628 Worked-Out Solutions to Exercises: Chapters 11 to 19
- This mapping, like the one in the Prob. 2, is not one-to-one, so it can’t be an injection
or a bijection. But it’s a surjection, because it’s onto the entire set of boys. - Our mapping was not an injection, because it was not one-to-one. If it wasn’t an
injection, it couldn’t have been a bijection. It was a surjection from A to B, because
we mapped our message onto the entire set B, the set of all 175,000 subscribers to
Internet Network Beta. The maximal domain was A, the set of all 60,000 subscribers to
Internet Network Alpha (before we were kicked out). The essential domain was the set
containing only you and me. The co-domain and the range were both the whole set B. - This relation is one-to-one. For every integer q, there is exactly one even integer r, which
we can get by doubling q. Conversely, for every even integer r, there is exactly one
integerq, which we get when we divide y by 2. Our relation is therefore an injection
from Q to R. It is not onto R, however, because there are plenty of real numbers that are
not even integers. That means we do not have a surjection from Q onto R. If it’s not a
surjection, then it can’t be a bijection. - This relation is not one-to-one. For example, q= 3/7 and q= 3/11 both map into z= 3.
That means it’s not an injection, so it cannot be a bijection, either. It is a surjection,
because it’s onto the entire set Z of integers. We can choose any integer z, place it into
the numerator of a fraction, choose a denominator such that the fraction is in lowest
terms, and have a rational number q that maps to z. - This relation is not quite one-to-one. For every nonzero integer z, there is exactly one
rational number q= 1/z. But if z= 0, there is no q such that q= 1/z. The relation
is therefore not injective nor bijective. It is not surjective, either. There are plenty of
rational numbers that are not reciprocals of integers; 3/7 is an example. - This relation is one-to-one. We “patched the hole” in the relation defined in Prob. 7.
We declared that if z= 0, then q= 0. There is no nonzero integer z such that 1/z= 0,
so we don’t get a “dupe” by making this declaration. We have an injective relation now!
But as with the relation in Prob. 7, we do not have a surjection. That means the relation
is not bijective. - This relation is a function. For any value of x we choose, there is exactly one value of y
such that y=x^4. It’s not one-to-one between the set of reals and the set of nonnegative
reals, but two-to-one except when x= 0. That means the function is not injective, so
it can’t be bijective, either. It’s surjective, because it’s onto the entire set of nonnegative
reals. For any nonnegative real number y, we can find a real number x such that y=x^4. - When we transpose the values of the independent and dependent variables while
leaving their names the same, we get
x=y^4which is equivalent toy=±(x1/4)The plus-or-minus symbol indicates that for every nonzero x, there are two values of y,
one positive and the other negative. This relation is one-to-two except when x= 0, so it