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A function of from the set of natural numbers to integers defined by \( \Large f \left(n\right) = \frac{\frac{n-1}{2},\ when\ n\ is\ odd}{-\frac{n}{2},\ when\ n\ is\ even} \) is:
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A) one-one but not onto |
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B) onto but not one-one |
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C) one-one and onto both |
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D) neither one-one nor onto |
Correct Answer:
| C) one-one and onto both |
Description for Correct answer:
We have \( \Large f \left(n\right) = \bigg\{ \frac{n-1}{2},\ when \ n \ is odd\ , {-\frac{n}{2},\ when\ n\ is\ even} \)
and \( \Large f : N \rightarrow I \) where N is the set of natulal numbers and I is the set of integers.
Let \( \Large x,\ y\ \epsilon\ N \) and both are even.
then \( \Large f \left(x\right) = f \left(y\right) \)
\( \Large -\frac{x}{2} = -\frac{y}{2} => x = y \)
Again \( \Large x,\ y\ \epsilon\ N \)
then \( \Large f \left(x\right)=f \left(y\right)\frac{x-1}{2}=\frac{y-1}{2} =>x=y \)
i.e. function is one-one.
Since, each negative integer is an image of even natural; and each positive integer is an image of odd natural number so function is onto.
Hence, function is one-one onto.
We have \( \Large f \left(n\right) = \bigg\{ \frac{n-1}{2},\ when \ n \ is odd\ , {-\frac{n}{2},\ when\ n\ is\ even} \)
and \( \Large f : N \rightarrow I \) where N is the set of natulal numbers and I is the set of integers.
Let \( \Large x,\ y\ \epsilon\ N \) and both are even.
then \( \Large f \left(x\right) = f \left(y\right) \)
\( \Large -\frac{x}{2} = -\frac{y}{2} => x = y \)
Again \( \Large x,\ y\ \epsilon\ N \)
then \( \Large f \left(x\right)=f \left(y\right)\frac{x-1}{2}=\frac{y-1}{2} =>x=y \)
i.e. function is one-one.
Since, each negative integer is an image of even natural; and each positive integer is an image of odd natural number so function is onto.
Hence, function is one-one onto.
Part of solved Set theory questions and answers : >> Elementary Mathematics >> Set theory
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