Consider the following statements:Statement-I | vedclass

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(D) Statement-$I$: $A$ function $f: A \rightarrow B$ is one-one (injective) if distinct elements in $A$ have distinct images in $B$. This is equivalen

Vedclass · Accepted Answer

Neither Statement-$I$ nor Statement-$II$ is true

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(D) Statement-$I$: $A$ function $f: A ightarrow B$ is one-one (injective) if distinct elements in $A$ have distinct images in $B$. This is equivalent to the contrapositive statement: $f(x) = f(y) \Rightarrow x = y$. The given statement $f(x) 
eq f(y) \Rightarrow x 
eq y$ is the contrapositive of the definition $x = y \Rightarrow f(x) = f(y)$,which is the definition of a function. However,the definition of one-one is $x 
eq y \Rightarrow f(x) 
eq f(y)$. The statement $f(x) 
eq f(y) \Rightarrow x 
eq y$ is logically equivalent to $x = y \Rightarrow f(x) = f(y)$,which is the definition of a function,not one-one. Thus,Statement-$I$ is false.Statement-$II$: $A$ relation $f: A ightarrow B$ is a function if every element in $A$ has a unique image in $B$. The condition $x 
eq y \Rightarrow f(x) 
eq f(y)$ defines an injective (one-one) function,not the definition of a function itself. $A$ function can map different inputs to the same output (many-one). Thus,Statement-$II$ is false.Therefore,neither statement is true.

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