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I know I am supposed to prove this using contradiction, but I am not sure how to proceed.

Here's what I have thus far:

Assume that there exists a smallest rational number $x$ such that $x > 1$.

$X = \dfrac{a}{b}$ where $a$ and $b$ are integers.

user729424
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nocheinmalbitte
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    Suppose $x>1$. Consider the average between $x$ and $1$. What is the average explicitly? Is it rational? Is it greater than $1$? Is it less than $x$? – JMoravitz Apr 22 '20 at 15:22
  • Can you prove there's no smallest rational number greater than zero? – Angina Seng Apr 22 '20 at 15:22

2 Answers2

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If $1<\frac{a}{b}$ and $b$ is a positive integer then $0 < b < a$ and $$1 < \frac{a+1}{b+1} < \frac{a}{b}$$ and $$1 = \frac{b+b}{2b} < \frac{a+b}{2b} < \frac{a+a}{2b}=\frac{a}{b}$$

so for any rational greater than $1$ there are smaller rationals greater than $1$ and thus no smallest rational greater than $1$

Henry
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If $1$ is a rationaly number and $\frac ab;\frac ab > 1$ is the smallest rational number that is larger $1$, then that would mean there are not any rational numbers between $1$ and $\frac ab$.

Find one.

$1 < \frac ab$

$1 = \frac bb < \frac ab$ so

$b < a$. $b$ and $a$ are unequal integers so

$b < b+\frac 12 < a$

$2b < 2b + 1 < 2a$

$\frac {2b}{2b} < \frac {2b+1}{2b} < \frac {2a}{2b}$

$1 < \frac {2b+1}{2b} < \frac ab$.

And that's it. $\frac ab$ is not the smallest rational that is greater than $1$.

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... or ....

What about the average of $1$ and $q=\frac ab$. That is $1 = \frac {1+1}2 < \frac {1+q}2 < \frac {q+q}2 = q$.

Is the average of $1$ and $q$ rational? It can't be if $q$ is the smallest rational larger than $1$.

But if $q= \frac ab; a,b$ both integers then the average of $1$ and $q$ is $\frac {1+q}2=\frac {1 +\frac ab}2 = \frac bb\frac {1+\frac ab}2=\frac {b+a}{2b}$. But then $a+b$ is an integer. And $2b$ is an integer. So $\frac {1+q}2$ is rational.

.....

We can extend that to: The average between any two rationals is rational.

......

We can go further (but we have to do a little real analysis) to show that for any real number $x$ (irrational or rational) for any distance $d> 0$ (irrational or rational) there will be rationals $q$ so that $x-d < q < x+d$.

So for any two real numbers, $x,y; x<y$ we can find $d$ and $e$ small enough and rational $q_1, q_2$ so that $x-d < q_1 < x+d < \frac {q_1 + q_2}2 < y-e < q_2 < y_e$.

So conclusion: Between any two real numbers (rational or irrational) there will be a rational number.

That means given any real number $x$ there can't be any rational number, not equal to $x$, that is closest to $x$.

Which is important to know.

fleablood
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