Prove that for every irrational real number $a$, there are irrational real numbers $b$ and $b'$ so that $a+b$ and $ab'$ are both rational while $ab$ and $a+b'$ are both irrational.
Problem
Source: APMO 2005 Problem 1
Tags:
mecrazywong
23.03.2005 17:33
one of (1-a)a,(2-a)a must be irrational; one of a+1/a,a+2/a must be irrational.
yptsoi
23.03.2005 18:08
mecrazywong wrote: one of (1-a)a,(2-a)a must be irrational; one of a+1/a,a+2/a must be irrational. Our solutions are exactly the same!
erdos
24.03.2005 15:04
mecrazywong wrote: one of (1-a)a,(2-a)a must be irrational; one of a+1/a,a+2/a must be irrational. Can you please post a more detailed solution. I didn't understand yours very well.
mecrazywong
24.03.2005 15:13
erdos wrote: mecrazywong wrote: one of (1-a)a,(2-a)a must be irrational; one of a+1/a,a+2/a must be irrational. Can you please post a more detailed solution. I didn't understand yours very well. Isn't that obvious? (2-a)a-(1-a)a=a, which is irrational, so (2-a)a and (1-a)a cannot be both rational. Similar approach for the other one.
Soarer
24.03.2005 20:08
My solution is different. 1. I take b=1-a 2. I showed that if a(1-a) is rational, it must be of the form p+sqrt(q), where p,q are rational. 3. I take b=-sqrt q if a is of the form p+sqrt(q). Similar for b'.
mahbub
04.04.2005 03:48
My solution is almost same to the crazy. But a bit generalized. p, 1/kp p*1/kp=1/k Rational. If p+1/kp is irrational then we will be done. So let for all rational k p+1/kp is rational. Then: p+1/kp – p-1/tp=1/p*(1/k-1/t) which is irrational. But rational-rational should be rational. So we get a contradiction. So we can get such rational k for which p+1/kp is irrational. Now take, p, k-p p+k-p=k Rational. Say for all rational k p(k-p) is rational. Then: pk-p^2-pt+p^2=p(k-t) contradiction! That ends my task! Siuhochung, can you show yours?
Soarer
04.04.2005 10:15
I have showed mine. 1. I take b=1-a 2. I showed that if a(1-a) is rational, it must be of the form p+sqrt(q), where p,q are rational. 3. I take b=-sqrt q if a is of the form p+sqrt(q). Similar for b'. All these steps are easy to be shown.
Konigsberg
25.05.2014 11:55
Also MOP 2005 Homework - Red Group #3
SMOJ
25.05.2014 16:57
Also APMO 2005 Problem 1
Ali3085
16.02.2020 11:13
Take $b=z-a : z \in Z$ if $a(z-a) \in Q \forall z \in Z$ then $a(z-a) -a(z+1-a) \in Z\implies a \in Q $ a contradiction Similar argument for$ b'= \frac{z}{a}$
Keith50
13.10.2020 19:47
Let $b=n_1-a$ and $b'=\frac{n_2}{a}$ where $n_1,n_2\in \mathbb{Z}-\{0\}$ (it is obvious that $b,b'\not\in \mathbb{Q}$), we claim that $\forall n_1,n_2\in \mathbb{Z}-\{0\}$ we can find a pair $(n_1,n_2)$ that makes $a+b, ab'\in \mathbb{Q}$ and $ab, a+b' \not\in \mathbb{Q}$. Assume for contrary, we then have $a+b, ab'\in \mathbb{Q}$, $ab=a(n-a)\in \mathbb{Q} \ \forall n\in \mathbb{Z}-\{0\}$, let $s,t \in \mathbb{Z}-\{0\}$ such that \[a(s-a), a(t-a) \in \mathbb{Q}\]subtracting one from the another gives \[a(s-t)\in \mathbb{Q} \implies a\in \mathbb{Q}\]a contradiction. Similarly, we have $a+\frac{n}{a}\in \mathbb{Q} \ \forall n\in \mathbb{Z}-\{0\}$, let $u,v \in \mathbb{Z}-\{0\}$ such that \[a+\frac{u}{a}, a+\frac{v}{a} \in \mathbb{Q}\]subtracting one from another gives \[\frac{u-v}{a}\in \mathbb{Q}\implies a\in \mathbb{Q}\]also a contradiction. Thus, we have that there must exist $(n_1,n_2)$ where $b=n_1-a$ and $b'=\frac{n_2}{a}$ such that $a+b, ab'\in \mathbb{Q}$ and $ab, a+b' \not\in \mathbb{Q}$. $\blacksquare$
trying_to_solve_br
29.04.2021 02:00
Gosh, my solution was more sofisticaded, don't have much sure if it is correct, but here it is: We can take $q$ irrational and $p$ rational, such that $p^2-4q$ is not a square of another rational. Hence we can create a second degree equation as $x^2-px+q=0$, with roots $a$ and $b$, which satisfies our claim. Same for $a$ and $b'$, $x^2-nx+m$, with $n$ irrational and $m$ rational.
bora_olmez
29.06.2021 23:02
Posting for storage
$\textbf{Part 1)}$
We shall firstly prove the existence of some irrational $b$ satisfying the condition.
Notice that if $a < 0 $, we can take $a \rightarrow -a$, $b \rightarrow -b$ which once again satisfies the condition so assume that $a>0$.
Then let $a = \sqrt{x}$ for some $x \in \mathbb{R}$.
If $x$ is irrational, take $b = -a$ and notice that $a+b = 0$ is rational, meanwhile; $ab = -a^2 = -x$ is irrational and consequently, $b=-a$ satisfies the condition.
If on the other hand $x$ is rational, take $x=a^2-a$, then $a+b = a^2 = x$ is rational, meanwhile; $ab = a^2(a-1)$ which must be irrational $a^2$ is rational and $a-1$ is irrational. $\square$
$\textbf{Part 2}$
We claim that either $b' = \frac{2}{a}$ or $b' = \frac{1}{a}$ works.
Assume that neither do, we clearly have that $ab' = a \cdot \frac{2}{a} = 2$ or $ab' = a \cdot \frac{1}{a} = 1$ both of which are rationals, on the other hand $a+b' \in \frac{a^2+1}{a}, \frac{a^2+2}{a}$. Assume that both of these reals are also rationals, we then have that their difference which is $\frac{a^2+2}{a} - \frac{a^2+1}{a} = \frac{1}{a}$ must also be a rational which is a contradiction. $\blacksquare$
john0512
03.07.2023 10:30
Claim: Either $b=1-a$ or $b=2-a$ works. Suppose not. Then, $a(1-a)$ and $a(2-a)$ are both rational. However, they differ by $a$, an irrational number, contradiction. Claim 2: Either $b'=1/a$ or $b'=2/a$ works. Suppose not. Then, $a+1/a$ and $a+2/a$ are both rational, but they differ by $1/a$, which is irrational, contradiction. Hence, we are done.