[Math] Prove there is an integer larger than a given real number

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A homework problem guides me to prove that there is a rational number $\frac{m}{n}$ between every two real numbers $x$ and $y$. The first step requires me to prove that there exists an integer $n$ such that $\frac{1}{y-x} < n$. That you can always find a natural number larger than any given real number seems trivial obvious to me, but I can't figure out how to prove it. I thought of trying to use the fact that there is no largest natural number, but that doesn't prove that there is always one larger than a given real number. The assignment says that I can use the fact that $\frac{1}{y-x}$ has a decimal expansion to prove this first step, but I don't see how that helps.

Can someone please help me figure out how to prove that there exists an integer $n$ such that $\frac{1}{y-x} < n$ for all real $x$ and $y$?

Best Answer

Since $y-x>0$, by Archimedean property, there exits $n \in \Bbb{N}$ such that $n(y-x)>1$

[For, suppose it is false. Then we have $n \leq \frac{1}{y-x}, \forall n.$ which means $\frac{1}{y-x}$ is an upper bound for $\Bbb{N}$, a contradiction! ]

Question : Why we need that ?

Answer:

Assuming the existence of such an $\frac{m}{n}$ with $n > 0$. So, we have $x < \frac{m}{n}<y$. That is, $nx < m < ny$. Thus we are claiming that the interval $(nx, ny)$ contains an integer. It is geometrically obvious that a sufficient condition for an interval $J = (nx, ny)$ to have an integer in it is that its length $ny-nx$ is greater than $1$.That is , $n(y-x)>1$. Archimedean property assures of such $n$'s.

Related Solutions

[Math] find the largest integer less than a number

This is a particularly simple variant of what is known as the coin problem. Here's a dynamic programming solution that runs in time linear in $t$. We use an array $m[0\dots t]$ of booleans, which I'll refer to as either T or F. The value of $m[i]$ will represent whether or not $i$ can be represented as the sum of zero or more copies of $a, b, c$. We will also keep a variable $w$, representing the current largest value not in $\text{PCOM}(a, b, c)$.

First, we initialize $m[0] = \text{T}$ (since we can obviously get a sum of zero by not using any of the values) and for $0 < i < a$ we set $m[i]=\text{F}$. We also initialize $w=a-1$.

Having done that, the algorithm is

for i = a to t
   m[i] = m[i - a] OR m[i - b] OR m[i - c]
   if m[i] = F
      w = i
return w

The only tweak is in the second line: if $i < b$ or $i < c$ we don't try to access the corresponding array element (since it would make no sense to do so here). There are some obvious improvements we can make in the algorithm, but I hope the idea is clear. If not, just ask.

[Math] Proof that there are infinitely many positive rational numbers smaller than any given positive rational number.

Your proof does not work. Indeed, subtracting $1$ from $\frac p q$ will give you a rational number, but it will be negative by assumption, so this doesn't help you (since it doesn't give you a contradiction).

A simpler approach: Explicitly state what the infinitely-many positive rationals less than $x$ are.

Hint: If $y$ is a positive rational, what can you say about $\frac{y}2$? About $\frac{y}4$? $\frac{y}8$? ...

Best Answer

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[Math] find the largest integer less than a number

[Math] Proof that there are infinitely many positive rational numbers smaller than any given positive rational number.

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