fact that a set is a subset of another set is a relation. In computing, for example in the database in it which the words are arranged in alphabetical order, we are dealing with the relation that one word precedes another word in alphabetical order. Other examples of relationships can be found in the family ties of family members: one person is the sister of another, is married to..., etc. Relations are classified according to the number of "objects" that connect them into: binary relations, ternary relations, etc. Machine Translated by Google is the relation from set A to set B if It is said that ÿ ÿ ÿ ×AB product A xisB,the sosubset of the Cartesian Example : The sets A={1,4,5}, B={1,2,3,6} are given. ÿA x B} ÿ={(x,y) | x < y, (x,y) ÿwith Venn diagram and system Show the relation Solution : coordinate. b= {(4,1),(4,2),(4,3),(5,1),(5,2),(5,3)} 35 Inverse of Relation Let ÿ ={x,y | x ÿ A and x ÿ B} , ÿ ÿ1= {(y,x) | x ÿ A and y ÿ B} is the inverse relation of the relation ÿ. Example: form of the Basquesite A = {ÿ3,ÿ1,0,1} , ÿ= {(ÿ3,ÿ2),(ÿ3, x B = {ÿ1,ÿ2,ÿ1}. For ÿ={(x,y) | x ÿ Let the ÿ A , y ÿ B} y , ,ÿ1) , (ÿ1,ÿ1)} ÿ ÿ1=? to be found Solution: Machine Translated by Google ÿ ÿ1= {(-2,-3),(-1,-3),(-1,-1)} 36 Relationship Properties Example : Let the set A ={1,2,3} be given. Discuss whether the following relations are reflexive! Machine Translated by Google a). B1 = {(1,1),(2,2),(3,3)} ÿ is a reflexive relation. b). B2 = {(1,2),(2,2),(3,3),(2,3)} ÿ is not reflexive because (1,1) is not an element of B2. c). B3 = {(1,1),(1,2),(2,2)} ÿ is not reflexive because (3,3) is not an element of B3 d). B4 = {(1,1),(2,2),(1,3),(3,3)} ÿ is reflexive. Example : Discuss whether the following relations are reflexive! a). B1={(l,d) | lÿd , l,d are straight lines in the plane} ÿ it is not a reflective relation since a straight line cannot be normal to itself. b). B2={(l,d) | l ÿ d, l,d line in the plane} ÿ yes reflective since it is assumed that each line is parallel to itself. c). B3={(l,d) | l congruent (congruent) with d, where l and d are triangles in the plane } ÿ yes reflexive because every triangle is congruent with itself. d). B4={(A,B) | AcB, A,B--set } ÿ is not a reflective relation since A ÿ A. 37 Example: Check if the following relations are symmetric! a). Example (Preliminary example) is a symmetric relation that l ÿ d ÿ ÿÿ ÿ b ). symmetric relation since ÿÿ ÿ ÿÿ ÿ ÿÿ ÿ c). symmetrical for the same reasons. d). is not a symmetric relation since ÿÿ ÿ ÿÿ then ÿÿ ÿ ÿÿ does not hold The antisymmetric relation does not mean that it is not like some relations having both properties, but the symmetric relation is not an antisymmetric relation. Example: Let the set A={1,2,3} be given. Discuss the properties of the following relations: a). B1={(1,2)} ÿ it is not reflexive, it is not symmetric, but it is antisymmetric. b). B2={(1,1),(2,2),(3,3)} ÿ is reflexive, is symmetric, and is antisymmetric. Machine Translated by Google c). B3={(1,1),(1,2),(3,2),(2,3)} ÿ it is not reflexive, it is not symmetric, but it is antisymmetric. d). B4={(1,2),(2,1),(3,2),(3,3)} ÿ is not reflexive, is not symmetric and is antisymmetric. e). B5={(1,2),(2,3),(3,1),(4,2)} ÿ is not a relation. (a,b) ÿ B and (b,c ) ÿ B ÿ (a,c) ÿ B- transitive property Example: Let the set A={1,2,3,4} be given. Discuss the transitive properties of the following relations: B1={(1,1),(2,3),(3,4),(2,4)} B1={(1,1),(2,2),(3,3),(3,4)} B1={(2,1),(1,2)} Solution: 38 B1 is transitive relation , B2 is transitive relation , B3 since (1,4) is not I B1 is a meets B3 then B3 is not a transitive relation , transitive relation. Example: Discuss the properties of the relation B={(4,4),(3,3),(1,3),(3,1),(2,2),(2,4)} defined on the set A={ 1,2,3,4} Solution: a). Since (1,1) does not belong to B, then B is not reflexive. b). Since (4,2) does not belong to B, then B is not symmetric. c). It is not antisymmetric since (3,3) ÿ ÿÿ ÿÿ ( 3,1 ) for 3 ÿ 1 d). It is not transitive since (1,3) ÿ ÿÿ ÿÿ ( 3,1) by (1,1) ÿ B. Example: Machine Translated by Google 2 2 Discuss the properties of the relation B={(x,y ) ÿ ÿÿ = 0, (x,y) ÿ } Solution: ÿÿ ÿÿ ÿ ÿÿ a). The reflexive property (x,x) ÿ ÿÿ, with dog than = ÿÿ ÿ(x,x) ÿ , so it is reflexive. b). Symmetric property (x,y) ÿ ÿÿ and (y,x) ÿ ÿÿ ÿÿ if (x,y) ÿ ÿÿ ÿ ÿ ÿÿ = 0|(-1) ÿÿÿ 2 + ÿÿ 2 =0 2 2 ÿÿ ÿ ÿÿ = 0 , (y,x) ÿ 22ÿ2 c). It is not an antisymmetric relation 2 = 0 ÿ (2,2) ÿ ÿÿ d). (x,y) ÿ ÿÿ ,(y,z) ÿ ÿÿ ÿ (x,z) ÿ ÿÿ 2 2 ÿ ÿÿ =0 {ÿÿ 2 2 ÿÿ ÿ ÿÿ =0 39 2 2 ÿÿ ÿ ÿÿ = 0 ÿ (x,z) ÿ ÿÿ ÿ so it is a Transitive relation. Equivalence Relation Let B be a relation on the set A. If B is reflexive, symmetric and transitive, then it is said that B is an equivalence relation in the set A. Example: Show that the relation ÿÿ = {(ÿÿ, ÿÿ), 3| ÿÿ ÿ ÿÿ, (ÿÿ, ÿÿ) ÿ Machine Translated by Google } equivalence relation Solution:.Reflexive property ÿ x ÿ ÿÿ,(x,x) ÿ B , because 3|xx , 3|0- so it is reflexive.. Vetia simetrike (x,y) ÿ B => (y,x) ÿ B (?) 3|xy, we have that x-y=3A,A ÿ Z -(x-y)=-3A yx=-3A : 3=> 3|yx =>(x,y) ÿ B is symmetric in this sense.. Transitive property 3/ xy and 3/ yz We have xy=3T, yz=3S, respectively xz=3(TS)=3L, where S,T,L ÿ That is, 3/ xz-so the transitive properties apply, which means that relation B is an equivalence relation. 40 Order relation Let B be a relation in the set A, if B is reflexive, antisymmetric and transitive then it is said that B is an ordering relation. Example: Let the set A={1,2,3,4} and the relation be given B={(ÿÿ, ÿÿ)| ÿÿ ÿy,(x,y) ÿ A x A} Defined in the set A. To examine if the given relation is an order relation! Solution: B={(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)}; So B is a reflexive relation, it is an antisymmetric relation but not transitive. So B is not an ordering relation Machine Translated by Google function Let f be a relation from the set A to the set B. If the relation satisfies the following two conditions, then we say that f is a function. ÿ x is A , ÿ y ÿ B ashtu që (x,y) ÿ f ÿ x ÿ A, ÿ y1 ,y2 ÿ B nëse (x,y1 ) ÿ f dhe (x,y2 ) ÿf atëherë y1=y2 A curve is a function if every line, x=n where xi corresponds to the domain of the function, cuts its curve in one and only one point. 41 is a function It is not a function Example: Consider whether the following relations represent functions: 2 a) B1={(x,y)| |x|+|y| ÿÿ. Cakto f(3), f(0)=? Solution: 6 f(3)=(2 ÿ 3) = (ÿ1) 6=1 00=2 f(0) = (2 ÿ 0) =1 43 Example: The function is given ÿÿ+1 2 ÿÿ(2 ) = ÿÿ + 2ÿÿ ÿ 5, set ÿÿ(32), f(0) Solution: 5 4+1 ÿÿ(32) = ÿÿ(2 ) = ÿÿ(2 )=4 2 + 2 ÿ 4 ÿ 5 = 16 ÿ 8 ÿ 5 = ÿÿ+1 3 f(0) has no solution, because 2 > 0, for any real number x Machine Translated by Google Graph of the function The set of all points of the coordinate plane corresponding to the elements of the function f:AB are called the graph of the function, and are symbolically denoted f(x)={(x,y)/ and x ÿ A goes to y ÿ B}. Example: 1. draw the graph of the functionÿÿ(ÿÿ) = 2ÿÿ 2ÿ1 , A={-2,-1,0,1,2}. Solution: ÿÿ(ÿ2) = 2 ÿ (2) 2ÿ1=2ÿ4ÿ1=7 2 ÿÿ(ÿ1) = 2 ÿ (ÿ1) ÿ 1 = ÿ1 ÿÿ(0) = 2 ÿ 0 2 ÿ 1 = ÿ1 2 ÿÿ(2) = 2 ÿ 2 ÿ1=7 ÿÿ(1) = 2 ÿ 1 ÿ 1 = 1 44 Machine Translated by Google 2 Draw the graph: a) y=3x-1 x01 3 and -1 0 45 2 3) Draw the graph of the function ÿÿ = ÿ Machine Translated by Google 2 4ÿÿ ÿ 5 1. the point of intersection of the parabola with Ox ÿ 1 4ÿÿ ÿ 5 = 0 ÿÿ 2ÿ = 4±ÿ16+20 4±6 2= ÿÿ1=5 2 |ÿÿ2=ÿ1 2. the vertex of the parabola 2)) = ((5 + (ÿ1) (ÿÿ1 + ÿÿ2 2, ÿÿ (ÿÿ1 + ÿÿ2 2 ) ÿ ÿÿ(2)) = (2, ÿ9) 2 ÿÿ(2) = 2 ÿ 4 ÿ 2 ÿ 5 = 4 ÿ 8 ÿ 5 = ÿ9 3.pikprerja e parables me Oy (0, ÿÿ(0)) = (0, ÿ5) ÿÿ(0) = 0 2 ÿ 4 ÿ 0 ÿ 5 = ÿ5 46 Machine Translated by Google Example: If from the following figure f(f(x))+f(2)=4, then set x=? Solution: f(f(x))+1=4 f(f(x))=3 f(x)=-1 x=-4 47 Types of Functions 1. Injective function (one-to-one) If the relation fe satisfies the condition ÿÿ1 ÿ ÿÿ2 => f(x1 ) ÿf(x2 ) or f(x1 )=f(x2 )=>x1=x2 then it is said that the function f is the Injective function. Example: Which of the following functions is Injective? Machine Translated by Google It is Injective, since each line parallel to the abscissa intersects the graph of the function at only one point It is not Injective that y=3 bisects long. 48 Example: Which of the following functions is Injective? a) F(x)=3x-1 , f:R =>12 2 b) F(x)= ÿÿ + 1 , f: ÿÿ =>12 + c) F(x)= ÿÿ 2 + 1 , f: ÿÿ =>R 3 d) F(x)= ÿÿ , f:R =>R Solution: a) x1 ÿx2 =>f(X1) ÿf(x2) Machine Translated by Google x1 ÿx2 / ÿ3 3x1-1 ÿ3x2-1/-1 f(x1) ÿ f(x2) =>f-Injektiv. b) f(x1)=f(x2)=>x1=x2? 2 x1 + 1 = x2 2+1 2 2 x1 ÿ x2 =1ÿ1 (x1-x2)(x1+x2)=0 x1-x2=0 ose x1+x2=0 x1=x2 ose x1=-x2 It is not injectable. c) and d) injective functions. Surjective function: If for the function f:A=>B the value f(A)=B then it is said that f is a surjective function 3 a) ÿÿ = ÿÿ , f:R =>R 49 Since ÿÿÿ ÿ ÿÿ, ÿÿ ÿÿ is y=f(x) =>f(x) is 2 surjetiv/ b) B) ÿÿ = ÿÿ , f:R =>R Machine Translated by Google 2 so for example => y=-10 -10= ÿÿ ÿ R, , x there is no such xi, so f(x) is not surjective. 50 inverse function The function which is Injective and Surjective is called Bijective Function. Example: f(x)=x+1, f:Z=>Z is the bijective function since it is an Injective and Surjective function. Every bijective function has an inverse function. ÿ1 Function is called the inverse of the function f if f is bijective and (x)= ÿ1 if f(x)={(x,y)x ÿA , y ÿB} then {(x,y)x ÿA , y ÿB} F F(x)=y ÿ1 ÿÿ (y)=x Machine Translated by Google ÿ1 ÿÿ Finding the Inverse: Example: 2ÿÿ+1 c) y= 3ÿÿÿ4 ÿÿ = 2ÿÿ+1 a) y=3x-1 a) x=3y-1 b) y=x+3 1 3ÿÿÿ4 b) y=x+3 3y=x+1 x=-y+3 2ÿÿ+1 c) y= +1 3ÿÿÿ4 y= 3 y=-x+3 x(3y-4)=1(2y+1) ÿ1 ÿ1 8+1 x=3 , f(x)=3x3-1=8 ÿÿ (x)=-x+3 3xy-4x=2y+1 ÿÿ (8)= 9 3= x=2 , f(2)=-2+3=1 3xy-2y=4x+1 3=3 ÿ1 4ÿÿ+1 => ÿÿ (1)=-1+3=2 y=(3x-2)=4x+1 y= 3ÿÿÿ2 ÿ1 = 4ÿÿ+1 ÿÿ 3ÿÿÿ2 51 2. For the bijective function f: [1, ÿ) =>[2, ÿ), f(x) = ( ÿÿ ÿ 1) 2 + 2. ÿ1 ÿ1 ÿÿ (x) , ÿÿ (3)=? Solution: y=( ÿÿ ÿ 1) 2+2 2 x=(ÿÿ ÿ 1) +2 2 x-2= (ÿÿ ÿ 1) ÿÿÿ ÿ 2 = (ÿÿ ÿ 1) ÿ ÿÿ ÿ 1 ÿ1 ÿÿ (ÿÿ) = ÿÿÿ ÿ 2 + 1 ÿ1 ÿÿ (3) = ÿ3 ÿ 2 + 1 = 1+1=2 3 ÿ1 a)f:R =>R , f(x)=ÿÿ 3. ÿ 2ÿÿÿÿ + 1 find m if passes through 3 2 ÿ 2ÿÿ ÿ1 the point (-2,1) b) f:R=>R , f(x) = + 2 find (2)=? ÿ1 ÿ1 3 a) ÿÿ (-2)=1 =>f(1)=-2 b) ÿÿ (2)=a =>f(a)=-2 f(1)= 1 ÿ 2ÿÿ + 1 = Machine Translated by Google 3 2 ÿ 2ÿÿ ÿÿ ÿ 2ÿÿ 2+2=2 3 2 2-2m=z ÿÿ ÿ 2ÿÿ = 0 ÿ1 ÿ1 2 -2m= -2-2ÿÿ (a-2)=0=>a=0 axis (2)=0 or (2)=a -2m=-4 m=2 52 Function composition A+BgC 2 f(x)=x-1 , g(x) ÿÿÿ -1 and 2 2 2 ÿ 1 ÿ1= ÿÿ ÿ 2 ÿÿÿÿÿÿ = ÿÿ[ÿÿ(ÿÿ)] = ÿÿ[ ÿÿ ÿ 1]= ÿÿ 2 2 ÿÿÿÿÿÿ=g[f(x)]=g[x-1]= (ÿÿ ÿ 1) ÿ1= ÿÿ ÿ 2ÿÿ + 1 Machine Translated by Google 2 = 1 ÿ ÿÿ ÿ 2ÿÿ Properties of the composition of functions: 1. ÿÿÿÿÿÿ ÿ ÿÿÿÿÿÿ 2. (ÿÿÿÿÿÿ)ÿÿÿ = ÿÿÿÿ(ÿÿÿÿÿ) ÿ1 3. ÿÿÿÿÿÿ = ÿÿ ÿ1ÿÿÿÿ => I(x)=x-identical function 4.foI=Iÿÿf ÿ1 ÿ1 5.(ÿÿ ) = ÿÿ Example: If f(x) = 3ÿÿ ÿ 2 and (ÿÿÿÿÿÿ)(ÿÿ) = 3ÿÿ ÿ 3 ÿÿÿÿÿÿÿÿÿÿÿÿÿ ÿÿÿÿÿÿÿÿÿÿ ÿÿ( ÿÿ) =? Solution : ÿÿ ÿ1ÿÿ(ÿÿÿÿÿÿ) = (ÿÿ ÿ1ÿÿÿÿ)ÿÿÿÿ = ÿÿÿÿÿÿ = ÿÿ 53 ÿ1 ÿ1 ÿÿ+2 g(x) = ÿÿ (ÿÿÿÿÿÿ) = 3, ÿÿ (ÿÿ) = ÿ1 ÿÿ+2 ÿ1 3 ÿÿ 3 , (x)= ÿÿ (3ÿÿ ÿ 5) =3ÿÿ ÿ 5 + 2 3=3(ÿÿ ÿ 1) 3= ÿÿ ÿ 1 DSH : Y=3x-2 g(x)=3x-2 , f(ÿÿÿÿÿÿ)=3x-3 X=3y-2 3=3ÿÿ ÿ 3 3y=x+2 ÿÿ+2 ÿÿ = Machine Translated by Google ÿ1 direction : (ÿÿ0ÿÿ) ÿ = ÿÿ as a preliminary task: Example: 2 Let f,g:R=>R , f(x)= 2ÿÿ be given ÿ 1 and g(x)=2x-3 ÿ1 ÿ1 Find: ( ÿ ÿÿ) (3) Solution : ÿ1 (ÿÿ ÿ ÿÿ) = ÿÿ ÿ1ÿÿÿÿ ÿ1 ÿ1 (ÿÿ ÿ1ÿÿÿÿ) ÿ1 = ÿÿ ÿ1ÿÿ(ÿÿ ÿ 1) ÿÿ => (ÿÿ = ÿÿ ÿÿ ÿ1 ÿ1ÿÿÿÿ) ÿ1 (3) = (ÿÿ ÿ1ÿÿÿÿ)(3) = ÿ1 ÿ1 2 ÿ1 17+3 ÿÿ (ÿÿ(3)) = ÿÿ (2 + 3 ÿ 1) = ÿÿ (17) = 20 2= 2= 10 y=2x-3 x=2y-3 +3 y= ÿ1 2= ÿÿ (ÿÿ) 54 Chapter IV-Mathematical induction, the sum symbol and Machine Translated by Google PROdUCER Mathematical Induction The given statement p(n),nÿN is proved to be true for every natural number if: 1. P(n) It is true. 2. For all natural numbers nÿ1 the implication from p(n)=>P(n+1) applies Example: By means of mathematical induction prove that: 1+2+3+....+n= ÿÿ(ÿÿ+1) 2 Solution: 1(1+1) for n=1, ÿÿ = 2 1ÿ2 ÿÿ = 2 i=1 + 2 + 3 + ÿ + ÿÿ = ÿÿ(ÿÿ+1) 1 It is assumed to hold for n=k , 2 + 2 + 3 + ÿ + ÿÿ + ÿÿ + 1 = (ÿÿ+1)(ÿÿ+1+1) 1 (ÿÿ+1)(ÿÿ+2) 2= 2(? ) hahahaha +++ ÿ ÿ 1 1 2 ( )( ) () () () 123... 1 1 1 · 1 ++++++=++= ++=ÿÿÿ ÿ hahahaha 222 Consequently, the equality holds for every natural number. Machine Translated by Google 55 Example: Prove that the equation holds: 2 ÿÿ+ nn () 1123.... 3333 ++++=ÿÿ ÿÿ n 2 + ÿÿ ÿÿ 1112 32 2 () 2 2 Proof for n=1, ÿÿ+ ÿ = ÿ = 111111= ÿ = kk ÿÿ () ÿÿ 1 333 ÿÿÿÿ22 We assume for n=k , +++=ÿÿ 12..... ÿÿ k ÿÿ++ 2 3 333 12 kk ( )( ) =ÿÿ 12... 1?+++++= ÿÿ kk For n=k+1 ,() () 2 Machine Translated by Google ÿÿ +ÿÿ kkk 2 () 2 22 1 111 ÿ ÿÿÿÿÿÿÿ ÿ++=+++= ÿÿ kkk ()() 24 kkÿÿ++++++ÿÿ hahahaha 44212 22 ()()() 2 = ÿÿ ÿ+=+ ÿÿ ÿÿÿÿ 11 22 ()() 422 2 Example: n 1 x 21 n ÿ ÿ To prove the equality: for n 1..... , 1 1. + + + + = ÿ = xxxx 1 ÿ x 11 110 , 11 ÿÿ ÿ xx xx is valid 1 == ÿ =11 ÿÿ xx k + () 1 1 21 Machine Translated by Google ÿ kx ÿ nxxxx++++=ÿ... , 1 We assume that for n=k, x ÿ 1 56 ÿ 1 + 21 We try for n=(k+1) , k + kkx 1 1...... ?+++++== xxxx 1 + x 1111· ÿÿÿ+ÿ kxxxkxkkk xx correct kk () ++ 11 2 1 Example: 2 = +== x 11ÿÿxx 11ÿÿxk +x ÿÿ. 1111 n ++++=..... To prove that: 1·2 2·3 3·4 1 1 n n n ++ () 1111 (). For n=1 () = ÿ = is valid 1111. For n=k 1111122 ++ k Machine Translated by Google ++++=.... 1·2 2·3 3·4 1 1 k k k ++ () 11111. For n=k+1 k + ++++=.... = +++++ 1·3 2·3 3·4 1 1 1 1 1 k k k ÿ+=== ( )( ) ÿÿ+ kkkkkk 1 2 1 2 1 1 +++++ 2 2 () () k 1 kkkkkk =ÿÿ +ÿÿ+ +++++++ k +1 ( )( ) 11·21·212 ·22 ( )( ) ( )( ) ( )( ) k k Example: ÿ is divisible by 4 for every natural number. To prove that n 51 () 1 for n=1 1 Machine Translated by Google 5 1 4 4·1. ÿ = = is valid kÿ We assume for n=k , 514 , = ÿ on k 514 , Tested for SSN + n=k+1, 1 ÿ=ÿ 5 1 5 ·5 5 4 5 5 1 4 5·4 4 4 5 1 4 5 1 4 kkk + ÿ=ÿ+=ÿ+ÿ+ as of now with full respect =+=ÿ ()() 1 57 Example: And ÿ ÿS plotpjesoh in 5, for each n To certify: 1 natural number ( ) 83: n COUNTS a) For n=1 , 10 8 3 5 5·1 ÿ = = We assume that it applies to n=k+1 , kk n=k , 835 ÿ=a () Let's try if it applies to 11 kk 835? b ++ ÿ= ++11 aÿ ÿ=ÿ=+ÿ=+ÿ 8 3 8 ·3 3 ·3 5 3 ·8 3 ·3 40 8·3 3·3 kkkkkkkk () n k 40 5·3 = +a 583 () k =+a = 5B therefore applies to any natural number. Machine Translated by Google 58 The Symbol of the Many n AMOUNT 1 2.... n = AND aaa + + + ÿSymbolic status can be marked +++ ÿ aaaa 12 in..... aa value that: i agate = == 1 11 nn =. ÿÿ Set the amounts: 4 1234 1. = =+++ 1 ÿ