Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Here's what I have done:

I think it's false, so I set out to prove the negation. Which is: $$\forall y \in \mathbb{N}, \exists x \in \mathbb{N}; 2x > y + 1$$ I then let $y$ be an arbitrary natural number. After this, I do not know how to proceed.

share|improve this question
1  
$\Bbb{N}$ is unbounded..... –  user38268 Sep 20 '12 at 6:07

3 Answers 3

up vote 3 down vote accepted

Suppose that I tell you that I’m thinking of some $y\in\Bbb N$, but I don’t tell you what it is. Could you give me a recipe for calculating an $x$ such that $x>y+1$? Sure: just set $x=y+2$. You don’t need an $x>y$: you just need an $x$ such that $2x>y+1$, and that ought to be even easier to specify by a recipe. I’ve included a couple of possible recipes below, spoiler-protected; mouse-over to see them.

The smallest $x$ that works is the smallest natural number $n>\frac{y+1}2$; that’s $\frac{y}2+1$ if $y$ is even and $\frac{y+1}2+1$ if $y$ is odd. But working that out is a waste of energy, at least as far as the immediate problem is concerned, because if $x>y+1$, then certainly $2x\ge x>y+1$. Thus, you might as well use the same recipe that I gave for the simpler problem: set $x=y+2$. That certainly ensures that $2x=2y+4=2(y+1)+2>y+1$!

share|improve this answer
    
Ok, so now I have 2(y+1) = y + 1. So that confirms the negation, right? Because I have shown a value of x that makes the statement 2x > y + 1 true no matter what value of Y I plug in. What does this mean then? If the negation is true, does that mean that the original statement is false? So I have disproven the original statement? –  William Sep 20 '12 at 6:15
    
@Willy: You have indeed. –  Brian M. Scott Sep 20 '12 at 6:16
    
If $x \in \mathbb{N} $ then $2x \in \mathbb{N} $, for any large $2x$ we can always find $y$ that is greater than $2x$ (since $\mathbb{N}$ is unbounded), am I correct? –  Vikram Sep 20 '12 at 6:55
    
@Vikram: Sure: $2x+1$ works. –  Brian M. Scott Sep 20 '12 at 6:58

For any $y\in \mathbb{N}$, $2(y+1)\gt y+1$.

share|improve this answer

Hint $\rm\ 2\,\Bbb N\, = $ even naturals, so if they are bounded above then there are only finitely many evens, hence only finitely many odds $\rm\,1 + 2\,\Bbb N,\:$ hence only finitely many naturals, contradiction.

Alternatively, finitely many odds implies finitely many primes, contra Euclid's classical proof.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.