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Anestetic [448]
2 years ago
10

Which of these relations on{0,1,2,3}are partial orderings? Determine the properties of a partial ordering that the others lack.

a) {(0,0),(2,2),(3,3)} b) {(0,0),(1,1),(2,0),(2,2),(2,3),(3,3)} c) {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)} d) {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)} e) {(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2), (1,3),(2,0),(2,2),(3,3)
Mathematics
1 answer:
omeli [17]2 years ago
3 0

Step-by-step explanation:

A = {0,1,2,3}

a): R = {(0,0),(2,2),(3,3)}

R is antisymmetric, because if the (a,b)∈R, than a=b.

R is not reflexive, because (1,1) ∉ R while 1 ∈ A.

R is transitive, because if the (a,b)∈R and (b, c) ∈ R, than a=b=c and (a,c)=(a,a)∈R.

R is not portable ordering because R is not reflexive.

b): R = {(0,0),(1,1),(2,0),(2,2),(2,3),(3,3)}

R is antisymmetric, because if the (a,b)∈R and if the (b, a) ∈ R, than a=b (since (2,0) ∈ R and (0,2) ∉ R; and (2,3) ∈ R and (3,2) ∉ R )

R is reflexive, because (a,a) ∈ R of every element a ∈ A.

R is transitive , because if the (a,b)∈R and if the ( b , c )∈R . then a = b or b = c ( since there are only two element not of the form ( a , a ) and that pair does not satisfy ( a,b ) ∈ R and ( b , a ) ∈ R ), which implies ( a , c ) = ( b , c ) ∈ R or ( a , c ) = ( a , b ) ∈ R.

R is a partial ordering, because R is reflexive, antisymmetric and transitive.

c): R =  {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)}

R is reflexive, because (a,a)∈R of every element a ∈ A.

R is antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R . then a = b ( since ( 1 , 2 )∈R and ( 2 , 1 ) ∉ R; ( 3 , 1 ) ∈ R and ( 1 , 3 ) ∉ R ).  

R is not transitive , because ( 3 , 1 ) ∈ R and ( 1 , 2 )∈R, while ( 3 , 2 ) ∈ R.

R is not a partial ordering. because R is not transitive .

d): R =  {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)}

R is the reflexive, because ( a , a )∈R of every elements∈A.

R is the antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R, then a = b ( since ( 1 . 2 )∈R and ( 2 . 1 )∉R; similarly, all other elements not of the form (a,a) ).

R is not transitive, because ( 1 , 2 )∈R and ( 2 , 0 )∈R, while ( 1 . 0 )∉R.

R is not a partial ordering, because R is not transitive,

e):  R = { ( 0 , 0 ) , ( 0, 1 ) , ( 0 , 2 ) , ( 0 , 3 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 2 , 0 ) , ( 2 , 2 ) , ( 3 , 3 ) }

R is the reflexive , because ( a , a )∈R of every element a∈A .

R is not antisymmetric, because ( 1 , 0 )∈R and ( 0 , 1 )∈R while 0 is not equal to 1.

R is not transitive, because ( 2 , 0 )∈Rand ( 0 , 3 )∈R, while ( 2 , 3 )∉R .

R is not a partial ordering, because R is not the antisymmetric and not the transitive.

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Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

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