1.3 3.5 3 n2 2, then n3-n is always divisible by 3. 5.7 (2n-1)(2n+1) 2n+1 Exerci

Question

1.3 3.5 3 n2 2, then n3-n is always divisible by 3. 5.7 (2n-1)(2n+1) 2n+1 Exercise 2: For each of the binary relations described below Write the pairs that satisfy this relation. this relation have? (reflexive, symmetric, anti-symmetric, transitive a) y x+1 on the set S3(1, 2, 3, 4, 5} b) y s x on the set S (0, 2,3,8, 9) c) y is multiple of x on the set S (1,3, 4, 6, 8, 9, 12) Exercise 3 Using a direct proof, show that: 1. Vx x2 +3x is an even number. 2, n positive prime number such that n+1, n+2, n+3, n+4 are not prime. Exercise 4: p: You have the flu (F) q:You miss the final examination (T) r: You pass the course (T). Express each of these propositions as an English sentence and find their truth 1. p q

Explanation / Answer

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Explaination:

reflexive------->for an element w, wRw is true for every w belonging to set T

symmetric--------->for elements w,s if wRs then sRw for all w ,s belonging to set T

antisymmetric-------->for elements w,s,g ,if wRs and sRw then x = y for all w,s belonging to set T

transitive---------->if wRs and sRg then wRg for all w,s,g belonging to set T.

a)

answer: y=x+1 , try put x as every value from set A and then find value of y

put them in a set R with pairs (x,y) as follows

R={(1,2), (2,3),(3,4), (4,5)}

we just need to find one case to proof that is not reflexive,symmetric,antisymmetric and transitive

There is no (1,1) so not reflexive

there (1,2) but not (2,1) so not symmetric

There are no counterexamples for antisymmetric hence The relation is antisymmetric

There is (1,2) (2,3) but not (1,3) so not transitive

b)

answer: y<=x on S={0,2,3,8,9}

put them in a set R with pairs (x,y) as follows

y<=x

i.e x>=y

R={(2,0) (3,0) (8,0) (9,0) (3,2) (8,2) (9,2) (8,3) (9,3) (9,8)}

no (0,0 ) hence not reflexive

(2,0) but not (0,2) hence not symmetric

No counterexamples for antisymmetric in the relation hence it is considered as antisymmetric

In order to check trasnsitivity take evry element find out w,s and s,g in relation and check if there is w,g for all. e.g (2,0) now [is there (0, something) --->no] so take next element ,keep on doing until you get a counter example or all elements are examined.

It is transitive because it satifies it's condition as explained above.

c)

asnwer: y is multiple of x in s={1,3,4,6,8,9,12}

put them in a set R with pairs (x,y) as follows

y = x * integer

x * integer=y

R={ (1,3) (1,4) (1,6) (1,8) (1,9) (1,12) (3,6) (3,9) (3,12) (4,8) (4,12) (6,12)}

no (1,1) so it not reflexive

(1,3) but not (3,1) so not symmetric

No counterexamples for antisymmetric in the relation hence it is considered as antisymmetric

It is transitive since for every wRs and sRg there is wRg  

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