SOLUTION: Decide whether or not the following proposition is true. If it is false, demonstrate this by presenting a counterexample. If it is true, decide whether or not the proof is correct.

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Question 574686: Decide whether or not the following proposition is true. If it is false, demonstrate this by presenting a counterexample. If it is true, decide whether or not the proof is correct. If the proof is not correct, rewrite the proof so that it is correct. If the proof is correct, decide whether or not it is well-written and revise the proof, as needed, so that it is well-written.
Proposition. For each natural number n with n > 2, it is true that 2n > 1 + n.
Proof. We let k be a natural number and assume that 2k > 1 + k. Multiplying both sides of this inequality by 2, we see that 2k+1 > 2 + 2k. However, 2 + 2k > 2 + k, and hence,
2k+1 > 1 + (k + 1).
By mathematical induction, we conclude that 2n > 1 + n.

Found 2 solutions by solver91311, richard1234:
Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Proposition correct, but proof is crap -- you only have half of it, and the half you have demonstrates insufficient rigor. A math induction proof requires demonstration of truth of the proposition for a base number.

Since is a condition of the proposition, demonstrate the truth of the proposition for the smallest possible , namely 3.

So it is true for

Assume that the proposition is true for some arbitrary natural number .

Then demonstrate that the proposition is true for

and

And given , it is clear that

Hence, if true for some natural number it is true for . It is true for 3, so it must be true for 4. Then true for 4 means true for 5...and so on forever.

John

My calculator said it, I believe it, that settles it

Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website!
The proof is a bit weak, because there is no base case, and I don't really see how they obtained 2k + 1 > 2 + 2k (multiplying by 2, you get 4k > 2+2k. You can say 4k = 2k+2k > 2k+(1+k) > 2k+1, but you can't claim that 2k+1 > 2+2k [which *cannot* be true]).

An induction proof would assert that the base case, n = 3, is true (which it is). If the inequality holds for some k >= 3, then we can show it holds for k+1.

However, we already know that 2k - (k+1) > 0 because 2k > k+1 (this is assumed to be true for our induction) so the above statement must be true. ∎

Of course, the simplest way to prove it is to say that

if and only if (subtracting both sides by n). However this is definitely true since n > 2, so we're done. ∎