Kenneth Rosen Edition 7 Exercise 2.3 Question 27 (Page No. 153)

Question

1. Prove that a strictly decreasing function from $R$ to it-self is one-to-one.
2. Give an example of an decreasing function from $R$ to itself that is not one-to-one

Answer 1

For a decreasing function --if x1 < x2 then f(x1) ≥ f(x2) .Here two values of a domain may have same values in codomain. So a decreasing function may or may not be a one-one function..

For a strictly decreasing function--if x1 < x2 then f(x1) > f(x2). Here no two values of domain can have same values in codomain . One may be less than or greater than other but not equal. So a strictly decreasing function is always one -one.

F(x)=|x| is a decreasing function that is not one-one. For x=-1 and x =1 ,f(x)=1 . So it is decreasing function but it is not one-one(since two domain values have same value in codomain)

Comments

F(x)=|x| is a decreasing function that is not one-one. For x=-1 and x =1 ,f(x)=1 . So it is decreasing function but it is not one-one(since two domain values have same value in codomain)

Incorrect!$f(x) = | x| $ is not a decreasing function with domain $A$ and co-domain $B$ being subsets of $R$ 

proof

To be decreasing function it has to satisfy $$\forall x\in A ~~\forall y \in A ( x < y \to f(x) \geq f(y))$$

But for $x=1$ and $y=2$, $f(x)=|1|=1$ and $f(y) = |2|=2$

Here $x<y$ but $f(x)\ngeq f(y)$

$\therefore$ By proof by counter example, f(x)=|x| from $R \to R$ is not a decreasing function 

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The graph of $f(x)=|x|$

$f(x) = |x| $ is strictly decreasing(thus decreasing) if is defined from $\mathbb{R^-\to R} $ 

$f(x) = |x| $ is strictly increasing(thus increasing) if is defined from $\mathbb{R^+\to R} $ 

Answer 2

Let $f:A \to B $ is a strictly decreasing function and $A \subseteq R$, $B \subseteq R$ Then

we have to prove $f$ is one to one function

Comments

The other way i.e proving its contrapositive