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Relations Test 7

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Relations Test 7
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  • Question 1
    1 / -0
    Let $$R$$ be the relation on $$Z$$ defined by $$R = \{(a, b): a, b \in z, a - b$$ is an integer$$\}$$. Find the domain and Range of $$R$$.
    Solution
    Given:

    $$R=\{ (a, b) : a, b, \in z, a-b \text{  is an integer}\}$$

    As difference of integers are also integers so,

    Domain of $$R = z$$

    Range of $$R = z$$, as

    $$a-b$$ spans the whole integer values.
  • Question 2
    1 / -0
    Find x and y, if $$(x+3,5)=(6,2x+y)$$.
    Solution
    By the definition of equality of ordered pairs
    $$(x+3,5)=(6,2x+y)$$
    $$x+3=6$$ and $$5=2x+y$$
    $$x=3$$ and $$5=2x+y$$
    $$x=3$$ and $$5=6+y$$
    $$x=3$$ and $$y=-1$$
  • Question 3
    1 / -0
    A relation $$R$$ is defined from $$\left\{ 2,3,4,5 \right\} $$ to $$\left\{ 3,6,7,10 \right\} $$ by:
    $$xRy\Leftrightarrow x$$ is relatively prime to $$y$$. Then, domain of $$R$$ is
    Solution
    Let's first write R as a set of ordered pairs,

    $$R=\{(2,3),(2,7),(3,7),(3,10),(4,3),(4,7),(5,3),(5,6),(5,7)\}$$

    So, Domain of R is $$\{2,3,4,5\}$$
  • Question 4
    1 / -0
    The relation $$R$$ in $$N\times N$$ such that $$(a,b)R(c,d)\Leftrightarrow a+d=b+c$$ is
    Solution
    If (a,b)R(c,d) then (c,d)R(a,b) as if a+d=b+c then b+d=a+c .
    Thus relation R is symmetric .

    For some a,b in N , (a,b)R(a,b) as a+b=b+a.
    Thus every element of $$N\times N$$ is related to itself,
    So relation R is also reflexive .

    For some a,b,c,d,e,f in N, if (a,b)R(c,d) and (c,d)R(e,f) then (a,b)R(e,f),
    As if a+d=b+c...1 and c+f=d+e….2 then a+f=b+e [since eq.1- eq.2 gives a-e=b-f$$\rightarrow$$a+f=b+e ]
    So, R is also symmetric relation .

    Thus R is a equivalence relation.So, C is correct answer .
  • Question 5
    1 / -0
    Let $$R$$ be the relation over the set of all straight lines in a plane such that $${l}_{1}$$ $$R$$ $${l}_{2}\Leftrightarrow {l}_{1}\bot {l}_{2}$$. Then, $$R$$ is
    Solution
    For some lines $$l_1,l_2$$, We know that $$l_1\perp l_2 \leftrightarrow l_2\perp l_1$$
    So, given relation is symmetric .

    For some lines $$l_1,l_2,l_3$$, We know that if $$l_1\perp l_2 \ and \ l_2\perp l_3$$, then $$l_1\parallel l_3$$ 
    So, given relation is not transitive .

    For some line $$l$$, We know that it is never possible that $$l\perp l$$.
    So, given relation is not reflexive .

    Thus $$A$$ is correct answer .
  • Question 6
    1 / -0
    Let $$R$$ be a relation on the set $$N$$ given by $$R=\left\{ \left( a,b \right) :a=b-2,b>6 \right\}$$. Then
    Solution
    $$a=b-2,b>6$$

    Above relation also imply $$a>6-2\rightarrow a>4$$

    Among the options 

    $$C$$ is correct answer as $$6>4$$ and aslo $$6=8-2$$ .
  • Question 7
    1 / -0
    If $$A=\left\{ a,b,c \right\} $$, then the relation $$R=\left\{ \left( b,c \right)  \right\} $$ on $$A$$ is
    Solution
    The relation R contains (b,c) but do not contain (c,b) , so it is not symmetric .
    The relation R do not relates any element of set A with itself, so it is not reflexive .
    The relation R contains only one pair (b,c), so it is transitive .

    Thus option C is correct .
  • Question 8
    1 / -0
    Which of the following is not an equivalence relation on $$Z$$?
    Solution
    Among the Options

    $$Opt:[C]$$

    $$aRb\Leftrightarrow a< b$$

    As, $$a<b$$ doesn't imply $$b<a$$ ,which make the relation asymmetric and thus it is not an equivalence relation .
  • Question 9
    1 / -0
    If $$A=\left\{ 1,2,3 \right\} , B=\left\{ 1,4,6,9 \right\} $$ and $$R$$ is a relation from $$A$$ to $$B$$ defined by $$x$$ is greater than $$y$$. The range of $$R$$ is
    Solution
    First Let's write R as set of ordered pairs,
    $$R=\{(2,1),(3,1)\}$$
    We can simply see from set that range of $$R = \{1\}$$
  • Question 10
    1 / -0
    A relation $$\phi$$ from $$C$$ to $$R$$ is defined by $$x\phi y\Leftrightarrow \left| x \right| =y$$. Which one is correct?
    Solution
     Given, a relation $$\phi$$ from $$C$$ to $$R$$ is defined by $$x\phi y\Leftrightarrow \left| x \right| =y$$.

    Now, $$|(2+3i)|=\sqrt{2^2+3^2}=\sqrt{13}$$.

    So $$(2+3i)$$ and $$13$$ are not related.

    Again,  since $$y>0$$ this gives $$3$$ not related to $$-3$$.

    $$|1+i|=\sqrt{1+1}=\sqrt 2$$

    So $$1+i$$ and $$2$$ are not related.

    Also $$|i|=1$$ so $$i\phi 1$$ holds.
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