As, it is mentioned here
f : R → R be defined as f(x) = 10x + 7
To, prove the function one-one
Let's take f(x) = f(y)
10x + 7 = 10y + 7
x = y
Hence f is one-one.
To, prove the function onto
y ∈ R, y = 10x+7
x = \frac{y-7}{10} ∈ R
So, it means for y ∈ R, there exists x = \frac{y-7}{10}
f(x) = f(\frac{y-7}{10}) = 10(\frac{y-7}{10})+7 = y - 7 + 7 = y
Hence f is onto.
As, f is one-one and onto. This f is invertible function.
Let's say g : R → R be defined as g(y) = \frac{y-7}{10}
g o f = g(f(x)) = g(10x+7) = \frac{(10x+7)-7}{10} = \frac{10x}{10} = x
f o g = f(g(x)) = f(\frac{x-7}{10}) = 10(\frac{x-7}{10})+7 = x - 7 + 7 = x
Hence, g : R → R such that g o f = f o g = 1R.
g : R → R is defined as g(y) = \frac{y-7}{10}
The function f is defined as
f(n)= \begin{cases} n-1, \hspace{0.2cm}n \hspace{0.2cm}is\hspace{0.2cm} odd\\ n+1,\hspace{0.2cm}n \hspace{0.2cm}is\hspace{0.2cm} even \end{cases}
As, we know f is invertible, if and only if f is one-one and onto.
ONE-ONE
For the pair of number, we will deal with three cases:
Case 1: When both numbers p and q are odd numbers.
f(p) = p-1
f(q) = q-1
f(p) = f(q)
p-1 = q-1
p - q = 0
Case 2: When both numbers p and q are even numbers.
f(p) = p+1
f(q) = q+1
f(p) = f(q)
p+1 = q+1
p - q = 0
Case 3: When p is odd and q is even
f(p) = p-1
f(q) = q+1
f(p) = f(q)
p-1 = q+1
p - q = 2
Subtracting an odd number and even always gives a odd number, not even. Hence, the case 3 result is impossible.
So, the function f is one-one, for case 1 and case 2 only.
ONTO
Case 1: When p is odd number
f(p) = p-1
y = p-1
p = y+1
Hence, when p is odd y is even.
Case 2: When p is even number
f(p) = p+1
y = p+1
p = y-1
Hence, when p is even y is odd.
So, it means for y ∈ W, there exists p = y+1 and y-1 for odd and even value of p respectively.
Hence f is onto.
As, f is one-one and onto. This f is an invertible function.
Let's say g : W → W be defined as g(y)= \begin{cases} y-1, \hspace{0.2cm}y \hspace{0.2cm}is\hspace{0.2cm} odd\\ y+1,\hspace{0.2cm}y \hspace{0.2cm}is\hspace{0.2cm} even \end{cases}
f = g
Hence, The inverse of f is f itself
As, it is mentioned here
f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = \frac{x}{1+|x|} , x ∈ R
As, we know f is invertible, if and only if f is one-one and onto.
ONE-ONE
For the pair of number, we will deal with three cases:
Case 1: When both numbers p and p are positive numbers.
The function f is defined as
Case 1: When both numbers p and q are positive numbers.
f(p) = \frac{p}{1+|p|}
f(q) = \frac{q}{1+|q|}
f(p) = f(q)
\frac{p}{1+|p|} = \frac{q}{1+|q|}
\frac{p}{1+p} = \frac{q}{1+q}
p(1+q) = q(1+p)
p = q
Case 2: When number p and q are negative numbers.
f(p) = \frac{p}{1+|p|}
f(q) = \frac{q}{1+|q|}
f(p) = f(q)
\frac{p}{1+|p|} = \frac{q}{1+|q|}
\frac{p}{1-p} = \frac{q}{1-q}
p(1-q) = q(1-p)
p = q
Case 3: When p is positive and q is negative
f(p) = \frac{p}{1+|p|}
f(q) = \frac{q}{1+|q|}
f(p) = f(q)
\frac{p}{1+|p|} = \frac{q}{1+|q|}
\frac{p}{1+p} = \frac{q}{1-q}
p(1-q) = q(1+p)
p + q = 2pq
Here, RHS will be negative and LHS will be positive. Hence, the case 3 result is impossible.
So, the function f is one-one, for case 1 and case 2.
ONTO
Case 1: When p>0.
f(p) = \frac{p}{1+|p|}
y = \frac{p}{1+p}
p = \frac{y}{1-y} (y≠1)
Case 2: When p <0
f(p) = \frac{p}{1+|p|}
y = \frac{p}{1-p}
p = \frac{y}{1+y} (y≠-1)
Hence, p is defined for all the values of y, p∈ R
Hence f is onto.
As, f is one-one and onto. This f is an invertible function.
As, it is mentioned here
f : R → R defined by f(x) = x3, x ∈ R
To prove f is injective (or one-one).
ONE-ONE
The function f is defined as
f(x) = x3
f(y) = y3
f(x) = f(y)
x3 = y3
x = y
The function f is one-one, so f is injective.
Two functions, f : N → Z and g : Z → Z
Taking f(x) = x and g(x) = |x|
Let's check, whether g is injective or not
g(5) = |5| = 5
g(-5) = |-5| = 5
As, we can see here that
Taking two integers, 5 and -5
g(5) = g(-5)
but, 5 ≠ -5
So, g is not an injective function.
Now, g o f: N → Z is defined as
g o f = g(f(x)) = g(x) = |x|
Now, as x,y∈ N
g(x) = |x|
g(y) = |y|
g(x) = g(y)
|x| = |y|
x = y (both x and y are positive)
Hence, g o f is an injective.
Two functions, f : N → N and g : N → N
Taking f(x) = x+1 and g(x)= \begin{cases} x-1, \hspace{0.2cm}x>1\\ 1,\hspace{0.2cm}x=1 \end{cases}
As, f(x) = x+1
y = x+1
x = y-1
But, when y=1, x = 0. Which doesn't satiny this relation f : N → N.
Hence. f is not an onto function.
Now, g o f: N → N is defined as
g o f = g(f(x)) = g(x+1)
When x+1=1, we have
g(x+1) = 1 (1∈ N)
And, when x+1>1, we have
g(x+1) = (x+1)-1 = x
y = x, which also satisfies x,y∈ N
Hence, g o f is onto.
Given, A and B are the subsets of P(x), A⊂ B
To check the equivalence relation on P(X), we have to check
As, we know that every set is the subset of itself.
Hence, A⊂ A and B⊂ B
ARA and BRB is reflexive for all A,B∈ P(X)
As, it is given that A⊂ B. But it doesn't make sure that B⊂ A.
To be symmetric it has to be A = B
ARB is not symmetric.
When A⊂ B and B⊂ C
Then of course, A⊂ C
Hence, R is transitive.
So, as R is not symmetric.
R is not an equivalence relation on P(X).
Given, P(X) × P(X) → P(X) is defined as A*B = A∩B ∀ A, B ∈ P(X)
This implies, A⊂ X and B ⊂ X
So, A∩X = A and B∩X = B ∀ A, B ∈ P(X)
⇒ A*X = A and B*X = B
Hence, X is the identity element for intersection of binary operator.