Suppose x and y are two arbitrary, distinct (unequal) real numbers. Prove that there exists a rational number q between x and y.

Question

I thought this proof was much simpler than it actually is. I used an Axiom that states, "if $p$ and $q$ are real numbers, then there is a number between them, i.e: $$\frac{(p + q)}{2}$$ My attempt at the proof was:

By the above Axiom, there exists a number, $q$, between $x$ and $y$, i.e: $$\frac{(x+y)}{2}$$ We know $q$ is rational because it is in the form $\frac ab$, where $a = x + y$ and $b = 2$. Thus, there exists a rational number, $q$ such that $q$ is between $x$ and $y$.

However, I realized that if $x=\pi$ and $y = e$, then $\frac{(x+y)}{2}$ will not be rational.

I know that to prove this, I need to use the Lemma: If $y-x>1$, ∃ n ∈ $\mathbb{Z}$ such that $x<n<y$.

I really don't know how to approach this differently because I did think it would be simpler.

Answer 1

• Find the minimum integer value of $k$ for which $$|x-y| < 10^k$$

• Round $\min(x,y)$ to the smallest multiple of $10^{k-1}$ greater than $\min(x,y)$.

• Add $10^{k-2}$. The result is a rational number strictly between $x$ and $y$.

Answer 2

Since $x$ and $y$ are different you can choose an integer $N$ large enough so that $1/N$ is less than their (positive) difference. Now imagine a ruler (number line) with marks every $1/N$th. One of those marks will have to fall strictly between $x$ and $y$. It's rational since it looks like $k/N$ for some integer $k$.

Answer 3

If the difference between two real numbers is greater than 1, then there will be an integer between the numbers. If the difference is less than , take the difference of first decimal number from where the numbers start to differ. And that decimal number at its position has to exist between the real numbers.

Eg: $0.12345......$ and $0.123378.....$ are such real numbers then, $0.1234$ has to exist between them which is a rational number.

Similar trick Can be applied for negative numbers also.