please full explanation and clear and not only the results ---prove they are cou
ID: 2901409 • Letter: P
Question
please full explanation and clear and not only the results ---prove they are countable
a) set of negative integers;
b) set of integers that are multiples of 5;
For a) and b) write the one to one correspondence between
Z+and the specified set.
2) R is the set of real numbers. Let S = { x?R | -3 < x < 0 }.
Show that the set S is not countable.
For questions 3,4 draw graphs with specified properties
or explain why no such graph exists.
3) Graph with 4 vertices of degrees 1,2,3,4.
4) Graph with 5 vertices of degrees 1,2,3,3,5.
Explanation / Answer
1
a) set of negative integers;---> Countable
mapping between negative interger and z+ as follow
f(x) = -x;
f(-1)=1, f(-2)=2,....
means there exists one to one mapping between set of negative integer and z+ so set is countable.
b) set of integers that are multiples of 5; ---> countable
mapping between interger multiple of 5 and z+ as follow
f(x) = x/5;
f(5)=1, f(10)=2, f(15)=3....
means there exists one to one mapping between set of integer multiple of 5 and z+ so set is countable.
2 Set is not countable.
i have given hint for proof
first assume that it is countable.
since it is contable we can map with z+
lets assume mapping is as follow
1 -- -3.a1a2,a3a4.....
2 -- -3.a1+1a2,a3a4.....
3 ----3.a1+2a2a3.....
but if we add +1 to each ith digit of ith row, the we will get one new no and for this no there is no mpping so our assumpution is wrong and set is not countable.
3) Graph with 4 vertices of degrees 1,2,3,4.
valid graph
A----B----C---2 parallel egde---D --selfloop
A is connected to B, B is connected with C, between C and D we have to parallel edge, and D has self loop also.
4) Graph with 5 vertices of degrees 1,2,3,3,5.
valid graph
since in any graph sum of the degree of virtices is always twice the no of edges
1+2+3+3+5= 14 so no of edges=.7
e1 A---B
e2 C--D
e3 C-- B
e4 E--B
e5 E---B
e6 E--D
e7 B--D
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