Tagged Questions
0
votes
2answers
88 views
Flawed proof that all positive integers are equal
Suppose that we are trying to prove that for every positive integer n, if x and y are positive integers with max(x, y) = n, then x = y. For the base case, we suppose n = 1. If max(x, y) = 1 and x and ...
15
votes
8answers
2k views
There are no bearded men in the world - What goes wrong in this proof?
Several years ago in a textbook I read this example as a faulty use of proof by induction. I never really realized why it fails. Here it goes:
Theorem. There are no bearded men in the world.
...
42
votes
12answers
3k views
All natural numbers are equal.
I saw the following "theorem" and its "proof".
I can't explain well why the argument is wrong. Could you give me clear explanation so that kids can understand.
Theorem: All natural numbers are ...
2
votes
2answers
107 views
Find the demonstration error for the statement “All positive integers are equal”
All positive integers are equal, that is, for each $n \in \mathbb{N}$ the assertion $P(N): 1 = \cdots = n$ is true.
(i) $P(1)$ is true because $1 = 1$
(ii) Suppose that $P(n)$ is true, then $1 = ...
3
votes
1answer
89 views
Find the fallacy in the following treatment
Claim:
any two positive integers are equal
Proof:
Let $A(n)$ be statement:
if $a$ and $b$ are any two positive integers such that $\max(a,b)=n$ then $a=b$
Suppose $A(r)$ is true. Let ...
4
votes
1answer
202 views
Explain what’s bogus about the proof.
I couldn't find what is wrong with this strong induction proof, any one knows ?
Question:
A sequence of numbers is weakly decreasing when each number in the sequence is
$\geq$ the numbers after it. ...
3
votes
3answers
516 views
Fake induction proof
Using the induction method:
$(\forall P)[[P(0) \land ( \forall k \in \mathbb{N}) (P(k) \Rightarrow P(k+1))] \Rightarrow ( \forall n \in \mathbb{N} ) [ P(n) ]]$
Why this proof is wrong?
$P(x)\equiv ...
38
votes
2answers
2k views
Proof of 1=0 by mathematical induction?
I got stuck with a problem that pop up in my mind while learning limits. I am still a high school student.
$\lim\limits_{n\to\infty}(\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{m})=0$
...
