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Home  >>  CBSE XII  >>  Math  >>  Relations and Functions
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Let \(A\) and \(B\)be sets. Show that\(f : A × B \to\; B×A\)such that \( f(a,b)\,=\,(b,a)\) is bijective function

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  • A function $f:A \to B$ where for every $x_1,x_2 \in X,f(x_1)=f(x_2)⇒x_1=x_2$ is called a one-one or injective function.
  • A function $f:X \to Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x)=y.$
Given: \(f : A × B \to\; B×A\) such that \( f(a,b)\,=\,(b,a)\)
 
Step1 - Injective or one-one
 
Consider two elements $x_1= (a,b)$ and and $x_2= (c,d)$
 
$f(x_1) =f(x_2) => f(a,b) = f(c,d)$
 
This implies that $(b,a) =(d,c)$
 
This is possible only if a=c and b=d
 
$=> x_1=x_2$
 
hence the function f defined by \(f : A × B \to\; B×A\) such that \( f(a,b)\,=\,(b,a)\) is one-one
 
Step2- Surjective or onto
 
Let the fuction f be defined by \(f : A × B \to\; B×A\)such that \( f(a,b)\,=\,(b,a)\)
 
the function is onto if for every element $y= (b,a) \in B \times A$ there exist an element $x=(a,b) \in A \times B$ such that $f(x) = y$
 
This is always possible for all $a\in A , and , b\in B $
 
Hence the function f is onto
 
Solution- The function f defined by \(f : A × B \to\; B×A\)such that \( f(a,b)\,=\,(b,a)\) is both one one and onto

 

 

answered Mar 18, 2013 by thagee.vedartham
edited Mar 19, 2013 by meena.p
 

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