Wednesday, June 23, 2010

how to prove it - ch5, sec5.3(Inverses of Functions) ex

This section is collection of some theorems about inverses of a funcion. Here they are, their proofs can be looked up in the book.

1. Suppose f:A -> B. f is one-to-one and onto iff f-1:B -> A

2. Suppose f is a function from A to B and f-1 is a function from B to A. Then f-1f=iA and ff-1=iB

3. Suppose f:A -> B.
- If there is a function g:B -> A s.t. gf=iA then f is one-to-one.
- If there is a function g:B -> A s.t. fg=iB then f is onto

All of the above can be combined into this one.

4. Suppose f:A -> B. Then the following statements are equivalent.
- f is one-to-one and onto
- f-1:B -.A
- There is a function g:B->A s.t. gf=iA and fg=iB

Now, this one is used in proofs alot.

5. Suppose f:A->B, If there exists a function g:B->A s.t. gf=iA and fg=iB. Then f is one-to-one and onto and g=f-1

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Ex-1: the person sitting immediately to the right of p

Ex-2: A\X

Ex-3: Consider, g(x) = 3x-52.
Then (gf)(x)
= g(f(x))
= g(2x+53)
= 3(2x+53)-52
= x

Thus gf=iR

Also, (fg)(x)
= f(g(x))
= f(3x-52)
= 2(3x-52)+53
= x

Thus fg=iR

It follows from theorem#5 above that f is one-to-one and onto. and f-1(x) = g(x) = 3x-52


Ex-4: Let g be a function from R to R defined as g(x) = (x+32)13

Then, (gf)(x)
= g(f(x))
= g(2x3-3)
= ((2x3-3)+32)13
= (x3)13
= x

Thus gf=iR

Also, (fg)(x)
= f(g(x))
= f((x+32)13)
= 2((x+32)13)3-3
= 2(x+32)-3
= x

Thus fg=iR

Hence f is one-to-one and onto and f-1(x) = g(x) = (x+32)13


Ex-5: Scratchwork:
y = 102-x
taking log of both sides with base 10
log(y) = (2-x)log(10)
=> log(y) = 2-x
=> x = 2 - log(y)

Formal Proof:
Let g:R+R defined by g(x) = 2 - log(x)

It is easy to check that gf=iR and fg=iR+. Hence f is one-to-one and onto and f-1 = g(x) = 2-log(x)


Ex-6: Scratchwork:
y = 3xx-2
=> xy - 2y = 3x
=> x(y-3) = 2y
=> x = 2yy-3

which is not defined for y = 3, clearly B = R\{3}

(a),(b) B = R\{3} and consider g:B -> A defined as g(x) = 2xx-3.

Let x be an arbitrary element of A.
Then (gf)(x)
= g(f(x))
= g(3xx-2)
= 2(3xx-2)(3xx-2)-3
= 6x3x-3(x-2)
= 6x6
= x

Thus gf=iA

Its trivial to check that fg=iB

Hence f is one-to-one and onto and f-1 = g(x) = 2xx-3


Ex-7:
(a) Let x be an arbitrary real number.
Then, (f2f1)(x)
= f2(f1(x))
= f2(x+7)
= x+75
= f(x)

Since x is arbitrary, so f2f1 = f

(b) Note: g means f-inverse-1 and h means f-inverse-2.
It is trivial to compute that
g(x) = x - 7
h(x) = 5x

(gh)(x) = g(h(x)) = g(5x) = 5x-7 = f-1(x)


Ex-8: Prove that ff-1=iB
(a) Let b be an arbitrary element of B. Then we can choose some aA s.t. f(a) = b. Then f-1(b)=a
Now, (ff-1)(b) = f(f-1(b)) = f(a) = b

Since b is arbitrary, so ff-1=iB

(b) It is given that f-1f=iA. Let b be an arbitrary element of B.
Then ((f-1f)f-1)(b) = (iAf-1)(b)
=> f-1((ff-1)(b)) = f-1(b)

Thus, (ff-1)(b)=b

Since b is arbitrary, so ff-1=iB


Ex-9: Suppose g:A->B s.t. fg=iB. Let b be an arbitrary element of B. Then (b,b)iB. Then (b,b)fg. Then, we can choose aA s.t. (b,a)g and (a,b)f. Thus f(a) = b. Since b is arbitrary, so f is onto.


Ex-10: () Let (b,a) be an arbitrary ordered pair in B X A s.t. (b,a)g. Since (b,b)iB, then (b,b)fg. Then we can choose some a0A s.t. (b,a0)g and (a0,b)f. Since (b,a)g, (b,a0)g and g is a function, so a=a0. Hence (a,b)f and then (b,a)f-1.

() Let (b,a) be an arbitrary ordered pair in B X A s.t. (b,a)f-1. Then (a,b)f. Since (a,a)iA, then (a,a)gf. Then we can choose some b0B s.t. (a,b0)f and (b0,a)g. Since (a,b)f, (a,b0)f and f is a function, so b=b0. Hence (b,a)g.


Ex-11:
(a) Since fg=iB, so by theorem 5.3.3(2) f is onto. Since f is one-to-one and onto, so f-1f=iA.
Then, g = iAg = (f-1f)g = f-1(fg) = f-1iB = f-1

(b) Since gf=iA, so by theorem 5.3.3(1) f is one-to-one. Since f is one-to-one and onto, so ff-1=iB.
Then, g = giB = g(ff-1) = (gf)f-1 = iAf-1 = f-1

(c) Since fg=iB, so by theorem 5.3.3(2) f is onto. We will prove by contradiction that f is not one-to-one. Assume f is one-to-one. Since f is one-to-one and fg=iB, using part(a) g = f-1. Since f is one-to-one and onto, so f-1f=iA. Then gf=iA, but this contradicts with the given. Hence the assumption is false and f is not one-to-one.

Since fg=iB, so by theorem 5.3.3(1) g is one-to-one. We will prove by contradiction that g is not onto. Assume g is onto. Then using part(2) f = g-1. Since g is one-to-one and onto, so gg-1=iA. Then gf=iA, but this contradicts with the given. Hence the assumption is false and g is not onto.


Ex-12: Let B' = Ran(f). We'll prove that f-1:B' -> A.
Existence:
Let b be an arbitrary element of B'. Since B' = Ran(f), so we can choose some aA s.t. (a,b)f. Then (b,a)f-1.

Uniqueness:
Let b be an arbitrary element of B' and a1, a2 be two arbitrary elements of A s.t. (b,a1)f-1 and (b,a2)f-1. Then (a1,b)f and (a2,b)f. Since f is one-to-one, so a1=a2.

Hence bB!aA(f-1(b)=a). Thus f-1:B' -> A


Ex-13:
(a) Its trivial to see that xAyA(xRyf(x)=f(y)). So f is compatible with R. By using result of ex-18(a) in section-5.1 there exists a function h:A/R -> B s.t. for all xA, h([x]R) = f(x)

(b) It follows straightforward from the result of ex-16, sec-5.2 that h is one-to-one.

Proof for h is onto:
Let b be an arbitrary element of B. Since f is onto, so we can choose aA s.t. f(a) = b. clearly a[a]R, so h([a]R) = f(a) = b. Hence, h is onto.

(c) Let g(b) = {xAf(x)=b}
Let a be an arbitrary element of A. Then
(gh)([a]R) = g(h([a]R)) = g(f(a)) = {xAf(x)=f(a)} = {xA(x,a)R} = [a]R. Thus, gh = iA/R

Let b be an arbitrary element of B. Then
(hg)(b) = h(g(b)) = h({xAf(x)=b})

Since f is onto, so we can choose some aA s.t. f(a) = b. Then
{xAf(x)=b} = {xAf(x)=f(a)} = {xAxRa} = [a]R

Then (hg)(b) = h([a]R) = f(a) = b. Thus hg = iB

Since gh = iA/R and hg = iB, so h-1(b) = g(b) = {xAf(x)=b}


NOTE: In all the exercises below, f|A denotes restriction of f to set A.


Ex-14:
(a) Let a be an arbitrary element of A'. Then we can choose some bB s.t. g(b) = a.
Now (gf)(a) = (gf)(g(b)) = (g(fg))(b) = (giB)(b) = g(iB(b)) = g(b) = a. Since a is arbitrary, so for any xAʹ, (gf)(x)=x

(b) Its easy to see that (fAʹ)g=iB and g(fAʹ)=iAʹ. Hence by theorem 5.3.4, f|A' is one-to-one and onto function from A' to B. And g = (fAʹ)-1


Ex-15: Let A' = Ran(g). It follows from result of ex-14(b) that g = (fAʹ)-1. Also, its trivial to find that Ran(g) = B, hence A' = B. Thus g = (fB)-1


Ex-16:
(a) Let b be some real number s.t. f(x) = b.
Then 4x-x2=b
=> x2-4x+b=0

This quadratic equation has real roots if (-4)2-4(1)(b)0 16-4b0 b4.

Hence B = {xRx4}

(b) Scratchwork:
Using ex-14 result, basically we want to find a function g:R -> R s.t. fg=iR. Then A = Ran(g) and g = (fA)-1.

(fg)(x)=x
=> f(g(x)) = x

Let g(x) = y, then
f(y) = x
=> 4y-y2=x

one root of above quadratic equation is, y = g(x) = 2+4-x

Formal Proof:
Let g(x) = 2+4-x.

(fg)(x) = f(g(x)) = f(2+4-x) = 4(2+4-x)-(2+4-x)2 = 8+44-x-(4+4-x+44-x) = x

Hence fg=iR

It follows from ex-14 result that A = Ran(g) = {xRx2} and (fA)-1(x) = g(x) = 2+4-x


Ex-17: Scratchwork:
If we prove that gf=iP and fg=iS, then it follows from theorem 5.3.3 and 5.3.5 that f is one-to-one and onto and g = f-1

Formal Proof:
Proof for gf=iP:
Let R be an arbitrary element of P.
Then, (gf)(R) = g(f(R)) = g(R\iA) = (R\iA)iA = R. Since R is arbitrary, so gf=iP

Proof for fg=iS:
Let R be an arbitrary element of S.
Then, (fg)(R) = f(g(R)) = f(RiA) = (RiA)\iA

Since R is strict partial order on A, hence irreflexive and hence R and iA are disjoint. Then (RiA)\iA = R. Then (fg)(R)=R. Since R is arbitrary, so fg=iS

Then it follows from theorem 5.3.3 and 5.3.5 that f is one-to-one and onto and g = f-1


Ex-18:
(a) First, we prove that R is reflexive on F
Let f be an arbitrary function in F. Let h = iA. Then h-1fh = f. Hence fRf. Since f is arbitrary, so R is reflexive.

Second, we prove that R is symmetric on F
Let (f,g) be arbitrary ordered pair in FxF s.t. (f,g)R. Then we can choose a function hP s.t. f = h-1gh. Then hf=gh. Then g=hfh-1. Let j = h-1. Then g = j-1fj. Hence (g,h)R. Since (f,g) is arbitrary, so R is symmetric.

Third, we prove that R is transitive on F
Let (f,g) and (g,h) be arbitrary elements of R. So we can choose functions u,v in P s.t.
f = u-1gu and g = v-1hv.

Then f = u-1gu = u-1(v-1hv)u = (vu)-1h(vu). Let j = vu. Then f = j-1hj. Hence (f,h)R. Hence R is transitive.


Thus R is equivalence relation on F.

(b) Suppose fRg, then we can choose hP s.t. f = h-1gh.

Then ff
= (h-1gh)(h-1gh)
= h-1(gg)h

Hence (ff)R(gg)

(c) Suppose f has a fixed point and fRg. Then we can choose some aA s.t. f(a) = a. We can choose hP s.t. f = h-1gh.
So f(a) = (h-1gh)(a)

Let h(a) = b. Then
f(a) = (h-1g)(b)
=> a = h-1(g(b))

Thus (g(b),a)h-1
=> (a,g(b))h.

we assumed h(a) = b, that is (a,b)h

Since h is one-to-one, hence g(b) = b. So g has a fixed point.

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