For the function below which maps fourbit binary numbers to

For the function below, which maps four-bit binary numbers to three-bit binary numbers, indicate whether the function is onto, one-to-one, or both. If it is not onto, specify its range by listing the elements using set notation. If the function is not one-to-one, give an example showing why.

f:{0,1}4{0,1}3 where the output of f is obtained by taking the input string and dropping the first bit. For example, f(1011)=011.

Solution

f:{0,1}⁴{0,1}³ is no doubt an onto function since element in {01}³ can be mapped from {01}⁴ since the function only removes the 1st bit from the 4-bit binary number. Or if we consider a 3-bit binary number X₃X₂X₁ (where X₁ is the LSB and X₃ is the MSB and Xn {0,1}) can be mapped from both 1X₃X₂X₁ or 0X₃X₂X₁. We know that both 1X₃X₂X₁ and 0X₃X₂X₁ are elements of {01}⁴ So we can easily prove that f:{0,1}⁴{0,1}³ is no doubt an onto function.

But f:{0,1}⁴{0,1}³ is not one-to-one function which we shall prove by contradictory example:
we know from the statement that

f(1011) --> 011 and also f(0011) --> 011
so two distinct elements of {01}⁴ maps so a particular element in {01}³, and hence it is not one-to-one function