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I'm working on trying to prove the following equivalence for all x in real numbers:

$x$ is rational $\leftrightarrow x - 5$ is rational $\leftrightarrow x/3$ is rational

I know i need to prove each one individually such as: $$x\,\,is\,\,rational \rightarrow x - 5 \,\,is\,\,rational\\ x-5\,\,is\,\,rational \rightarrow x/3\\ x/3 \,\,is\,\,rational \rightarrow x \,\,is\,\,rational\\$$

  1. I started by assuming if $x = \frac{p}{q} $ then $x-5= \frac{p}{q}-5$ which is still a rational.
  2. Then if $x - 5 = \frac{p}{q}$ then $\frac{x}{3} = \frac{\frac{p}{q}}{3}$ which is still a rational (Is there a proper way to prove this algebraically, I feel like it might be insufficient proof to just keep saying it is a rational).
  3. After this I am very unsure what to do to get from step 2 to 3. I've looked at a bunch of examples but I suck at proving rationals. Would it be something like if $3*\frac{x}{3} = \frac{\frac{p}{q}}{3}*3$ then $x = \frac{p}{q} $

Edit:

  1. Assume x is rational   $p \rightarrow q$
    x = $\frac{p}{q}$
    Prove x-5 is rational
    substitute x :  $x-5 = \frac{p-5q}{q}$   : x-5 is a rational
  2. Assume x-5 is rational   $q \rightarrow r$
    x-5 = $\frac{p}{q}$
    solve for x:   $x=\frac{p-5q}{q}$
    Prove $\frac{x}{3}$
    substitude for x:   $\frac{x}{3} = \frac{\frac{p-5q}{q}}{3}$   : $\frac{x}{3}$   is rational because the numerator and denominator are rationals
  3. Assume $\frac{x}{3}$ is rational   $r \rightarrow p$
    $\frac{x}{3} = \frac{p}{q}$
    solve for x = $\frac{\frac{p-5q}{q}}{3}$
    Prove x is a rational
    we already proved x is a rational in the second step, because the quotient of two rationals is always a quotient.
below_avg_st
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  • You don't need to prove each individually in both directions. It suffices to show the three one-way implications: $x$ rational $\implies x-5$ rational, $x-5$ rational $\implies x/3$ rational, and $x/3$ rational $\implies x$ rational. – JMoravitz Feb 10 '18 at 01:02
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    You need to be careful to not confuse what $\frac{p}{q}$ represents. In your second statement for example, you said that $x-5=\frac{p}{q}$ and then $\frac{x}{3}=\frac{\frac{p}{q}}{3}$, but that is incorrect. Given your assumption that $x-5=\frac{p}{q}$ that would mean that $\frac{x-5}{3}=\frac{\frac{p}{q}}{3}$. Another nitpick is that at this level of difficulty what you should end with is very specifically an integer divided by another nonzero integer in order to show that a number is rational. $\frac{p}{q}-5$ for example should be written $\frac{p-5q}{q}$ – JMoravitz Feb 10 '18 at 01:06
  • @JMoravitz I took what you said and edited my work. Can you please tell me if I did it right now that I changed some stuff around? – below_avg_st Feb 10 '18 at 04:02
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    You still have errors and you have not in several places fully reduced to a simple fraction, where there is only one fraction bar. Yes, a rational divided by a nonzero rational is rational, but given that this is such an introductory question, you should show it by first principles and not rely on results like that. – JMoravitz Feb 10 '18 at 04:05

1 Answers1

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$(i)\implies (ii)$

Suppose $x$ is rational. Then $x=\frac{p}{q}$ for some integers $p,q$ with $q\neq 0$.

We have then $x-5=\frac{p}{q}-5=\frac{p}{q}-\frac{5q}{q}=\frac{p-5q}{q}$ and noting that $p-5q$ is an integer (since integers are closed under addition and multiplication) and $q\neq 0$ as before, we have successfully written $x-5$ as a ratio of two integers with the integer in the denominator nonzero and is therefore rational.

$(ii)\implies (iii)$

Suppose $x-5$ is rational. Then $x-5=\frac{p}{q}$ for some integers $p,q$ with $q\neq 0$.

Then $\frac{x}{3}=\frac{1}{3}\left((x-5)+5\right)=\frac{1}{3}\left(\frac{p}{q}+5\right)=\frac{1}{3}(\frac{p+5q}{q})=\frac{p+5q}{3q}$ and noting that $p+5q$ is an integer since integers are closed under addition and multiplication and that $3q$ is a nonzero integer since both $3$ and $q$ are both nonzero integers and the product of two nonzero integers is again a nonzero integer, we see then that we can write $\frac{x}{3}$ as a ratio of two integers with the integer in the denominator nonzero and is therefore rational.

$(iii)\implies (i)$

Similar to the other cases.

JMoravitz
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