234k views
0 votes
Prove by induction the product of n odd integers is odd for n >1. Recall a number k is odd if there exists an integer m such that k= 2m+ 1.

1 Answer

2 votes

Step-by-step explanation: Induction:

Suppose that for n odd numbers is true that their product is odd.

Lets find for n+1 numbers:

The pruduct of n+1 odd numbers can be written as:

(2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1)*(2kₙ₊₁ + 1)

where the numbers "k" can be any integer number.

Because of the hipotesis, we know that:

(2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1) is an odd number, so we can write this as:

(2k + 1)

Then we have: (2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1)*(2kₙ₊₁ + 1) = (2k + 1)*(2kₙ₊₁ + 1)

now we have to solve this product:

(2k + 1)*(2kₙ₊₁ + 1) = 2k*2kₙ₊₁ + 2k + 2kₙ₊₁ + 1 = 2*(2k*kₙ₊₁ + k + kₙ₊₁) + 1

Because k and kₙ₊₁ are integers, the number (2k*kₙ₊₁ + k + kₙ₊₁) is also an integer, that we can write as L.

So we have that:

(2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1)*(2kₙ₊₁ + 1) = 2L + 1

So the product of n + 1 odd integers is also an odd number.

Now you can test that this is true for n = 2 (the same procedure that I did in the product of (2k + 1)*(2kₙ₊₁ + 1) ) and there you have your demonstration by induction.

User Kemen Paulos Plaza
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.