[Math] Proof: Is there a line in the xy plane that goes through only rational coordinates

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Question: Is there a line in the XY plane that has all rational coordinates. Prove your answer.

Idea: There is most certainly not. I believe it can be shown that between any 2 rational points that there is at least one irrational coordinate. Therefore, there can not be a line that contains only rational points. The issue is that I am not sure how to show this. Any ideas? I am also open to any other ideas of how to do this. Thanks.

Note: this is for an intro to proofs study guide. So, I would prefer not to use advanced theorems.

Best Answer

Two proofs.

First one based on cardinality. A line has the cardinality of the continuum like $\mathbb R$, while $\mathbb Q$ is countable.

Second one. A line has an equation $ax+by+c=0$ with $(a,b) \neq (0,0)$. If $a \neq 0$ then $(-\frac{b \pi +c}{a},\pi)$ belongs to the line and the second coordinate of that point is not rational. While if $a=0$, $(\pi,-\frac{c}{b})$ belongs to the line (thanks to immibis comment).

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[Math] Show that the set of polynomials with rational coefficients is countable.

Let us begin by stating that the set of rationals is countable, as Z² is countable and every rational number can be expressed as an ordered pair of integers.

Then we can enumerate the polynomials as follows:

1.Start with n=1
2.For every m on [1,n] list the next order-m polynomial
3.Increase n by 1 and return to step 2.

For a proper enumeration of each order of polynomials (which must exist as, Q being countable, Qⁿ is countable for all finite n and every order-n rational-coefficient polynomial can be expressed as an ordered n-tuple), every polynomial of rational coefficients will eventually appear exactly once.

[Math] Prove that there is an irrational number and a rational number between any two distinct real numbers

Nice attempt, but unfortunately your proof is wrong. $y-10^{-n}y=y(1-10^{-n})$. Since $1-10^{-n}$ is rational and $y$ is irrational, $y(1-10^{-n})$ is irrational. Also, as pointed out by Mees de Vries in comments, $\frac{x+y}{2}$ may be rational.

In this link, you can find a proof by joeA that there is a rational between two real numbers. He uses the Archemidean Property of the real numbers, which can be stated as follows:

For every number $x\in\mathbb{R}$, there exists a natural number $n$ such as $x<n$.

for this purpose (even the intuitive fact that there is an integer between two numbers $x,y$ satisfying $y-x>1$ can be proved using this property) . Then you can conclude the result for irrationals; indeed if $x,y\in\mathbb{R}$, take any irrational number of your choice, say $\sqrt{2}$. Suppose that $x<y$. Then $x-\sqrt{2}<y-\sqrt{2}$. Thus there exists a rational number $q$ such as $x-\sqrt{2}<q<y-\sqrt{2}$, that it $x<q+\sqrt{2}<y$ and $q+\sqrt{2}$ is irrational.

Best Answer

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[Math] Show that the set of polynomials with rational coefficients is countable.

[Math] Prove that there is an irrational number and a rational number between any two distinct real numbers

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