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There are 197 different non-zero real numbers, and the sum of any two numbers is a rational number or the product is a rational number. Prove that: 197 numbers, each square is a rational number.
edit:
There are 197 different non-zero real numbers, and the sum of any two distinct numbers is a rational number or the product is a rational number. Prove that: 197 numbers, each square is a rational number.
The number 197 is not important, so I can assume that there are only 10 numbers to make the question sounds easier:
when all of the 10 numbers are rational, then their square must be rational.
Now I assume that there is a irrational number a . Then, the remaining 9 numbers can be either p - a or p / a, while p is a rational number.
However, if there will not be three number like p - a. let b1 = p1 - a, b2 = p2 - a and b3 = p - a, then b1 + b2 = p1 + p2 - 2a, which is not a rational number, so b1 b2 = p1p2 - a(p1 + p2) + a2 must be rational number. b2b3 and b1b3 must also be rational number.
Therefore, p1p2 - a(p1 + p2) + a2, p2p3 - a(p2 + p3)+ a2 , p1p3 - a(p1 + p3)+ a2 are rational numbers. p * p is absolutely a rational number, so A0 = - a(p1 + p2) + a2, A1 = - a(p2 + p3)+ a2 , A2 = - a(p1 + p3)+ a2, A is rational number, A2 -A1 = a(p2 - p1), p2 - p1 can only be 0, which contradicts that every number is different.
This shows that there are at most 2 numbers like p - a, and there must be at least 8 numbers like p / a. However I didn't know how to continue the proof, anyone can help me?
elementary-number-theory
asked Jul 27 at 6:16
Thomas Yeung Sin Chun
83
 |Â
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There are 197 different non-zero real numbers, and the sum of any two numbers is a rational number or the product is a rational number. Prove that: 197 numbers, each square is a rational number.
edit:
There are 197 different non-zero real numbers, and the sum of any two distinct numbers is a rational number or the product is a rational number. Prove that: 197 numbers, each square is a rational number.
The number 197 is not important, so I can assume that there are only 10 numbers to make the question sounds easier:
when all of the 10 numbers are rational, then their square must be rational.
Now I assume that there is a irrational number a . Then, the remaining 9 numbers can be either p - a or p / a, while p is a rational number.
However, if there will not be three number like p - a. let b1 = p1 - a, b2 = p2 - a and b3 = p - a, then b1 + b2 = p1 + p2 - 2a, which is not a rational number, so b1 b2 = p1p2 - a(p1 + p2) + a2 must be rational number. b2b3 and b1b3 must also be rational number.
Therefore, p1p2 - a(p1 + p2) + a2, p2p3 - a(p2 + p3)+ a2 , p1p3 - a(p1 + p3)+ a2 are rational numbers. p * p is absolutely a rational number, so A0 = - a(p1 + p2) + a2, A1 = - a(p2 + p3)+ a2 , A2 = - a(p1 + p3)+ a2, A is rational number, A2 -A1 = a(p2 - p1), p2 - p1 can only be 0, which contradicts that every number is different.
This shows that there are at most 2 numbers like p - a, and there must be at least 8 numbers like p / a. However I didn't know how to continue the proof, anyone can help me?
elementary-number-theory
asked Jul 27 at 6:16
Thomas Yeung Sin Chun
83
2
"Proof-verification" is a tag used for questions about the validity of a certain proof. Since you have provided no proof, this would hardly be an appropriate tag. Also, it is customary here to convince the readers that you have put in some effort in order to solve the problem, so that the general spirit of the site is a true discussion of mathematical ideas, and not a "do-my-homework" service taken advantage of by lazy students who will never get anywhere anyway.
– uniquesolution
Jul 27 at 6:22
@MattiP. - and it fulfills the conclusion too, so what are you saying?
– uniquesolution
Jul 27 at 6:23
It's not very clear what you're asking. If I have a list $1, 2, 3, ldots , 196, 197$, it fulfills the requirements ...
– Matti P.
Jul 27 at 6:23
@MattiP. ditto .The assumption is that in this set of $197$ numbers, given any two, either their product or their sum is rational. You are required to prove that for each of the $197$ numbers, its square is rational.
– uniquesolution
Jul 27 at 6:24
1
@LordSharktheUnknown Obviously $i=j$ is excluded.
– uniquesolution
Jul 27 at 6:28
 |Â
show 6 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
There are 197 different non-zero real numbers, and the sum of any two numbers is a rational number or the product is a rational number. Prove that: 197 numbers, each square is a rational number.
edit:
There are 197 different non-zero real numbers, and the sum of any two distinct numbers is a rational number or the product is a rational number. Prove that: 197 numbers, each square is a rational number.
The number 197 is not important, so I can assume that there are only 10 numbers to make the question sounds easier:
when all of the 10 numbers are rational, then their square must be rational.
Now I assume that there is a irrational number a . Then, the remaining 9 numbers can be either p - a or p / a, while p is a rational number.
However, if there will not be three number like p - a. let b1 = p1 - a, b2 = p2 - a and b3 = p - a, then b1 + b2 = p1 + p2 - 2a, which is not a rational number, so b1 b2 = p1p2 - a(p1 + p2) + a2 must be rational number. b2b3 and b1b3 must also be rational number.
Therefore, p1p2 - a(p1 + p2) + a2, p2p3 - a(p2 + p3)+ a2 , p1p3 - a(p1 + p3)+ a2 are rational numbers. p * p is absolutely a rational number, so A0 = - a(p1 + p2) + a2, A1 = - a(p2 + p3)+ a2 , A2 = - a(p1 + p3)+ a2, A is rational number, A2 -A1 = a(p2 - p1), p2 - p1 can only be 0, which contradicts that every number is different.
This shows that there are at most 2 numbers like p - a, and there must be at least 8 numbers like p / a. However I didn't know how to continue the proof, anyone can help me?
elementary-number-theory
asked Jul 27 at 6:16
Thomas Yeung Sin Chun
83
There are 197 different non-zero real numbers, and the sum of any two numbers is a rational number or the product is a rational number. Prove that: 197 numbers, each square is a rational number.
edit:
There are 197 different non-zero real numbers, and the sum of any two distinct numbers is a rational number or the product is a rational number. Prove that: 197 numbers, each square is a rational number.
The number 197 is not important, so I can assume that there are only 10 numbers to make the question sounds easier:
when all of the 10 numbers are rational, then their square must be rational.
Now I assume that there is a irrational number a . Then, the remaining 9 numbers can be either p - a or p / a, while p is a rational number.
However, if there will not be three number like p - a. let b1 = p1 - a, b2 = p2 - a and b3 = p - a, then b1 + b2 = p1 + p2 - 2a, which is not a rational number, so b1 b2 = p1p2 - a(p1 + p2) + a2 must be rational number. b2b3 and b1b3 must also be rational number.
Therefore, p1p2 - a(p1 + p2) + a2, p2p3 - a(p2 + p3)+ a2 , p1p3 - a(p1 + p3)+ a2 are rational numbers. p * p is absolutely a rational number, so A0 = - a(p1 + p2) + a2, A1 = - a(p2 + p3)+ a2 , A2 = - a(p1 + p3)+ a2, A is rational number, A2 -A1 = a(p2 - p1), p2 - p1 can only be 0, which contradicts that every number is different.
This shows that there are at most 2 numbers like p - a, and there must be at least 8 numbers like p / a. However I didn't know how to continue the proof, anyone can help me?
elementary-number-theory
asked Jul 27 at 6:16
Thomas Yeung Sin Chun
83
edited Jul 27 at 9:13
asked Jul 27 at 6:16
Thomas Yeung Sin Chun
83
asked Jul 27 at 6:16
Thomas Yeung Sin Chun
83
asked Jul 27 at 6:16
Thomas Yeung Sin Chun
83
83
2
"Proof-verification" is a tag used for questions about the validity of a certain proof. Since you have provided no proof, this would hardly be an appropriate tag. Also, it is customary here to convince the readers that you have put in some effort in order to solve the problem, so that the general spirit of the site is a true discussion of mathematical ideas, and not a "do-my-homework" service taken advantage of by lazy students who will never get anywhere anyway.
– uniquesolution
Jul 27 at 6:22
@MattiP. - and it fulfills the conclusion too, so what are you saying?
– uniquesolution
Jul 27 at 6:23
It's not very clear what you're asking. If I have a list $1, 2, 3, ldots , 196, 197$, it fulfills the requirements ...
– Matti P.
Jul 27 at 6:23
@MattiP. ditto .The assumption is that in this set of $197$ numbers, given any two, either their product or their sum is rational. You are required to prove that for each of the $197$ numbers, its square is rational.
– uniquesolution
Jul 27 at 6:24
1
@LordSharktheUnknown Obviously $i=j$ is excluded.
– uniquesolution
Jul 27 at 6:28
 |Â
show 6 more comments
2
"Proof-verification" is a tag used for questions about the validity of a certain proof. Since you have provided no proof, this would hardly be an appropriate tag. Also, it is customary here to convince the readers that you have put in some effort in order to solve the problem, so that the general spirit of the site is a true discussion of mathematical ideas, and not a "do-my-homework" service taken advantage of by lazy students who will never get anywhere anyway.
– uniquesolution
Jul 27 at 6:22
@MattiP. - and it fulfills the conclusion too, so what are you saying?
– uniquesolution
Jul 27 at 6:23
It's not very clear what you're asking. If I have a list $1, 2, 3, ldots , 196, 197$, it fulfills the requirements ...
– Matti P.
Jul 27 at 6:23
@MattiP. ditto .The assumption is that in this set of $197$ numbers, given any two, either their product or their sum is rational. You are required to prove that for each of the $197$ numbers, its square is rational.
– uniquesolution
Jul 27 at 6:24
1
@LordSharktheUnknown Obviously $i=j$ is excluded.
– uniquesolution
Jul 27 at 6:28
2
2
"Proof-verification" is a tag used for questions about the validity of a certain proof. Since you have provided no proof, this would hardly be an appropriate tag. Also, it is customary here to convince the readers that you have put in some effort in order to solve the problem, so that the general spirit of the site is a true discussion of mathematical ideas, and not a "do-my-homework" service taken advantage of by lazy students who will never get anywhere anyway.
– uniquesolution
Jul 27 at 6:22
"Proof-verification" is a tag used for questions about the validity of a certain proof. Since you have provided no proof, this would hardly be an appropriate tag. Also, it is customary here to convince the readers that you have put in some effort in order to solve the problem, so that the general spirit of the site is a true discussion of mathematical ideas, and not a "do-my-homework" service taken advantage of by lazy students who will never get anywhere anyway.
– uniquesolution
Jul 27 at 6:22
@MattiP. - and it fulfills the conclusion too, so what are you saying?
– uniquesolution
Jul 27 at 6:23
@MattiP. - and it fulfills the conclusion too, so what are you saying?
– uniquesolution
Jul 27 at 6:23
It's not very clear what you're asking. If I have a list $1, 2, 3, ldots , 196, 197$, it fulfills the requirements ...
– Matti P.
Jul 27 at 6:23
It's not very clear what you're asking. If I have a list $1, 2, 3, ldots , 196, 197$, it fulfills the requirements ...
– Matti P.
Jul 27 at 6:23
@MattiP. ditto .The assumption is that in this set of $197$ numbers, given any two, either their product or their sum is rational. You are required to prove that for each of the $197$ numbers, its square is rational.
– uniquesolution
Jul 27 at 6:24
@MattiP. ditto .The assumption is that in this set of $197$ numbers, given any two, either their product or their sum is rational. You are required to prove that for each of the $197$ numbers, its square is rational.
– uniquesolution
Jul 27 at 6:24
1
1
@LordSharktheUnknown Obviously $i=j$ is excluded.
– uniquesolution
Jul 27 at 6:28
@LordSharktheUnknown Obviously $i=j$ is excluded.
– uniquesolution
Jul 27 at 6:28
 |Â
show 6 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
If the set - call it $S$ - contains one rational number, say $r$, then all of the numbers are rational, for if $x in Ssetminus r$ such that $r + x = p in mathbbQ$, then $x = p-r in mathbbQ$, and if $x in Ssetminus r$ is such that $rcdot x = p in mathbbQ$, then $x = fracpr in mathbbQ$. In that case, it is clear that all squares are rational.
So we can assume that all elements of $S$ are irrational. For each $alpha in S$, there are at most two elements of $S$ whose sum with $alpha$ is rational. For if there were three or more, say $alpha + beta = r in mathbbQ$, $alpha + gamma = s in mathbbQ$ and $alpha + delta = t in mathbbQ$, then $beta + gamma = r+s - 2alpha notin mathbbQ$, $beta + delta = r+t-2alpha notin mathbbQ$ and $gamma + delta = s+t - 2alpha notin mathbbQ$, so by the assumption we would have $betadelta in mathbbQ$ and $gammadelta in mathbbQ$, and consequently
$$(beta - gamma)delta = (r-s)delta in mathbbQ,,$$
which by the irrationality of $delta$ implies $r = s$ and thus $beta = gamma$, contradicting our assumption that $beta,gamma,delta$ are distinct.
Thus, if $S$ contains at least $7$ elements, then for every $alpha in S$ there are $beta,gammain S$ such that $alphabeta, alphagamma, betagamma$ are all rational. (Picking $alpha$ excludes at most two numbers, and then choosing $beta$ excludes at most two more.) But then $beta = fracralpha$ for some $r in mathbbQ$, and $gamma = fracsbeta = fracsralpha$ for some $sin mathbbQ$, whence $$alpha^2 = fracrsalphagamma in mathbbQ,.$$
Actually, we can lower the cardinality required in this argument.
For $alpha in S$, let $E(alpha) = x in S : xalpha notin mathbbQ$. This is the set excluded by $alpha$ in the above argument, and before that we saw that $E(alpha)$ contains at most two elements, since $E(alpha) subseteq x in S : x + alpha in mathbbQ$. This yields the trivial bound $operatornamecardbigl(E(alpha) cup E(beta)bigr) leqslant 4$ if $beta in Ssetminus E(alpha)$ used above. But in fact, we have the bound $operatornamecardbigl(E(alpha) cup E(beta)bigr) leqslant 3$. For if $alphabeta in mathbbQ$ and $alpha + x = r in mathbbQ$, then $xbeta = (r-alpha)beta = rbeta - alphabeta$ is irrational unless $r -alpha= 0$, i.e. $x = -alpha$. So if $E(alpha)$ contains two elements, at least one of them also belongs to $E(beta)$. And $E(alpha) cup E(beta)$ can only contain three elements if $-alpha in E(alpha)$ and $-beta in E(beta)$. So we can unconditionally lower the required cardinality to $6$, and we can lower it to $5$ under the condition that $S$ contains at most one pair of negatives.
We cannot lower the required cardinality further, because for any irrational $x$ whose square is also irrational, the set
$$S = x, -x, x^-1, -x^-1$$
has the property that for each pair of distinct elements either the sum or the product is rational, but none of the squares of the elements is rational.
edited Jul 28 at 14:02
Thomas Yeung Sin Chun
83
answered Jul 27 at 15:03
Daniel Fischer♦
171k16154274
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Let $a$ be one of these numbers. Then $a^2 in mathbb Q$ or $2a in mathbb Q$.
In the first case we are done. In the second case we have $4a^2 in mathbb Q$, hence $a^2 in mathbb Q$.
answered Jul 27 at 6:28
Fred
37.2k1237
Sorry my question is not clear since the original question is in Chinese and I use google translate to translate it. In the original question, the two numbers chosen must not be the same number, I have already updated the question
– Thomas Yeung Sin Chun
Jul 27 at 8:56
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If the set - call it $S$ - contains one rational number, say $r$, then all of the numbers are rational, for if $x in Ssetminus r$ such that $r + x = p in mathbbQ$, then $x = p-r in mathbbQ$, and if $x in Ssetminus r$ is such that $rcdot x = p in mathbbQ$, then $x = fracpr in mathbbQ$. In that case, it is clear that all squares are rational.
So we can assume that all elements of $S$ are irrational. For each $alpha in S$, there are at most two elements of $S$ whose sum with $alpha$ is rational. For if there were three or more, say $alpha + beta = r in mathbbQ$, $alpha + gamma = s in mathbbQ$ and $alpha + delta = t in mathbbQ$, then $beta + gamma = r+s - 2alpha notin mathbbQ$, $beta + delta = r+t-2alpha notin mathbbQ$ and $gamma + delta = s+t - 2alpha notin mathbbQ$, so by the assumption we would have $betadelta in mathbbQ$ and $gammadelta in mathbbQ$, and consequently
$$(beta - gamma)delta = (r-s)delta in mathbbQ,,$$
which by the irrationality of $delta$ implies $r = s$ and thus $beta = gamma$, contradicting our assumption that $beta,gamma,delta$ are distinct.
Thus, if $S$ contains at least $7$ elements, then for every $alpha in S$ there are $beta,gammain S$ such that $alphabeta, alphagamma, betagamma$ are all rational. (Picking $alpha$ excludes at most two numbers, and then choosing $beta$ excludes at most two more.) But then $beta = fracralpha$ for some $r in mathbbQ$, and $gamma = fracsbeta = fracsralpha$ for some $sin mathbbQ$, whence $$alpha^2 = fracrsalphagamma in mathbbQ,.$$
Actually, we can lower the cardinality required in this argument.
For $alpha in S$, let $E(alpha) = x in S : xalpha notin mathbbQ$. This is the set excluded by $alpha$ in the above argument, and before that we saw that $E(alpha)$ contains at most two elements, since $E(alpha) subseteq x in S : x + alpha in mathbbQ$. This yields the trivial bound $operatornamecardbigl(E(alpha) cup E(beta)bigr) leqslant 4$ if $beta in Ssetminus E(alpha)$ used above. But in fact, we have the bound $operatornamecardbigl(E(alpha) cup E(beta)bigr) leqslant 3$. For if $alphabeta in mathbbQ$ and $alpha + x = r in mathbbQ$, then $xbeta = (r-alpha)beta = rbeta - alphabeta$ is irrational unless $r -alpha= 0$, i.e. $x = -alpha$. So if $E(alpha)$ contains two elements, at least one of them also belongs to $E(beta)$. And $E(alpha) cup E(beta)$ can only contain three elements if $-alpha in E(alpha)$ and $-beta in E(beta)$. So we can unconditionally lower the required cardinality to $6$, and we can lower it to $5$ under the condition that $S$ contains at most one pair of negatives.
We cannot lower the required cardinality further, because for any irrational $x$ whose square is also irrational, the set
$$S = x, -x, x^-1, -x^-1$$
has the property that for each pair of distinct elements either the sum or the product is rational, but none of the squares of the elements is rational.
edited Jul 28 at 14:02
Thomas Yeung Sin Chun
83
answered Jul 27 at 15:03
Daniel Fischer♦
171k16154274
add a comment |Â
up vote
1
down vote
accepted
If the set - call it $S$ - contains one rational number, say $r$, then all of the numbers are rational, for if $x in Ssetminus r$ such that $r + x = p in mathbbQ$, then $x = p-r in mathbbQ$, and if $x in Ssetminus r$ is such that $rcdot x = p in mathbbQ$, then $x = fracpr in mathbbQ$. In that case, it is clear that all squares are rational.
So we can assume that all elements of $S$ are irrational. For each $alpha in S$, there are at most two elements of $S$ whose sum with $alpha$ is rational. For if there were three or more, say $alpha + beta = r in mathbbQ$, $alpha + gamma = s in mathbbQ$ and $alpha + delta = t in mathbbQ$, then $beta + gamma = r+s - 2alpha notin mathbbQ$, $beta + delta = r+t-2alpha notin mathbbQ$ and $gamma + delta = s+t - 2alpha notin mathbbQ$, so by the assumption we would have $betadelta in mathbbQ$ and $gammadelta in mathbbQ$, and consequently
$$(beta - gamma)delta = (r-s)delta in mathbbQ,,$$
which by the irrationality of $delta$ implies $r = s$ and thus $beta = gamma$, contradicting our assumption that $beta,gamma,delta$ are distinct.
Thus, if $S$ contains at least $7$ elements, then for every $alpha in S$ there are $beta,gammain S$ such that $alphabeta, alphagamma, betagamma$ are all rational. (Picking $alpha$ excludes at most two numbers, and then choosing $beta$ excludes at most two more.) But then $beta = fracralpha$ for some $r in mathbbQ$, and $gamma = fracsbeta = fracsralpha$ for some $sin mathbbQ$, whence $$alpha^2 = fracrsalphagamma in mathbbQ,.$$
Actually, we can lower the cardinality required in this argument.
For $alpha in S$, let $E(alpha) = x in S : xalpha notin mathbbQ$. This is the set excluded by $alpha$ in the above argument, and before that we saw that $E(alpha)$ contains at most two elements, since $E(alpha) subseteq x in S : x + alpha in mathbbQ$. This yields the trivial bound $operatornamecardbigl(E(alpha) cup E(beta)bigr) leqslant 4$ if $beta in Ssetminus E(alpha)$ used above. But in fact, we have the bound $operatornamecardbigl(E(alpha) cup E(beta)bigr) leqslant 3$. For if $alphabeta in mathbbQ$ and $alpha + x = r in mathbbQ$, then $xbeta = (r-alpha)beta = rbeta - alphabeta$ is irrational unless $r -alpha= 0$, i.e. $x = -alpha$. So if $E(alpha)$ contains two elements, at least one of them also belongs to $E(beta)$. And $E(alpha) cup E(beta)$ can only contain three elements if $-alpha in E(alpha)$ and $-beta in E(beta)$. So we can unconditionally lower the required cardinality to $6$, and we can lower it to $5$ under the condition that $S$ contains at most one pair of negatives.
We cannot lower the required cardinality further, because for any irrational $x$ whose square is also irrational, the set
$$S = x, -x, x^-1, -x^-1$$
has the property that for each pair of distinct elements either the sum or the product is rational, but none of the squares of the elements is rational.
edited Jul 28 at 14:02
Thomas Yeung Sin Chun
83
answered Jul 27 at 15:03
Daniel Fischer♦
171k16154274
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If the set - call it $S$ - contains one rational number, say $r$, then all of the numbers are rational, for if $x in Ssetminus r$ such that $r + x = p in mathbbQ$, then $x = p-r in mathbbQ$, and if $x in Ssetminus r$ is such that $rcdot x = p in mathbbQ$, then $x = fracpr in mathbbQ$. In that case, it is clear that all squares are rational.
So we can assume that all elements of $S$ are irrational. For each $alpha in S$, there are at most two elements of $S$ whose sum with $alpha$ is rational. For if there were three or more, say $alpha + beta = r in mathbbQ$, $alpha + gamma = s in mathbbQ$ and $alpha + delta = t in mathbbQ$, then $beta + gamma = r+s - 2alpha notin mathbbQ$, $beta + delta = r+t-2alpha notin mathbbQ$ and $gamma + delta = s+t - 2alpha notin mathbbQ$, so by the assumption we would have $betadelta in mathbbQ$ and $gammadelta in mathbbQ$, and consequently
$$(beta - gamma)delta = (r-s)delta in mathbbQ,,$$
which by the irrationality of $delta$ implies $r = s$ and thus $beta = gamma$, contradicting our assumption that $beta,gamma,delta$ are distinct.
Thus, if $S$ contains at least $7$ elements, then for every $alpha in S$ there are $beta,gammain S$ such that $alphabeta, alphagamma, betagamma$ are all rational. (Picking $alpha$ excludes at most two numbers, and then choosing $beta$ excludes at most two more.) But then $beta = fracralpha$ for some $r in mathbbQ$, and $gamma = fracsbeta = fracsralpha$ for some $sin mathbbQ$, whence $$alpha^2 = fracrsalphagamma in mathbbQ,.$$
Actually, we can lower the cardinality required in this argument.
For $alpha in S$, let $E(alpha) = x in S : xalpha notin mathbbQ$. This is the set excluded by $alpha$ in the above argument, and before that we saw that $E(alpha)$ contains at most two elements, since $E(alpha) subseteq x in S : x + alpha in mathbbQ$. This yields the trivial bound $operatornamecardbigl(E(alpha) cup E(beta)bigr) leqslant 4$ if $beta in Ssetminus E(alpha)$ used above. But in fact, we have the bound $operatornamecardbigl(E(alpha) cup E(beta)bigr) leqslant 3$. For if $alphabeta in mathbbQ$ and $alpha + x = r in mathbbQ$, then $xbeta = (r-alpha)beta = rbeta - alphabeta$ is irrational unless $r -alpha= 0$, i.e. $x = -alpha$. So if $E(alpha)$ contains two elements, at least one of them also belongs to $E(beta)$. And $E(alpha) cup E(beta)$ can only contain three elements if $-alpha in E(alpha)$ and $-beta in E(beta)$. So we can unconditionally lower the required cardinality to $6$, and we can lower it to $5$ under the condition that $S$ contains at most one pair of negatives.
We cannot lower the required cardinality further, because for any irrational $x$ whose square is also irrational, the set
$$S = x, -x, x^-1, -x^-1$$
has the property that for each pair of distinct elements either the sum or the product is rational, but none of the squares of the elements is rational.
edited Jul 28 at 14:02
Thomas Yeung Sin Chun
83
answered Jul 27 at 15:03
Daniel Fischer♦
171k16154274
If the set - call it $S$ - contains one rational number, say $r$, then all of the numbers are rational, for if $x in Ssetminus r$ such that $r + x = p in mathbbQ$, then $x = p-r in mathbbQ$, and if $x in Ssetminus r$ is such that $rcdot x = p in mathbbQ$, then $x = fracpr in mathbbQ$. In that case, it is clear that all squares are rational.
So we can assume that all elements of $S$ are irrational. For each $alpha in S$, there are at most two elements of $S$ whose sum with $alpha$ is rational. For if there were three or more, say $alpha + beta = r in mathbbQ$, $alpha + gamma = s in mathbbQ$ and $alpha + delta = t in mathbbQ$, then $beta + gamma = r+s - 2alpha notin mathbbQ$, $beta + delta = r+t-2alpha notin mathbbQ$ and $gamma + delta = s+t - 2alpha notin mathbbQ$, so by the assumption we would have $betadelta in mathbbQ$ and $gammadelta in mathbbQ$, and consequently
$$(beta - gamma)delta = (r-s)delta in mathbbQ,,$$
which by the irrationality of $delta$ implies $r = s$ and thus $beta = gamma$, contradicting our assumption that $beta,gamma,delta$ are distinct.
Thus, if $S$ contains at least $7$ elements, then for every $alpha in S$ there are $beta,gammain S$ such that $alphabeta, alphagamma, betagamma$ are all rational. (Picking $alpha$ excludes at most two numbers, and then choosing $beta$ excludes at most two more.) But then $beta = fracralpha$ for some $r in mathbbQ$, and $gamma = fracsbeta = fracsralpha$ for some $sin mathbbQ$, whence $$alpha^2 = fracrsalphagamma in mathbbQ,.$$
Actually, we can lower the cardinality required in this argument.
For $alpha in S$, let $E(alpha) = x in S : xalpha notin mathbbQ$. This is the set excluded by $alpha$ in the above argument, and before that we saw that $E(alpha)$ contains at most two elements, since $E(alpha) subseteq x in S : x + alpha in mathbbQ$. This yields the trivial bound $operatornamecardbigl(E(alpha) cup E(beta)bigr) leqslant 4$ if $beta in Ssetminus E(alpha)$ used above. But in fact, we have the bound $operatornamecardbigl(E(alpha) cup E(beta)bigr) leqslant 3$. For if $alphabeta in mathbbQ$ and $alpha + x = r in mathbbQ$, then $xbeta = (r-alpha)beta = rbeta - alphabeta$ is irrational unless $r -alpha= 0$, i.e. $x = -alpha$. So if $E(alpha)$ contains two elements, at least one of them also belongs to $E(beta)$. And $E(alpha) cup E(beta)$ can only contain three elements if $-alpha in E(alpha)$ and $-beta in E(beta)$. So we can unconditionally lower the required cardinality to $6$, and we can lower it to $5$ under the condition that $S$ contains at most one pair of negatives.
We cannot lower the required cardinality further, because for any irrational $x$ whose square is also irrational, the set
$$S = x, -x, x^-1, -x^-1$$
has the property that for each pair of distinct elements either the sum or the product is rational, but none of the squares of the elements is rational.
edited Jul 28 at 14:02
Thomas Yeung Sin Chun
83
answered Jul 27 at 15:03
Daniel Fischer♦
171k16154274
edited Jul 28 at 14:02
Thomas Yeung Sin Chun
83
edited Jul 28 at 14:02
Thomas Yeung Sin Chun
83
edited Jul 28 at 14:02
Thomas Yeung Sin Chun
83
83
answered Jul 27 at 15:03
Daniel Fischer♦
171k16154274
answered Jul 27 at 15:03
Daniel Fischer♦
171k16154274
answered Jul 27 at 15:03
Daniel Fischer♦
171k16154274
171k16154274
add a comment |Â
add a comment |Â
up vote
1
down vote
Let $a$ be one of these numbers. Then $a^2 in mathbb Q$ or $2a in mathbb Q$.
In the first case we are done. In the second case we have $4a^2 in mathbb Q$, hence $a^2 in mathbb Q$.
answered Jul 27 at 6:28
Fred
37.2k1237
Sorry my question is not clear since the original question is in Chinese and I use google translate to translate it. In the original question, the two numbers chosen must not be the same number, I have already updated the question
– Thomas Yeung Sin Chun
Jul 27 at 8:56
add a comment |Â
up vote
1
down vote
Let $a$ be one of these numbers. Then $a^2 in mathbb Q$ or $2a in mathbb Q$.
In the first case we are done. In the second case we have $4a^2 in mathbb Q$, hence $a^2 in mathbb Q$.
answered Jul 27 at 6:28
Fred
37.2k1237
Sorry my question is not clear since the original question is in Chinese and I use google translate to translate it. In the original question, the two numbers chosen must not be the same number, I have already updated the question
– Thomas Yeung Sin Chun
Jul 27 at 8:56
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $a$ be one of these numbers. Then $a^2 in mathbb Q$ or $2a in mathbb Q$.
In the first case we are done. In the second case we have $4a^2 in mathbb Q$, hence $a^2 in mathbb Q$.
answered Jul 27 at 6:28
Fred
37.2k1237
Let $a$ be one of these numbers. Then $a^2 in mathbb Q$ or $2a in mathbb Q$.
In the first case we are done. In the second case we have $4a^2 in mathbb Q$, hence $a^2 in mathbb Q$.
answered Jul 27 at 6:28
Fred
37.2k1237
answered Jul 27 at 6:28
Fred
37.2k1237
answered Jul 27 at 6:28
Fred
37.2k1237
answered Jul 27 at 6:28
Fred
37.2k1237
37.2k1237
Sorry my question is not clear since the original question is in Chinese and I use google translate to translate it. In the original question, the two numbers chosen must not be the same number, I have already updated the question
– Thomas Yeung Sin Chun
Jul 27 at 8:56
add a comment |Â
Sorry my question is not clear since the original question is in Chinese and I use google translate to translate it. In the original question, the two numbers chosen must not be the same number, I have already updated the question
– Thomas Yeung Sin Chun
Jul 27 at 8:56
Sorry my question is not clear since the original question is in Chinese and I use google translate to translate it. In the original question, the two numbers chosen must not be the same number, I have already updated the question
– Thomas Yeung Sin Chun
Jul 27 at 8:56
Sorry my question is not clear since the original question is in Chinese and I use google translate to translate it. In the original question, the two numbers chosen must not be the same number, I have already updated the question
– Thomas Yeung Sin Chun
Jul 27 at 8:56
add a comment |Â
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2
"Proof-verification" is a tag used for questions about the validity of a certain proof. Since you have provided no proof, this would hardly be an appropriate tag. Also, it is customary here to convince the readers that you have put in some effort in order to solve the problem, so that the general spirit of the site is a true discussion of mathematical ideas, and not a "do-my-homework" service taken advantage of by lazy students who will never get anywhere anyway.
– uniquesolution
Jul 27 at 6:22
@MattiP. - and it fulfills the conclusion too, so what are you saying?
– uniquesolution
Jul 27 at 6:23
It's not very clear what you're asking. If I have a list $1, 2, 3, ldots , 196, 197$, it fulfills the requirements ...
– Matti P.
Jul 27 at 6:23
@MattiP. ditto .The assumption is that in this set of $197$ numbers, given any two, either their product or their sum is rational. You are required to prove that for each of the $197$ numbers, its square is rational.
– uniquesolution
Jul 27 at 6:24
1
@LordSharktheUnknown Obviously $i=j$ is excluded.
– uniquesolution
Jul 27 at 6:28