# Proving that things are or aren't rational numbers

My question comes in three parts:

• Suppose $x,y\in \Bbb Q$. Prove that $2x-5y\in \Bbb Q$
• Prove that $3^{1/2}\not\in \Bbb Q$
• Suppose $x\in \Bbb Q$. Prove that $x^2+3^{1/2}\not\in \Bbb Q$

In the third question I'm not sure how I should proceed to solve it. I'm aware of how to prove root 3 is irrational but that question I'm not sure of. Also the first question I don't understand how i can prove that $2x-5y$ is irrational.

• Visit this page for information on how to properly type mathematics on this site using MathJax and $\LaTeX$. Compare the readability of (m + n)2014 2 C to $(m+n)^{2014}\in C$ – JMoravitz May 31 '16 at 16:41
• The difference between " k E Z " and "m 2 A" is rather unclear. – user228113 May 31 '16 at 16:45
• I have removed all of the first half of the questions in order to narrow the scope of the post. When posting a question, it should only be one question instead of several. With too many questions posted in a single post it can be difficult to properly associate one post with another in terms of labeling things as duplicates and such. The questions that I removed should be posted as a separate question, but the approach for those is to try to express each as a multiple of three, one more than a multiple of three, or two more than a multiple of three depending on which set they're in. – JMoravitz May 31 '16 at 16:52
• @JMoravitz could you give me an example sorry i'm just struggling to understand these questions could you perhaps solve one for me if possible – Daniel May 31 '16 at 16:54

Hints: $1$. As $x,y \in \mathbb{Q}$ and $2,-5 \in \mathbb{Q}$, so $2x,-5y\in \mathbb{Q}$.

$2$. If possible consider $3^{1/2}$ is rational. Then $\exists p/q \in \mathbb{Q}$ such that $3=p^2/q^2$, then try to find a contradiction.

$3$. If $x\in \mathbb{Q}$ then $x^2\in \mathbb{Q}$. Then what can you say about $x^2+3^{1/2}$, if $3^{1/2} \not \in \mathbb{Q}$?

• if x^2 is rational and root 3 is irrational adding them together will surely result in me getting something irrational.Sorry im writing a maths exam this week and im stressing because i dont understand the questions – Daniel May 31 '16 at 17:04
• Yes you are right. We can also argue as $x^2 \in \mathbb{Q}$ and suppose $x^2+3^{1/2} \in \mathbb{Q}$ , then their subtraction should also be in $\mathbb{Q}$, but $x^2+3^{1/2}-x^2=\sqrt{3}$ which is not a rational number. Hence contradiction. – Kushal Bhuyan May 31 '16 at 17:09
• So basically we are showing that even adding x^2 to root three is still irrational? Jmovaritz posted a solution to the problem. This is my first time encountering these types of questions. – Daniel May 31 '16 at 17:15
• Yes adding $x^2$ to $3^{1/2}$ does not make it rational, it will still remain irrational as I have shown in the comment above. If you are having so much difficulty in understanding this type of problems then I would recommend you to go through about the rational and irrational numbers and their basic arithmetic from some books. – Kushal Bhuyan May 31 '16 at 17:20
• its the manner in which its asked which i am not used to – Daniel May 31 '16 at 17:34

For (a), this follows from the more general fact that a rational number plus another rational number is again rational. Still, it is worth going through the effort of proving this from first principles at least once.

Suppose that $x,y\in\Bbb Q$ are both rational numbers. We wish to prove then that $2x-5y\in \Bbb Q$ is also a rational number (you wrote the word irrational there presumably by mistake. Do not confuse the words as they mean completley opposite things)

Since $x\in \Bbb Q$, this means that there are integers $a$ and $b$ with $b\neq 0$ such that $x=\frac{a}{b}$.

Similarly, $y\in \Bbb Q$ implies that there are integers $c$ and $d$ with $d\neq 0$ such that $y=\frac{c}{d}$. (note: I used different letters to describe $x$ than I did for $y$ since they are able to be different)

Then we have $2x-5y = 2\frac{a}{b}-5\frac{c}{d} = \frac{2ad-5bc}{bd}$. We ask ourselves, is this a rational number? Is it the ratio of two integers with the bottom integer nonzero?

For (c), this follows from the more general fact that a rational number plus an irrational number is irrational. Again, it is worth approaching from first principles at least once.

We prove this via contradiction:

Suppose that $x\in \Bbb Q$ and that $x^2+\sqrt{3}\in \Bbb Q$

Then there exist integers $a,b,c,d$ with $b$ and $d$ nonzero such that $x=\frac{a}{b}$ and $x^2+\sqrt{3}=\frac{c}{d}$

But then, $\sqrt{3} = (x^2+\sqrt{3})-x^2 = \frac{c}{d}-(\frac{a}{b})^2 = \frac{b^2c - a^2d}{db^2}$. What do we know about the rationality of $\sqrt{3}$ from the previous part? What does this line say about the rationality of $\sqrt{3}$ though? Is this possible as a result? What does this imply about the original wording of the problem?

• @JMovaritz, thank you so much i'll try go over your answers until i understand them im being a bit slow. – Daniel May 31 '16 at 17:08