2(x+3)^2=49 A. 1.95 and 7.95 B. -1.95 and -7.95 C.1.95 and -7.95 D.4 and -10
1 answer:
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Answer:
a) Countably infinite
b) Countably infinite
c) Finite
d) Uncountable
e) Countably infinite
Step-by-step explanation:
a) Let S the set of integers grater than 10.
Consider the following correspondence:
defined by
for
.
Let's see that the function is one-to-one.
Suppose that f(10+k)=f(10+j) for k≠j. Then k-1=j-1. Thus k-j=1-1=0. Then k=j. This implies that 10+k=10+j. Then the correspondence is injective.
b) Let S the set of odd negative integers
Consider the following correspondence:
defined by
.
Let's see that the function is one-to-one.
Suppose that f(-(2k+1))=f(-(2j+1)) for k≠j. By definition, k=j. This implies that the correspondence is injective.
c) The integers with absolute value less than 1,000,000 are in the intervals A=(-1.000.000, 0) B=[0, 1.000.000). Then there is 998.000 integers in A that satisfies the condition and 999.000 integers in B that satifies the condition.
d) The set of real number between 0 and 2 is the interval (0,2) and you can prove that the interval (0,2) is equipotent to the reals. Then the set is uncountable.
e) Let S the set A×Z+ where A={2,3}
Consider the following correspondence:
defined by
Let's see that the function is one-to-one.
Consider three cases:
1. , then 2k=2j, thus k=j.
2. , then 2k+1=2j+1, then 2k=2j, thus k=j.
3. , then 2k=2j+1. But this is impossible because 2k is an even number and 2j+1 is an odd number.
Then we conclude that the correspondence is one-to-one.