Problem 5.58. supposef xy and g : y z are functions if g of is one-to-one, prove that fmust be one-to-one 2. find an example where g o f is one-to-one, but g is not one-to-one

  f : X → Y  and  g: Y → Z

Now we have to show:

If gof is one-to-one then f must be one-to-one.

Given:

gof is one-to-one

To prove:

f is one-to-one.

Proof:

Let us assume that f(x) is not one-to-one .

This means that there exist x and y such that x≠y but f(x)=f(y)

On applying both side of the function by the function g we get:

g(f(x))=g(f(y))

i.e. gof(x)=gof(y)

This shows that gof is not one-to-one which is a contradiction to the given statement.

Hence, f(x) must be one-to-one.

Let A={1,2,3,4}  B={1,2,3,4,5} C={1,2,3,4,5,6}

Let f: A → B

be defined by f(x)=x

and g: B → C be defined by:

g(1)=1,g(2)=2,g(3)=3,g(4)=g(5)=4

is not a one-to-one function.

since 4≠5 but g(4)=g(5)

Also, gof : A → C

is a one-to-one function.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}