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How do we prove that the three statements below about the real number $x$ are equivalent?

(i) $\displaystyle x$ is rational

(ii) $\displaystyle \frac{x}{2}$ is rational

(iii) $\displaystyle 3x-1$ is rational

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I think multiplication and addition operations on rational no's lead a rational no' indeed. –  Mr.ØØ7 Mar 24 '13 at 5:50
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What is your definition of rational? If you can answer that, the result is immediate. –  xylon97 Mar 24 '13 at 5:52

4 Answers 4

If $x$ is rational, then $x=\frac{p}{q}$ for some integer $p$ and some nonzero integer $q$. Then $\frac{x}{2}=\frac{p}{2q}$ with $p$ an integer and $2q$ a nonzero integer.

If $\frac{x}{2}$ is rational, then $\frac{x}{2}=\frac{m}{n}$ for some integer $m$ and some nonzero integer $n$. Then $3x-1=\frac{6m-n}{n}$ with $6m-n$ an integer and $n$ a nonzero integer.

If $3x-1$ is rational, then $3x-1=\frac{a}{b}$ for some integer $a$ and some nonzero integer $b$. Then $x=\frac{a+b}{3b}$ with $a+b$ an integer and $3b$ a nonzero integer.

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Hint: Any rational number is a fraction $\frac{a}{b}$ with $a, b \in Z, b \neq 0$.

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It is enough to prove that $$(i) \implies (ii) \implies (iii) \implies (i)$$

$1$. $(i) \implies (ii)$. Let $x = \dfrac{p}q$, where $p,q \in\mathbb{Z}$. We then have $\dfrac{x}2 = \dfrac{p}{2q}$ and we have $p,2q \in \mathbb{Z}$. Hence, $$(i) \implies (ii)$$

$2$. $(ii) \implies (iii)$. Let $\dfrac{x}2 = \dfrac{p}q$, where $p,q \in \mathbb{Z}$. This gives $x = \dfrac{2p}q$, which in-turn gives $$3x-1 = \dfrac{6p}q - 1 = \dfrac{6p-q}q$$ Since $p,q \in \mathbb{Z}$, we have $q,6p-q \in \mathbb{Z}$. Hence, $$(ii) \implies (iii)$$

$3$. $(iii) \implies (i)$. Let $3x-1 = \dfrac{p}q$, where $p,q \in \mathbb{Z}$. This gives $$3x = \dfrac{p}q + 1 \implies x = \dfrac{p+q}{3q}$$ Since $p,q \in \mathbb{Z}$, we have $3q,p+q \in \mathbb{Z}$. Hence, $$(iii) \implies (i)$$

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Rational numbers are closed under multiplication and addition.

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