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I am stuck while reading book FOUNDATIONS OF ANALYSIS by EDMUND LANDAU. I can't understand that how the number $\xi+X$ is always irrational whenever $\xi$ is irrational and $X$ is rational.

The book says:enter image description here

Please how the auther proves that $\xi+X$ is irrational.

Thank you.

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    $\begingroup$ If $\xi + X = Y$ with $Y$ rational, then $\xi = Y - X$ is the difference of two rational numbers, hence rational. $\endgroup$ – Daniel Fischer Sep 20 '13 at 13:55
  • $\begingroup$ But $\xi$ is already declared to be irrational. So how it now can be rational? $\endgroup$ – Silent Sep 20 '13 at 14:03
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    $\begingroup$ It can't. And that contradiction proves that $\xi + X$ is also irrational when $X$ is rational. $\endgroup$ – Daniel Fischer Sep 20 '13 at 14:08
  • $\begingroup$ Thanks a lot.It really helped me. $\endgroup$ – Silent Sep 20 '13 at 14:11
  • $\begingroup$ dupe math.stackexchange.com/questions/480369/… $\endgroup$ – Don Larynx Sep 20 '13 at 17:28
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The proof you cite is a proof by contradiction. I'm going to insert way more words of explanation than you will encounter in this kind of proof:

Premises: $\xi$ is irrational and $X$ is rational. Assume $\xi + X$ is rational, then $\xi + X = Y$ where $Y$ is rational. By arithmetic in real numbers, $\xi = Y-X$. But $Y-X$ is the difference of two rationals, hence also rational (have you seen the proof of this fact?). This contradicts the premise that $\xi$ is irrational. Therefore $\xi + X$ must be irrational.

Hope this helps!

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  • $\begingroup$ This really helps! Hope all the instructors were student-friendly like you! $\endgroup$ – Silent Sep 20 '13 at 14:13
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Field axiom (A1) states $x, y \in F \rightarrow x + y \in F$. It was used here to add $(-r)$ to the rational on each side.

Field axiom (A5) states that $\forall x \in F, \exists -x \in F : x + (-x) = 0$. It was used here to show $r + -r = 0.$

Field axiom (M5) states $x \in F, x \neq 0 \rightarrow \exists \frac{1}{x} \in F : x(\frac{1}{x}) = 1$.}

Field axiom (M4) states $F$ contains an element $1 \neq 0 : 1x = x$ for all $x \in F$.


If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove $r + x$ and $rx$ are irrational.

(Proof by contradiction) Suppose $r + x$ is rational. Then $$ \exists a, b \in \mathbb{Q} : \frac{a}{b} = r + x $$ Since $\mathbb{Q}$ is an (ordered) field, the field axioms are applicable. First using field axiom (A1) and then using field axiom (A5), the above equation can be re-written as $$ \frac{a}{b} + (-r) = (-r) + r + x \rightarrow \frac{a}{b} + (-r) = x $$ The absurdity is clear now: although the LHS is in $\mathbb{Q}$, the RHS isn't. RAA.

Similarly, suppose $rx$ is rational. Then $$ \exists a, b \in \mathbb{Q} : \frac{a}{b} = rx $$ which, by (M5) and (M4) should give us $$ (\frac{1}{r})ab = (\frac{1}{r})rx \rightarrow (\frac{1}{r})ab = 1x = x $$ The absurdity is clear now, just as before. RAA.

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  • $\begingroup$ +1, I like it. But I think you mean $(1/r)(a/b)$ in your last line and not $(1/r)(ab)$? $\endgroup$ – Numbersandsoon Sep 20 '13 at 19:05
  • $\begingroup$ yea, that was a mistake $\endgroup$ – Don Larynx Sep 20 '13 at 19:26
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Suppose it were true. Then let $\xi=Y-X$ and $Y=p/q$, $X=r/s$, then we can rewrite the equation as:

$\xi=p/q-r/s=\frac{ps-qr}{qs}$, a rational number, which contradicts the claim that $\xi$ is irrational.

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Rational numbers are (at least) closed under "+", "-" and "*", which is easy to prove. See Wiki. That means, the sum, difference, and product of two rational numbers are also rational.

If $\xi+X$ were rational, then $\xi+X - X=\xi$ must also be rational, contradiction!

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  • $\begingroup$ Yes, I know it. But how this fact proves that $\xi+X$ is irrational? $\endgroup$ – Silent Sep 20 '13 at 14:04
  • $\begingroup$ @Sush see edit. $\endgroup$ – ftfish Sep 20 '13 at 14:07

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