Show that the relation R on ℝ defined as R={(a , b):a ≤ b}, is reflexive, and transitive but not symmetric.

Or

Prove that the function f:ℕ→ℕ, defined by f(x)=x²+x+1 is one-one but not onto. Find inverse of f:ℕ→S, where S is range of f.

Given that R={(a ,b ):a ≤ b}

Since a=a ∀a ∈ ℝ,

therefore a ≤ a always.

Hence (a, a) always belongs to R ∀a ∈ ℝ. Therefore, R is reflexive.

If a ≤b

then b≥ a ⇏ b ≤a .

Example:

(2,4) ∈ ℝ as 2 ≤ 4.

But (4,2) ∉ ℝ as 4  is greater than 2.


Hence if (a, b) belongs to R, then (b, a) does not always belong to R.

Hence R is not symmetric.

If a ≤b  -----(1) and

b ≤c  -----(2)

Add (1) and (2) to get,

a + b ≤ b +c

Hence a ≤c.

Hence if (a, b)∈R and (b, c)∈R, then (a, c)∈R ∀a, b, c ∈ℝ.

Hence, R is transitive.

OR

f(x)=x²+x+1, f:ℕ→ℕ

A function is one-one if f(a)=f(b)

⇒a=b

f(a)=f(b)

⇒ a²+a+1=b²+b+1

⇒ a²- b²+a -b=0 or (a-b)(a+b+1)=0

Hence, a=b or a + b=-1

Since a, b ∈ℕ, therefore a + b=-1 is not possible.

Hence a=b.

Since f(a)=f(b)

⇒a=b

Therefore, f(x) is one-one.

Let y= x²+x+1

differentiating with respect to x, we get,

y’=2x+1>0 ∀x ∈ℕ, hence f is an increasing function.

The range of y={3,7,13,21…} which is not equal to ℕ.

Since the range is not equal to codomain, therefore, f is not onto.

Let S be the range of f.

Then f(x)=x²+x+1,

f:ℕ→S

y= x²+x+1

⇒ x²+x+1-y=0

Using the quadratic formula, we get,

There are two possibilities  for f^-1(x).




As f(1) = 3

So, f^-1(3) = 1

Hence, f^-1:S→ℕ