Solved problems on functions and sequences

Question 1
We check whether the domains and co-domains of f and g asbelow;
f : { 0 ,1 } → { 0 ,1 } ; f ( x ) =x2
When x=0 ,
f ( 0 )=02 =0
When x=1 ,
f ( 1 ) =12=1
Similarly,
g : { 0 , 1 } → { 0 , 1 } ; g ( x ) =x
When x=0 ,
g ( 0 )=0
When x=1 ,
g ( 1 ) =1

Therefore, they are equal functions in the given domain.
Question 2
To prove that a function if bijective, we prove that it is both injective and surjective (Frederic P.
Miller, 2010).
To prove that it is injective;
f ( x )=f ( y )
4 x+1=4 y+1
4 x=4 y
x= y

Thus, the function is injective.
To prove that it is surjective;
Taking x ∈ [ 0 , 1 ]∧ y ∈ [0 , 1.5]
From the function y=4 x +1 ,
x= y−1
4
Using the range given for y and comparing with the range of values of x, we find that the
function is surjective.
Since the function is both injective and surjective, it is therefore bijective.
Question 3
Similar to question 2, we start by checking if the function is injective (Kim, 2010).
x , y ∈[−1 ,1]
f ( x )=f ( y )
x2= y2
x= √ y2=± y
The function is not injective. Thus, the function is not bijective.
Question 4
First, we need to understand that ⌊ y ⌋ mean the closest integer less than y (Shoup, 2009).
Let’s take for instance x = 2.1 and y = 0.9;
⌊ x ⌋=2∧⌊ y ⌋=0
It means that
⌊ x ⌋ + ⌊ y ⌋ =2+0=2
And
⌊ x+ y ⌋=⌊ 2.1+0.9 ⌋=3
Therefore, ⌊ x+ y ⌋ ≠ ⌊ x ⌋ + ⌊ [ y ] ⌋