Help - my grandson came up with this question:
Prove that every natural number is either even or odd.
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
- his teacher suggested proof by induction
i gave a solution to this problem a few semesters ago in some math class, but my professor didnt like it <.< im going to post it to see what others think. one sec let me see if i can find it.
i saw what was wrong with my answer, never mind. I would just say that it is impossible to solve the equation:\[2n=2m+1\]with natural numbers n and m, so its impossible to have a natural number that is both even and odd. Not too sure how you would do that my induction, since a number being even or odd doesnt really depend on previous numbers being even or odd.
Not the answer you are looking for? Search for more explanations.
let n one number from set of natural numbers N ,so than
Alrighty, lets give induction a go.
First, we need to define even and odd. The standard definitions are as follows:
An integer n is even if it can be expressed in the form n = 2k, where k is some integer
An integer n is odd if it can be expressed in the form n = 2k + 1, where k is some integer
We are now equipped to prove the statement via induction. We assume that the natural number n is either even or odd. We seek to show that this implies that the natural number (n+1) is also either even or odd.
Case 1: n is an even number
If n is an even number, then it can be expressed as n = 2k, with k some integer. Thus, n+1 = 2k+1, and by the definition of an odd number, n+1 is odd.
Case 2: n is an odd number
If n is an odd number, then it can be expressed as n = 2k + 1, with k some integer. Thus, n+1 = 2k+1 +1 = 2k+2 = 2(k+1). Since k is an integer, k+1 is an integer, and thus by the definition of an even number, n+1 is even.
In either of these two cases, n+1 is either even or odd.
Finally, we show that n= 1 is either even or odd. Since k=0 is an integer, note that 1 can be expressed as n = 2k+1 -> 1 = 2(0) + 1. Thus n is odd, and more generally, n=1 is either even or odd.
By induction, we may say that any natural number n is either even or odd.