1. Tardigrade
  2. Question
  3. Mathematics
  4. Let f be a function from a set X to a set Y. Consider the following statements: P: For each x ∈ X t,here exists uniqueX ∈ Ysuch that f (x) Q: For each y ∈ Y ,there exists x ∈ X such that f (x) = y . R: There exist x1,x2 ∈ X such that x1 ≠ x2 and f (x1) = f (x2). The negation of the statement f is one-to-one and onto is

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Q. Let $f$ be a function from a set $X$ to a set $Y$. Consider the following statements:
$P:$ For each x $\in$ X t,here exists uniqueX $\in$ Ysuch that $f$ (x)
$Q:$ For each y $\in$ Y ,there exists x $\in$ X such that $f$ (x) = y .
$R:$ There exist x$_1$,x$_2$ $\in$ X such that x$_1$ $\ne$ x$_2$ and $f$ (x$_1$) = $f$ (x$_2$).
The negation of the statement "$f$ is one-to-one and onto " is

KEAMKEAM 2013Mathematical Reasoning

Solution:

We know that,
(i) A function $f: X \to Y$ is said to be one-one, if distinct elements of $X$ have distinct images in $Y$.
$f$ is one-one when $f\left(x_{1}\right)=f\left(x_{2}\right) $
$\Rightarrow x_{1}=x_{2}$
(ii) A function $f: X \to Y$ is said to be onto, if every element in $Y$ has atleast one pre-image in $X$. Thus, if $f$ is onto, then for each $y \in X \exists$ atleast one element $x \in X$ such that $y=f(x)$.
$\therefore $ Negative of the statement " $f$ is one-one and onto" iś
" $f$ is not one-to-one and onto".
which hold the logical statement, " $R$ or not $Q$ "