Question

Let $f$ be a function from a set $X$ to a set $Y$.
Consider the following statements
$P:$ For each $xϵX$, there exists unique $yϵY$ such that $f\left(x\right)=y$.
$Q:$ For each $yϵY$, these exists $xϵX$ such that $f\left(x\right)=y$.
$R:$ There exist $x_1,x_2ϵX$ such that $x_1\ne x_2$ and $f\left(x_1\right)=f\left(x_2\right)$
The negation of the statement $"f$ is one-to-one and onto" is

• A. $P$ or not $R$
• B. $R$ or not $P$
• C. $R$ or not $Q$
• D. $P$ and not $R$
• E. $R$ and not $Q$

Answer 1

The correct option is C $R$ or not $Q$

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 per-image in $X$.
Thus, if $f$ is onto, then for each $yϵX\exists$ atleast one element $xϵX$ such that $y=f\left(x\right)$.
Therefore, negative of the statement $"f$ is one-one and onto" is
$"f$ is not one-to-one and onto".
which hold the logical statement, $"R$ or not $Q"$.