Question

6. Prove that if 3n^3+13 is odd then n is even for all integers n, using the proof by contradiction.

Answer 1

Answer:

If 3n^3+13 is odd then n is even for all integers n, using the proof by contradiction.

Step-by-step explanation:

By contradiction method, we need to prove that if $3n^3+13$ is odd then n is even for all integers n.

Proof by contradiction:

Let as assume if $3n^3+13$ is odd then n is odd for all integers n.

$n=2k+1$

Substitute the value of n in the given expression.

$3(2k+1)^3+13$

Cube of any odd number is an odd number.

$(2k+1)^3=Odd$

Product of two odd numbers is an odd number.

$3(2k+1)^3=Odd$

$3(2k+1)^3=Odd$ is an odd number and 13 is an odd number. We know that addition of two odd numbers is an even number.

$3(2k+1)^3+13=Even$

Which is the contradiction of our assumption.

If 3n^3+13 is odd then n is even for all integers n, using the proof by contradiction.

Hence proved.