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If $f:\{a,b\}^{\ast}\rightarrow \{a,b\}^{\ast }$ be given by $f(n)=ax$ for every value of $n\in \{a,b\}$, then $f$ is

  1. one to one not onto
  2. one to one and onto
  3. not one to one and not onto
  4. not one to one and onto

1 Answer

2 2 votes
Considering x belongs to {a,b}*. Domain is {a,b,aa,ab,ba,bb,...} and range is also {a,b,aa,ab,ba,bb,...} . Suppose x1 and x2 are different elements in domain which map to the same element in range then f(x1)=f(x2)

implies a*x1=a*x2

implies x1=x2. This means x1 and x2 are same i.e. no two elements in domain will map to the same element in range. So f is one-one.

Now, a function is said to be onto if every element in range set has a pre image in domain set(no element in range set should be left out)

Here,f is not onto because for element b in range, there is no pre-image in domain.

So answer should be A
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