1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $(m+n)^{2014}\in C$ $\endgroup$ – JMoravitz May 31 '16 at 16:41
  • $\begingroup$ The difference between " k E Z " and "m 2 A" is rather unclear. $\endgroup$ – user228113 May 31 '16 at 16:45
  • $\begingroup$ 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. $\endgroup$ – JMoravitz May 31 '16 at 16:52
  • $\begingroup$ @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 $\endgroup$ – Daniel May 31 '16 at 16:54
0
$\begingroup$

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 $3^{1/2} \not \in \mathbb{Q}$?

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Daniel May 31 '16 at 17:04
  • $\begingroup$ 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. $\endgroup$ – Kushal Bhuyan May 31 '16 at 17:09
  • $\begingroup$ 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. $\endgroup$ – Daniel May 31 '16 at 17:15
  • $\begingroup$ 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. $\endgroup$ – Kushal Bhuyan May 31 '16 at 17:20
  • $\begingroup$ its the manner in which its asked which i am not used to $\endgroup$ – Daniel May 31 '16 at 17:34
0
$\begingroup$

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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.