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Artemon [7]
4 years ago
11

Let C be the collection of all lines in the plane. Define a relation on C by saying that two lines are equivalent if and only if

they are parallel or equal. Show that this an equivalence relation on L.
Mathematics
1 answer:
pav-90 [236]4 years ago
6 0

Answer:

Recall that a equivalence relation must satisfy three conditions: it must be symmetric, reflexive and transitive. Let us study each one of them.

<em>Reflexive</em>: Here we have to prove that an element is in relation with itself. In this particular case, we need to show that every line is parallel or equal to itself, which is obvious.

<em>Symmetric</em>: Here we have to show that if one element is in relation with another, the reciprocal is also true. So, in this case, we want to show that if the line r is in relation with s, then the line s is in relation with r. So, as r is in relation with s, they are equal or parallel.

If they are equal, r=s implies s=r directly. If they are parallel, from elementary geometry we know that r║s means that s║r too. So, the relation is <em>reflexive.</em>

<em>Transitive</em>: Here we want to show that if r is in relation with s, and s is in relation with t, then r is in relation with t. Let us assume that the three lines are parallel, because if we have equality between two of them the statement is true directly from the above example. If the three are parallel and distinct, we know from elementary geometry that if r║s and s║t then r║t. So, the relation is transitive.

Therefore, as the relation is transitive, symmetric and reflexive is an equivalence relation.

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goldenfox [79]
f(x)=\quad 2{ x }^{ 2 }+3x+5

Let's called the input 'z'

When we plug 'z' in the function we get ;

f(z)=\quad 2{ z }^{ 2 }+3z+5

And we know that, this is equal to 19, so ;

2{ z }^{ 2 }+3z+5=\quad 19

Let's rearrange the equation.

2{ z }^{ 2 }+3z+5=\quad 19\\ \\ 2{ z }^{ 2 }+3z=\quad 19-5\\ \\ 2{ z }^{ 2 }+3z=\quad 14\\ \\ 2{ z }^{ 2 }+3z-14=\quad 0

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We'll use this formula to solve it :

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The formula is used in equation formed like this :

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In our equation,

a=2 , b=3 and c=-14

Let's plug in the values in the formula to solve,

a=2\quad b=3\quad c=-14\\ \\ \frac { -3\pm \sqrt { 9-(4\cdot 2\cdot -14) }  }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9-(-112) }  }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9+112 }  }{ 4 } \\ \\ \frac { -3\pm \sqrt { 121 }  }{ 4 } \\ \\ \frac { -3\pm 11 }{ 4 }

So,

z=\quad \frac { -3+11 }{ 4 } \quad ,\quad \frac { -3-11 }{ 4 } \\ \\ z=\quad \frac { 8 }{ 4 } \quad ,\quad \frac { -14 }{ 4 } \\ \\ z=\quad 2,\quad -\frac { 7 }{ 2 }

So the input can be both, 2 and -\frac{7}{2}

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