Solve this following

Question:

Consider the statement : "For an integer $\mathrm{n}$, if $\mathrm{n}^{3}-1$ is even, then $\mathrm{n}$ is odd." The contrapositive statement of this statement is :

  1. For an integer $\mathrm{n}$, if $\mathrm{n}^{3}-1$ is not even, then $\mathrm{n}$ is not odd.

  2. For an integer $\mathrm{n}$, if $\mathrm{n}$ is even, then $\mathrm{n}^{3}-1$ is odd.

  3. For an integer $\mathrm{n}$, if $\mathrm{n}$ is odd, then $\mathrm{n}^{3}-1$ is even.

  4. For an integer $n$, if $n$ is even, then $n^{3}-1$ is even.

Correct Option: , 2

Solution:

Contrapositive of $(p \rightarrow q)$ is $\sim q \rightarrow \sim p$

For an integer $\mathrm{n}$, if $\mathrm{n}$ is even then $\left(\mathrm{n}^{3}-1\right)$ is odd

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