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Zolol [24]
2 years ago
13

Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:

Mathematics
1 answer:
professor190 [17]2 years ago
5 0

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

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Answer: smaller number = 5, bigger number = 8

<u>Step-by-step explanation:</u>

Let x represent the smaller number

and y represent the bigger number.

The sum of 2 times the smaller number and 3 times the bigger number is 34.

EQ1: 2x + 3y = 34

Two times the bigger number is subtracted from 5 times the smaller number is 9.

EQ2: 5x - 2y = 9

Solve the system of equations using the Elimination method:

EQ1: 2x + 3y = 34      →   2(2x + 3y = 34)     →   4x + 6y = 68

EQ2: 5x - 2y = 9       →    3(5x - 2y = 9 )      →  <u> 15x - 6y</u> = <u>27 </u>

                                                                         19x         = 95

                                                                      <u> ÷19          </u>  <u>÷19 </u>

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Substitute x = 5 into either equation to solve for y:

EQ2: 5x - 2y = 9

      5(5) - 2y = 9

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                  y =  8

The smaller number (x) is 5 and the bigger number (y) is 8.

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