Question

For each of the following functions, determine whether or not they are (i) one-to-one and i) onto. Justify your answers (a) f : R-{0} → R and f(x) = 3r-1/x (b) g : R _ {1} → R and g(x) = x + 1/(x-1) (c) l : S → Znon-reg and l(s) = number of 1's in s, for all strings s E S, where s is the set of all strings of O's and 1's. (d) 1 : S → Z and 1(s) number of 1's minus number of 0's in s, for all strings in S, where S is the same as above

Answer 1

f:R-10} → R
defined as
(a)-3r1

Then we have $f(x)=f(y)\Rightarrow 3x-\frac{1}{x}=3y-\frac{1}{y}$

So

1 (x -y)

If $x\neq y$ we have

ry

In particular we have $x=-\frac{1}{3},\,y=1\Rightarrow 3xy+1=0$

Checking, we have

f(-1/3)-1+3 2, f(1) 3-1-2

Thus, this function is not one to one

b) $f:R-\{1\}\to R$ defined as

f(x) = x +

Then $f(x)=f(y)\Rightarrow x+\frac{1}{x-1}=y+\frac{1}{y-1}$

So $x-y=\frac{1}{y-1}-\frac{1}{x-1}=\frac{(x-1)-(y-1)}{(x-1)(y-1)}$

And so $x-y=\frac{x-y}{(x-1)(y-1)}$

If $x\neq y$ we must have $(x-1)(y-1)=1$

So $x=3,y=\frac{3}{2}\Rightarrow (x-1)(y-1)=1$ and

$f(x)=3+\frac{1}{2}=\frac{7}{2},\,f(y)=\frac{3}{2}+\frac{1}{\frac{3}{2}-1}=\frac{3}{2}+2=\frac{7}{2}$

So the given function is not one to one

c) $l(110)=l(011)=2$ (each of these strings has 2 ones) but clearly $110\neq 011$ as binary strings

So this function is not one to one

d) $l(101)=1,\,l(011)=1$ (each of these strings has 2 ones and 1 zero) but clearly $110\neq 011$ as binary strings

So this function is also not one to one

$\blacksquare$

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