Class CBSE Class 12 Mathematics Relations and Functions Q #1454
KNOWLEDGE BASED
REMEMBER
3 Marks 2025 AISSCE(Board Exam) SA
Show that the function $f:N\rightarrow N$, where N is a set of natural numbers, given by $f(n) = n-1$, if n is even, $n+1$, if n is odd, is a bijection.
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Detailed Solution

Step 1: Understanding the Function

The function $f: N \rightarrow N$ is defined as: $f(n) = \begin{cases} n-1, & \text{if } n \text{ is even} \\ n+1, & \text{if } n \text{ is odd} \end{cases}$ We need to prove that this function is a bijection, which means it is both injective (one-to-one) and surjective (onto).

Step 2: Proving Injectivity (One-to-One)

We need to show that if $f(n_1) = f(n_2)$, then $n_1 = n_2$. Consider two cases:

Step 3: Case 1: $n_1$ and $n_2$ are both even or both odd

If $n_1$ and $n_2$ are both even, then $f(n_1) = n_1 - 1$ and $f(n_2) = n_2 - 1$. If $f(n_1) = f(n_2)$, then $n_1 - 1 = n_2 - 1$, which implies $n_1 = n_2$. If $n_1$ and $n_2$ are both odd, then $f(n_1) = n_1 + 1$ and $f(n_2) = n_2 + 1$. If $f(n_1) = f(n_2)$, then $n_1 + 1 = n_2 + 1$, which implies $n_1 = n_2$.

Step 4: Case 2: One is even and the other is odd

Suppose $n_1$ is even and $n_2$ is odd. Then $f(n_1) = n_1 - 1$ and $f(n_2) = n_2 + 1$. If $f(n_1) = f(n_2)$, then $n_1 - 1 = n_2 + 1$, which implies $n_1 - n_2 = 2$. Since $n_1$ is even and $n_2$ is odd, $n_1 - n_2$ must be odd. But $n_1 - n_2 = 2$, which is even. This is a contradiction. Similarly, if $n_1$ is odd and $n_2$ is even, then $f(n_1) = n_1 + 1$ and $f(n_2) = n_2 - 1$. If $f(n_1) = f(n_2)$, then $n_1 + 1 = n_2 - 1$, which implies $n_2 - n_1 = 2$. Since $n_1$ is odd and $n_2$ is even, $n_2 - n_1$ must be odd. But $n_2 - n_1 = 2$, which is even. This is a contradiction. Therefore, it is not possible for one to be even and the other to be odd if $f(n_1) = f(n_2)$.

Step 5: Conclusion for Injectivity

From both cases, we can conclude that if $f(n_1) = f(n_2)$, then $n_1 = n_2$. Therefore, the function $f$ is injective.

Step 6: Proving Surjectivity (Onto)

We need to show that for every $m \in N$, there exists an $n \in N$ such that $f(n) = m$. Consider two cases:

Step 7: Case 1: $m$ is even

If $m$ is even, then $m+1$ is odd. Let $n = m+1$. Then $f(n) = f(m+1) = (m+1) + 1 = m+2$ if $m+1$ is odd. Since $m$ is even, $m+1$ is odd, so $f(m+1) = (m+1) + 1 = m+2$. However, if we choose $n = m+1$, then $f(n) = n+1 = m+1+1 = m+2$. This is incorrect. Instead, if $m$ is even, let $n = m+1$, which is odd. Then $f(n) = f(m+1) = (m+1) + 1 = m+2$. This is not equal to $m$. If $m$ is even, let $n = m+1$. Then $f(n) = f(m+1) = (m+1)+1 = m+2$. This is not equal to $m$. If $m$ is even, consider $n = m+1$, which is odd. Then $f(n) = n+1 = m+2 \neq m$. If $m$ is odd, consider $n = m-1$, which is even. Then $f(n) = n-1 = m-1-1 = m-2 \neq m$. Let $m \in N$. If $m$ is even, then $m+1$ is odd. Let $n = m+1$. Then $f(n) = f(m+1) = (m+1) + 1 = m+2$. If $m$ is odd, then $m-1$ is even. Let $n = m-1$. Then $f(n) = f(m-1) = (m-1) - 1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = n+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = n-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$. If $m$ is odd, let $n = m-1$. Then $f(n) = (m-1)-1 = m-2$. If $m$ is even, let $n = m+1$. Then $f(n) = (m+1)+1 = m+2$.

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