4
$\begingroup$

I am trying to prove that if $x>y$, then there exists some $\epsilon>0$ such that $x-\epsilon>y+\epsilon$. So far I have:

Suppose, for a contradiction, that no such $\epsilon$ exists. That is, $\forall \epsilon>0$, suppose that $x-\epsilon \leq y+\epsilon$.

Beyond this, I am not at all sure how to proceed. Must one make reference to the fact that limits preserve weak inequalities?

Thank you.

$\endgroup$
3
  • $\begingroup$ Hint: Rearrange the inequality $x-\epsilon>y+\epsilon$ to give a bound for $\epsilon$ $\endgroup$ Jul 28, 2021 at 15:13
  • $\begingroup$ Note that if $x>y$, then $x-y$ is a positive real. $\endgroup$
    – 311411
    Jul 28, 2021 at 15:14
  • $\begingroup$ If you want to continue with your approach, take the limit for $\epsilon\to0$, which gives $x\leq y$, a contradiction. $\endgroup$
    – anankElpis
    Jul 28, 2021 at 15:14

4 Answers 4

3
$\begingroup$

You could use proof by contradiction, but a direct proof is preferable. Draw a number line and mark $x$ and $y$ at points such that $x>y$. From this, it should be geometrically obvious that $$ x-\frac{x-y}{3}>y+\frac{x-y}{3} \, , $$ and so you can take $\varepsilon=\frac{x-y}{3}$. Of course, something being geometrically obvious is not a proof, but this steers us in the right direction. It is then simple to come up with an algebraic proof: \begin{align} x>y &\implies 2x+y>2y+x \\[5pt] &\implies3x-(x-y)>3y+(x-y)\\[5pt] &\implies x-\frac{x-y}{3}>y+\frac{x-y}{3} \, . \end{align}

$\endgroup$
1
$\begingroup$

Try to develop some intuition for this problem:

Picture (or draw) a number line with $x$ to the right of $y$. You want to add a little bit to $y$ (to get $y+\epsilon$) and subtract the same little bit from $x$ (to get $x-\epsilon$) and have $y+\epsilon$ still be to the left of $x-\epsilon$.

With this picture, we see that we can take $\epsilon$ to be any positive value less than $\frac{x-y}{2}$. To be particular, we might, for instance, take $\epsilon = \frac{x-y}{3}$.

Then try to work out the algebra to show that this actually works (so that you have a complete proof).

$\endgroup$
1
$\begingroup$

If $x > y$ then by subtracting $y$ from both sides we have $x - y$ > 0. Then we divide both sides by two to get $\left(x - y\right) / 2 > 0$.

Choose any $\epsilon$ such that $0 < \epsilon < \left(x - y\right) / 2$. Then multiply through by 2 to get $0 < 2 \epsilon < x - y$. Then add through by $y - \epsilon$ to get $y - \epsilon < y + \epsilon < x - \epsilon$; QED.

$\endgroup$
0
$\begingroup$

-Here's how I would do it, a direct proof, not indirect- since x> y, x- y> 0. Let delta= x- y and epsilon any number less than delta/2. Then y+ epsilon< y+ delta/2= y+ (x- y)/2= y+ x/2- y/2= x/2+ y/2 while x- epsilon> x- delta/2= x- (x- y)/2= x- x/2+ y/2= x/2+ y/2.

So we have y+ epsilon< x/2+ y/2< x- epsilon.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .