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Prove there exist infinitely many real numbers $x$ such that $2x-x^2 \gt \frac{999999}{1000000}$.

I'm not really sure of the thought process behind this, I know that $(0,1)$ is uncountable but I dont know how to apply that property to this situation.

Any help would be appreciated,


Mrs. Clinton

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Are you sure that is the correct stactement? – ՃՃՃ Jul 16 '13 at 2:14
There are uncountably many real numbers between any two distinct real numbers. Here, there are two solutions to $2x-x^2=999999/1000000$ and all the reals between them solve your inequality. Note that I believe you have made a typo of either too many 9s or too few 0's. – alex.jordan Jul 16 '13 at 2:18
Oh, I get it, Thanks alot! my mistake with the typo – Hillary Clinton Jul 16 '13 at 2:20
up vote 1 down vote accepted

Firstly we are guessing you have one too many $9$ on the top. We proceed with $\frac{999999}{1000000}$

Define $f(x)=2x-x^2$ then $f$ has its maximum at $1$ and its maximum is $1$. Now as $f$ is continuous we can find a neighborhood (which means an open interval) around $x=1$ so that if $y$ is in the neighborhood, then $f(1)-f(y)<\frac{1}{1000000}$.

Thus $f(y)>f(1)-\frac{1}{1000000}=\frac{999999}{1000000}$.

Hence every point in our open interval satisfies the inequality and as open intervals have infinite cardinality we are done.

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If $f(x)$ is a continuous function, $a$ is a real number, and there is some real $b$ with $f(b)>a$, then there are infinitely many real numbers $c$ such that $f(c)>a$.

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HINT: If $f(x)=2x-x^2=x(2-x)$, you’re trying to show that the inequality $f(x)>0.999999$ has infinitely many solutions. One way to do this is to show that there is an $\epsilon>0$ such that $f(1-\epsilon)=f(1+\epsilon)=0.999999$, and then show that $f(x)>0.999999$ whenever $1-\epsilon<x<1+\epsilon$: the interval $(1-\epsilon,1+\epsilon)$ is not just infinite, but uncountable.

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If a problem seems hard to me, I will try to simplify it. In this case, the process is:
1. $$2x - x^2 > \frac{999999}{1000000} \Leftrightarrow 1 - (x-1)^2 > \frac{999999}{1000000} \Leftrightarrow \frac{1}{1000000} > (x-1)^2$$
2. Let $y = x - 1$, the problem is equivalent to prove that there exist infinitely real numbers $y$ such that: $$\frac{1}{1000000} > y^2.$$
3. With $y > 0$, we have: $$\frac{1}{1000000} > y^2 \Leftrightarrow \frac{1}{1000} > y.$$ 4. Here I solve the problem! As you know, $(0,1)$ is uncountable. So the same is $(0,\frac{1}{1000})$, which means that there is infinitely many real numbers $y$ such that: $$ 0 < y < \frac{1}{1000}.$$

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Typo: That should probably be "infinitely many real numbers $y$" – Eric Stucky Jul 16 '13 at 2:53
Fixed, thank you. – Du Phan Jul 16 '13 at 3:30

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