Mixing
Proving that if $x_1,dots,x_n$ are rational numbers and $sqrt{x_1}+dotssqrt{x_n}$ is rational, then each...
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I'm having a hard time with the following problem:
Let $x_1,x_2...x_n$ be rational numbers. Prove that if the sum $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational, then all $sqrt{x_i}$ are rational. Show, that the assumption for $x_i$ to be rational is necessary.
The only thing that I came up with is how to show this for n=2. Maybe there's some analogy for bigger n-s as well. If we assume that $sqrt{x_1}+sqrt{x_2}$ is rational then so must be $sqrt{x_1}-sqrt{x_2}$ (their product is rational). By adding those two together we get that both $sqrt{x_1}+sqrt{x_2}+sqrt{x_1}-sqrt{x_2} = 2sqrt{x_1}$ and $sqrt{x_1}+sqrt{x_2}-sqrt{x_1}+sqrt{x_2} = 2sqrt{x_2}$ are rational which implies the rationality of both $sqrt{x_1}$ and $sqrt{x_2}$
Any help is appreciated. Thank you.
BTW: I've also tried to prove by contradiction and induction. Both attempts didn't work..
rational-numbers
edited Jan 15 at 16:24
Asaf Karagila♦
305k32434764
asked Jan 14 at 19:30
The CatThe Cat
25112
$endgroup$
|
show 8 more comments
$begingroup$
I'm having a hard time with the following problem:
Let $x_1,x_2...x_n$ be rational numbers. Prove that if the sum $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational, then all $sqrt{x_i}$ are rational. Show, that the assumption for $x_i$ to be rational is necessary.
The only thing that I came up with is how to show this for n=2. Maybe there's some analogy for bigger n-s as well. If we assume that $sqrt{x_1}+sqrt{x_2}$ is rational then so must be $sqrt{x_1}-sqrt{x_2}$ (their product is rational). By adding those two together we get that both $sqrt{x_1}+sqrt{x_2}+sqrt{x_1}-sqrt{x_2} = 2sqrt{x_1}$ and $sqrt{x_1}+sqrt{x_2}-sqrt{x_1}+sqrt{x_2} = 2sqrt{x_2}$ are rational which implies the rationality of both $sqrt{x_1}$ and $sqrt{x_2}$
Any help is appreciated. Thank you.
BTW: I've also tried to prove by contradiction and induction. Both attempts didn't work..
rational-numbers
edited Jan 15 at 16:24
Asaf Karagila♦
305k32434764
asked Jan 14 at 19:30
The CatThe Cat
25112
$endgroup$
1
$begingroup$
Could you clarify your proof for the case of $n=2$?
$endgroup$
– ShapeOfMatter
Jan 14 at 19:32
1
$begingroup$
If you can show this for $n=2$, you should be able to prove by induction. Can you include your attempt?
$endgroup$
– Morgan Rodgers
Jan 14 at 19:33
1
$begingroup$
I tried using induction the following way: for n=1 this is trivial. Our induction hypothesis looks like this: If $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational then all $sqrt{x_i}$ are rational. Now i would like to that that if $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational then so are all $sqrt{x_i}$. I cannot see how the induction hypothesis would help to be honest. If the sum $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational we cannot conclude that $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational as well so as to use the hypothesis.
$endgroup$
– The Cat
Jan 14 at 19:49
2
$begingroup$
Related: math.stackexchange.com/q/30687/42969.
$endgroup$
– Martin R
Jan 14 at 19:51
5
$begingroup$
I would be legitimately surprised (and delighted) to see a good high school-level proof of this fact. Where did you come across the problem?
$endgroup$
– Steven Stadnicki
Jan 14 at 20:17
|
show 8 more comments
$begingroup$
I'm having a hard time with the following problem:
Let $x_1,x_2...x_n$ be rational numbers. Prove that if the sum $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational, then all $sqrt{x_i}$ are rational. Show, that the assumption for $x_i$ to be rational is necessary.
The only thing that I came up with is how to show this for n=2. Maybe there's some analogy for bigger n-s as well. If we assume that $sqrt{x_1}+sqrt{x_2}$ is rational then so must be $sqrt{x_1}-sqrt{x_2}$ (their product is rational). By adding those two together we get that both $sqrt{x_1}+sqrt{x_2}+sqrt{x_1}-sqrt{x_2} = 2sqrt{x_1}$ and $sqrt{x_1}+sqrt{x_2}-sqrt{x_1}+sqrt{x_2} = 2sqrt{x_2}$ are rational which implies the rationality of both $sqrt{x_1}$ and $sqrt{x_2}$
Any help is appreciated. Thank you.
BTW: I've also tried to prove by contradiction and induction. Both attempts didn't work..
rational-numbers
edited Jan 15 at 16:24
Asaf Karagila♦
305k32434764
asked Jan 14 at 19:30
The CatThe Cat
25112
$endgroup$
I'm having a hard time with the following problem:
Let $x_1,x_2...x_n$ be rational numbers. Prove that if the sum $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational, then all $sqrt{x_i}$ are rational. Show, that the assumption for $x_i$ to be rational is necessary.
The only thing that I came up with is how to show this for n=2. Maybe there's some analogy for bigger n-s as well. If we assume that $sqrt{x_1}+sqrt{x_2}$ is rational then so must be $sqrt{x_1}-sqrt{x_2}$ (their product is rational). By adding those two together we get that both $sqrt{x_1}+sqrt{x_2}+sqrt{x_1}-sqrt{x_2} = 2sqrt{x_1}$ and $sqrt{x_1}+sqrt{x_2}-sqrt{x_1}+sqrt{x_2} = 2sqrt{x_2}$ are rational which implies the rationality of both $sqrt{x_1}$ and $sqrt{x_2}$
Any help is appreciated. Thank you.
BTW: I've also tried to prove by contradiction and induction. Both attempts didn't work..
rational-numbers
rational-numbers
edited Jan 15 at 16:24
Asaf Karagila♦
305k32434764
asked Jan 14 at 19:30
The CatThe Cat
25112
edited Jan 15 at 16:24
Asaf Karagila♦
305k32434764
asked Jan 14 at 19:30
The CatThe Cat
25112
edited Jan 15 at 16:24
Asaf Karagila♦
305k32434764
edited Jan 15 at 16:24
Asaf Karagila♦
305k32434764
edited Jan 15 at 16:24
Asaf Karagila♦
305k32434764
305k32434764
asked Jan 14 at 19:30
The CatThe Cat
25112
asked Jan 14 at 19:30
The CatThe Cat
25112
asked Jan 14 at 19:30
The CatThe Cat
25112
25112
1
$begingroup$
Could you clarify your proof for the case of $n=2$?
$endgroup$
– ShapeOfMatter
Jan 14 at 19:32
1
$begingroup$
If you can show this for $n=2$, you should be able to prove by induction. Can you include your attempt?
$endgroup$
– Morgan Rodgers
Jan 14 at 19:33
1
$begingroup$
I tried using induction the following way: for n=1 this is trivial. Our induction hypothesis looks like this: If $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational then all $sqrt{x_i}$ are rational. Now i would like to that that if $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational then so are all $sqrt{x_i}$. I cannot see how the induction hypothesis would help to be honest. If the sum $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational we cannot conclude that $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational as well so as to use the hypothesis.
$endgroup$
– The Cat
Jan 14 at 19:49
2
$begingroup$
Related: math.stackexchange.com/q/30687/42969.
$endgroup$
– Martin R
Jan 14 at 19:51
5
$begingroup$
I would be legitimately surprised (and delighted) to see a good high school-level proof of this fact. Where did you come across the problem?
$endgroup$
– Steven Stadnicki
Jan 14 at 20:17
|
show 8 more comments
1
$begingroup$
Could you clarify your proof for the case of $n=2$?
$endgroup$
– ShapeOfMatter
Jan 14 at 19:32
1
$begingroup$
If you can show this for $n=2$, you should be able to prove by induction. Can you include your attempt?
$endgroup$
– Morgan Rodgers
Jan 14 at 19:33
1
$begingroup$
I tried using induction the following way: for n=1 this is trivial. Our induction hypothesis looks like this: If $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational then all $sqrt{x_i}$ are rational. Now i would like to that that if $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational then so are all $sqrt{x_i}$. I cannot see how the induction hypothesis would help to be honest. If the sum $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational we cannot conclude that $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational as well so as to use the hypothesis.
$endgroup$
– The Cat
Jan 14 at 19:49
2
$begingroup$
Related: math.stackexchange.com/q/30687/42969.
$endgroup$
– Martin R
Jan 14 at 19:51
5
$begingroup$
I would be legitimately surprised (and delighted) to see a good high school-level proof of this fact. Where did you come across the problem?
$endgroup$
– Steven Stadnicki
Jan 14 at 20:17
1
1
$begingroup$
Could you clarify your proof for the case of $n=2$?
$endgroup$
– ShapeOfMatter
Jan 14 at 19:32
$begingroup$
Could you clarify your proof for the case of $n=2$?
$endgroup$
– ShapeOfMatter
Jan 14 at 19:32
1
1
$begingroup$
If you can show this for $n=2$, you should be able to prove by induction. Can you include your attempt?
$endgroup$
– Morgan Rodgers
Jan 14 at 19:33
$begingroup$
If you can show this for $n=2$, you should be able to prove by induction. Can you include your attempt?
$endgroup$
– Morgan Rodgers
Jan 14 at 19:33
1
1
$begingroup$
I tried using induction the following way: for n=1 this is trivial. Our induction hypothesis looks like this: If $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational then all $sqrt{x_i}$ are rational. Now i would like to that that if $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational then so are all $sqrt{x_i}$. I cannot see how the induction hypothesis would help to be honest. If the sum $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational we cannot conclude that $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational as well so as to use the hypothesis.
$endgroup$
– The Cat
Jan 14 at 19:49
$begingroup$
I tried using induction the following way: for n=1 this is trivial. Our induction hypothesis looks like this: If $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational then all $sqrt{x_i}$ are rational. Now i would like to that that if $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational then so are all $sqrt{x_i}$. I cannot see how the induction hypothesis would help to be honest. If the sum $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational we cannot conclude that $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational as well so as to use the hypothesis.
$endgroup$
– The Cat
Jan 14 at 19:49
2
2
$begingroup$
Related: math.stackexchange.com/q/30687/42969.
$endgroup$
– Martin R
Jan 14 at 19:51
$begingroup$
Related: math.stackexchange.com/q/30687/42969.
$endgroup$
– Martin R
Jan 14 at 19:51
5
5
$begingroup$
I would be legitimately surprised (and delighted) to see a good high school-level proof of this fact. Where did you come across the problem?
$endgroup$
– Steven Stadnicki
Jan 14 at 20:17
$begingroup$
I would be legitimately surprised (and delighted) to see a good high school-level proof of this fact. Where did you come across the problem?
$endgroup$
– Steven Stadnicki
Jan 14 at 20:17
|
show 8 more comments
2 Answers
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Here I show how to generalize the argument you gave for $,n=2,$ to general n. It uses very simple field theory. Since you remark in a comment that you are in high-school so wish to avoid field theory I will explain what's needed below, and work through a special case of the linked proof for motivation.
As with many inductive proofs, they key is to strengthen the inductive hypothesis, which here means proving the statement not only for rational numbers $,Bbb Q,$ but also for larger "number systems" of real numbers that are obtained by adjoining square roots of positive numbers.
For example $,Bbb Q(sqrt 5),$ denotes the reals obtainable by (field) arithmetic on rationals $,Bbb Q,$ and $,sqrt 5,$, where field arithmetic consists of the operations of addition, multiplication and division $,a/b,, bneq 0.,$ It is easy to show that the reals obtainable by iterating these operations are exactly those writable in the form $,a+bsqrt{5},$ for $,a,bin Bbb Q,$ (for division we can rationalize the denominator). We can iterate this construction, e.g. adjoining $,sqrt 3,$ to $,F = Bbb Q(sqrt 5)$ to get $,F(sqrt 3),$ with numbers $,a+bsqrt 3,$ for $,a,bin Bbb Q(sqrt 5)$. This step-by-step construction of such towers of number systems proves very handy for inductive proofs (a special case of structural induction).
For motivation, we show how the induction step works to reduce the case $n=3$ to $n=2$ (your result). The induction step in the general proof works exactly the same way.
Suppose $sqrt 2 + sqrt 3 + sqrt 5 = qin Bbb Q.,$ It suffices to show one summand $in Bbb Q,$ since then the sum of the other two is in $,Bbb Q,$ so induction (your $n=2$ proof) shows they too are in $,Bbb Q$.
$,sqrt 2 + sqrt 3 = q-sqrt 5 in Bbb Q(sqrt 5) = { a + bsqrt 5 : a,binBbb Q} $ so by induction $,sqrt 2,sqrt 3in Bbb Q(sqrt 5),$ so
$$begin{align}
sqrt{2} = a_2 + b_2 sqrt{5}, a_2,b_2in Bbb Q\
sqrt{3} = a_3 + b_3 sqrt{5}, a_3,b_3in Bbb Q
end{align}$$
If $,b_3 < 0,$ then $, a_3 = sqrt 3 - b_3sqrt 5 = sqrt 3 +! sqrt{5b_3^2}in Bbb Q,Rightarrow, sqrt 3inBbb Q,$ by induction. Ditto if $,b_2 < 0,$
Else all $,b_i ge 0,$ so $,q = sqrt 2! +! sqrt 3! +! sqrt 5 = a_2!+!a_3+(b_2!+!b_3!+!1)sqrt 5,Rightarrow,sqrt 5 in Bbb Q $ by solving for $,sqrt 5,,$ using $,b_2!+!b_3!+!1 neq 0,$ by all $,b_ige 0. $
Thus in every case some summand $in Bbb Q,,$ which completes the proof.
answered Jan 15 at 0:57
Bill DubuqueBill Dubuque
211k29192645
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The case $n=3$ is actually very easy and requires no field theory, explicit or implicit.
If $sqrt{x_1}+sqrt{x_2}+sqrt{x_3}=a$ is rational, then moving $sqrt{x_3}$ to the right-hand side and squaring we get
$$ 2sqrt{x_1x_2} = a_1-2asqrt{x_3}, $$
where $a_1=a^2+x_3-x_1-x_2$ is rational. Squaring again,
$$ a_2 = -4aa_1sqrt{x_3} $$
with $a_2=4x_1x_2-4a^2x_3-a_1^2$. Since $a>0$, it follows that either $sqrt{x_3}$ is rational, or $a_1=a_2=0$. In the former case we are done, in the latter case $x_1x_2=a^2x_3$ and also $a^2+x_3-x_1-x_2=0$. Excluding $a^2$ we then get $x_3in{x_1,x_2}$. Thus, either $sqrt{4x_1}+sqrt{x_2}$, or $sqrt{x_1}+sqrt{4x_2}$ is rational, and the claim follows by induction.
In the same way it should be possible to show that if none of the products $x_ix_j$ is a square, and $alpha_1sqrt{x_1}+alpha_2sqrt{x_2}+alpha_3sqrt{x_3}$ is rational, then in fact all summands are rational. (Here $alpha_i$ and $x_ige 0$ are rational.)
In fact, I strongly suspect that one should be able to prove the general claim, with $n$ summands, using nothing but induction. The trick is, the induction should be by the rank of the group generated (multiplicatively) by $x_1,dotsc,x_n$, not by $n$. This, however, would be a little longer to describe in details.
answered Jan 15 at 15:43
W-t-PW-t-P
1,137611
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$begingroup$
Yes, with some (nontrivial) effort we can generalize elementary methods like that, e.g. see this elementary proof. But one pays a price for avoiding (nontrivial) field theory - namely increased induction complexity.
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– Bill Dubuque
Jan 15 at 16:54
add a comment |
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2 Answers
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Here I show how to generalize the argument you gave for $,n=2,$ to general n. It uses very simple field theory. Since you remark in a comment that you are in high-school so wish to avoid field theory I will explain what's needed below, and work through a special case of the linked proof for motivation.
As with many inductive proofs, they key is to strengthen the inductive hypothesis, which here means proving the statement not only for rational numbers $,Bbb Q,$ but also for larger "number systems" of real numbers that are obtained by adjoining square roots of positive numbers.
For example $,Bbb Q(sqrt 5),$ denotes the reals obtainable by (field) arithmetic on rationals $,Bbb Q,$ and $,sqrt 5,$, where field arithmetic consists of the operations of addition, multiplication and division $,a/b,, bneq 0.,$ It is easy to show that the reals obtainable by iterating these operations are exactly those writable in the form $,a+bsqrt{5},$ for $,a,bin Bbb Q,$ (for division we can rationalize the denominator). We can iterate this construction, e.g. adjoining $,sqrt 3,$ to $,F = Bbb Q(sqrt 5)$ to get $,F(sqrt 3),$ with numbers $,a+bsqrt 3,$ for $,a,bin Bbb Q(sqrt 5)$. This step-by-step construction of such towers of number systems proves very handy for inductive proofs (a special case of structural induction).
For motivation, we show how the induction step works to reduce the case $n=3$ to $n=2$ (your result). The induction step in the general proof works exactly the same way.
Suppose $sqrt 2 + sqrt 3 + sqrt 5 = qin Bbb Q.,$ It suffices to show one summand $in Bbb Q,$ since then the sum of the other two is in $,Bbb Q,$ so induction (your $n=2$ proof) shows they too are in $,Bbb Q$.
$,sqrt 2 + sqrt 3 = q-sqrt 5 in Bbb Q(sqrt 5) = { a + bsqrt 5 : a,binBbb Q} $ so by induction $,sqrt 2,sqrt 3in Bbb Q(sqrt 5),$ so
$$begin{align}
sqrt{2} = a_2 + b_2 sqrt{5}, a_2,b_2in Bbb Q\
sqrt{3} = a_3 + b_3 sqrt{5}, a_3,b_3in Bbb Q
end{align}$$
If $,b_3 < 0,$ then $, a_3 = sqrt 3 - b_3sqrt 5 = sqrt 3 +! sqrt{5b_3^2}in Bbb Q,Rightarrow, sqrt 3inBbb Q,$ by induction. Ditto if $,b_2 < 0,$
Else all $,b_i ge 0,$ so $,q = sqrt 2! +! sqrt 3! +! sqrt 5 = a_2!+!a_3+(b_2!+!b_3!+!1)sqrt 5,Rightarrow,sqrt 5 in Bbb Q $ by solving for $,sqrt 5,,$ using $,b_2!+!b_3!+!1 neq 0,$ by all $,b_ige 0. $
Thus in every case some summand $in Bbb Q,,$ which completes the proof.
answered Jan 15 at 0:57
Bill DubuqueBill Dubuque
211k29192645
$endgroup$
add a comment |
$begingroup$
Here I show how to generalize the argument you gave for $,n=2,$ to general n. It uses very simple field theory. Since you remark in a comment that you are in high-school so wish to avoid field theory I will explain what's needed below, and work through a special case of the linked proof for motivation.
As with many inductive proofs, they key is to strengthen the inductive hypothesis, which here means proving the statement not only for rational numbers $,Bbb Q,$ but also for larger "number systems" of real numbers that are obtained by adjoining square roots of positive numbers.
For example $,Bbb Q(sqrt 5),$ denotes the reals obtainable by (field) arithmetic on rationals $,Bbb Q,$ and $,sqrt 5,$, where field arithmetic consists of the operations of addition, multiplication and division $,a/b,, bneq 0.,$ It is easy to show that the reals obtainable by iterating these operations are exactly those writable in the form $,a+bsqrt{5},$ for $,a,bin Bbb Q,$ (for division we can rationalize the denominator). We can iterate this construction, e.g. adjoining $,sqrt 3,$ to $,F = Bbb Q(sqrt 5)$ to get $,F(sqrt 3),$ with numbers $,a+bsqrt 3,$ for $,a,bin Bbb Q(sqrt 5)$. This step-by-step construction of such towers of number systems proves very handy for inductive proofs (a special case of structural induction).
For motivation, we show how the induction step works to reduce the case $n=3$ to $n=2$ (your result). The induction step in the general proof works exactly the same way.
Suppose $sqrt 2 + sqrt 3 + sqrt 5 = qin Bbb Q.,$ It suffices to show one summand $in Bbb Q,$ since then the sum of the other two is in $,Bbb Q,$ so induction (your $n=2$ proof) shows they too are in $,Bbb Q$.
$,sqrt 2 + sqrt 3 = q-sqrt 5 in Bbb Q(sqrt 5) = { a + bsqrt 5 : a,binBbb Q} $ so by induction $,sqrt 2,sqrt 3in Bbb Q(sqrt 5),$ so
$$begin{align}
sqrt{2} = a_2 + b_2 sqrt{5}, a_2,b_2in Bbb Q\
sqrt{3} = a_3 + b_3 sqrt{5}, a_3,b_3in Bbb Q
end{align}$$
If $,b_3 < 0,$ then $, a_3 = sqrt 3 - b_3sqrt 5 = sqrt 3 +! sqrt{5b_3^2}in Bbb Q,Rightarrow, sqrt 3inBbb Q,$ by induction. Ditto if $,b_2 < 0,$
Else all $,b_i ge 0,$ so $,q = sqrt 2! +! sqrt 3! +! sqrt 5 = a_2!+!a_3+(b_2!+!b_3!+!1)sqrt 5,Rightarrow,sqrt 5 in Bbb Q $ by solving for $,sqrt 5,,$ using $,b_2!+!b_3!+!1 neq 0,$ by all $,b_ige 0. $
Thus in every case some summand $in Bbb Q,,$ which completes the proof.
answered Jan 15 at 0:57
Bill DubuqueBill Dubuque
211k29192645
$endgroup$
add a comment |
$begingroup$
Here I show how to generalize the argument you gave for $,n=2,$ to general n. It uses very simple field theory. Since you remark in a comment that you are in high-school so wish to avoid field theory I will explain what's needed below, and work through a special case of the linked proof for motivation.
As with many inductive proofs, they key is to strengthen the inductive hypothesis, which here means proving the statement not only for rational numbers $,Bbb Q,$ but also for larger "number systems" of real numbers that are obtained by adjoining square roots of positive numbers.
For example $,Bbb Q(sqrt 5),$ denotes the reals obtainable by (field) arithmetic on rationals $,Bbb Q,$ and $,sqrt 5,$, where field arithmetic consists of the operations of addition, multiplication and division $,a/b,, bneq 0.,$ It is easy to show that the reals obtainable by iterating these operations are exactly those writable in the form $,a+bsqrt{5},$ for $,a,bin Bbb Q,$ (for division we can rationalize the denominator). We can iterate this construction, e.g. adjoining $,sqrt 3,$ to $,F = Bbb Q(sqrt 5)$ to get $,F(sqrt 3),$ with numbers $,a+bsqrt 3,$ for $,a,bin Bbb Q(sqrt 5)$. This step-by-step construction of such towers of number systems proves very handy for inductive proofs (a special case of structural induction).
For motivation, we show how the induction step works to reduce the case $n=3$ to $n=2$ (your result). The induction step in the general proof works exactly the same way.
Suppose $sqrt 2 + sqrt 3 + sqrt 5 = qin Bbb Q.,$ It suffices to show one summand $in Bbb Q,$ since then the sum of the other two is in $,Bbb Q,$ so induction (your $n=2$ proof) shows they too are in $,Bbb Q$.
$,sqrt 2 + sqrt 3 = q-sqrt 5 in Bbb Q(sqrt 5) = { a + bsqrt 5 : a,binBbb Q} $ so by induction $,sqrt 2,sqrt 3in Bbb Q(sqrt 5),$ so
$$begin{align}
sqrt{2} = a_2 + b_2 sqrt{5}, a_2,b_2in Bbb Q\
sqrt{3} = a_3 + b_3 sqrt{5}, a_3,b_3in Bbb Q
end{align}$$
If $,b_3 < 0,$ then $, a_3 = sqrt 3 - b_3sqrt 5 = sqrt 3 +! sqrt{5b_3^2}in Bbb Q,Rightarrow, sqrt 3inBbb Q,$ by induction. Ditto if $,b_2 < 0,$
Else all $,b_i ge 0,$ so $,q = sqrt 2! +! sqrt 3! +! sqrt 5 = a_2!+!a_3+(b_2!+!b_3!+!1)sqrt 5,Rightarrow,sqrt 5 in Bbb Q $ by solving for $,sqrt 5,,$ using $,b_2!+!b_3!+!1 neq 0,$ by all $,b_ige 0. $
Thus in every case some summand $in Bbb Q,,$ which completes the proof.
answered Jan 15 at 0:57
Bill DubuqueBill Dubuque
211k29192645
$endgroup$
Here I show how to generalize the argument you gave for $,n=2,$ to general n. It uses very simple field theory. Since you remark in a comment that you are in high-school so wish to avoid field theory I will explain what's needed below, and work through a special case of the linked proof for motivation.
As with many inductive proofs, they key is to strengthen the inductive hypothesis, which here means proving the statement not only for rational numbers $,Bbb Q,$ but also for larger "number systems" of real numbers that are obtained by adjoining square roots of positive numbers.
For example $,Bbb Q(sqrt 5),$ denotes the reals obtainable by (field) arithmetic on rationals $,Bbb Q,$ and $,sqrt 5,$, where field arithmetic consists of the operations of addition, multiplication and division $,a/b,, bneq 0.,$ It is easy to show that the reals obtainable by iterating these operations are exactly those writable in the form $,a+bsqrt{5},$ for $,a,bin Bbb Q,$ (for division we can rationalize the denominator). We can iterate this construction, e.g. adjoining $,sqrt 3,$ to $,F = Bbb Q(sqrt 5)$ to get $,F(sqrt 3),$ with numbers $,a+bsqrt 3,$ for $,a,bin Bbb Q(sqrt 5)$. This step-by-step construction of such towers of number systems proves very handy for inductive proofs (a special case of structural induction).
For motivation, we show how the induction step works to reduce the case $n=3$ to $n=2$ (your result). The induction step in the general proof works exactly the same way.
Suppose $sqrt 2 + sqrt 3 + sqrt 5 = qin Bbb Q.,$ It suffices to show one summand $in Bbb Q,$ since then the sum of the other two is in $,Bbb Q,$ so induction (your $n=2$ proof) shows they too are in $,Bbb Q$.
$,sqrt 2 + sqrt 3 = q-sqrt 5 in Bbb Q(sqrt 5) = { a + bsqrt 5 : a,binBbb Q} $ so by induction $,sqrt 2,sqrt 3in Bbb Q(sqrt 5),$ so
$$begin{align}
sqrt{2} = a_2 + b_2 sqrt{5}, a_2,b_2in Bbb Q\
sqrt{3} = a_3 + b_3 sqrt{5}, a_3,b_3in Bbb Q
end{align}$$
If $,b_3 < 0,$ then $, a_3 = sqrt 3 - b_3sqrt 5 = sqrt 3 +! sqrt{5b_3^2}in Bbb Q,Rightarrow, sqrt 3inBbb Q,$ by induction. Ditto if $,b_2 < 0,$
Else all $,b_i ge 0,$ so $,q = sqrt 2! +! sqrt 3! +! sqrt 5 = a_2!+!a_3+(b_2!+!b_3!+!1)sqrt 5,Rightarrow,sqrt 5 in Bbb Q $ by solving for $,sqrt 5,,$ using $,b_2!+!b_3!+!1 neq 0,$ by all $,b_ige 0. $
Thus in every case some summand $in Bbb Q,,$ which completes the proof.
answered Jan 15 at 0:57
Bill DubuqueBill Dubuque
211k29192645
edited Jan 15 at 1:11
answered Jan 15 at 0:57
Bill DubuqueBill Dubuque
211k29192645
answered Jan 15 at 0:57
Bill DubuqueBill Dubuque
211k29192645
answered Jan 15 at 0:57
Bill DubuqueBill Dubuque
211k29192645
211k29192645
add a comment |
add a comment |
$begingroup$
The case $n=3$ is actually very easy and requires no field theory, explicit or implicit.
If $sqrt{x_1}+sqrt{x_2}+sqrt{x_3}=a$ is rational, then moving $sqrt{x_3}$ to the right-hand side and squaring we get
$$ 2sqrt{x_1x_2} = a_1-2asqrt{x_3}, $$
where $a_1=a^2+x_3-x_1-x_2$ is rational. Squaring again,
$$ a_2 = -4aa_1sqrt{x_3} $$
with $a_2=4x_1x_2-4a^2x_3-a_1^2$. Since $a>0$, it follows that either $sqrt{x_3}$ is rational, or $a_1=a_2=0$. In the former case we are done, in the latter case $x_1x_2=a^2x_3$ and also $a^2+x_3-x_1-x_2=0$. Excluding $a^2$ we then get $x_3in{x_1,x_2}$. Thus, either $sqrt{4x_1}+sqrt{x_2}$, or $sqrt{x_1}+sqrt{4x_2}$ is rational, and the claim follows by induction.
In the same way it should be possible to show that if none of the products $x_ix_j$ is a square, and $alpha_1sqrt{x_1}+alpha_2sqrt{x_2}+alpha_3sqrt{x_3}$ is rational, then in fact all summands are rational. (Here $alpha_i$ and $x_ige 0$ are rational.)
In fact, I strongly suspect that one should be able to prove the general claim, with $n$ summands, using nothing but induction. The trick is, the induction should be by the rank of the group generated (multiplicatively) by $x_1,dotsc,x_n$, not by $n$. This, however, would be a little longer to describe in details.
answered Jan 15 at 15:43
W-t-PW-t-P
1,137611
$endgroup$
$begingroup$
Yes, with some (nontrivial) effort we can generalize elementary methods like that, e.g. see this elementary proof. But one pays a price for avoiding (nontrivial) field theory - namely increased induction complexity.
$endgroup$
– Bill Dubuque
Jan 15 at 16:54
add a comment |
$begingroup$
The case $n=3$ is actually very easy and requires no field theory, explicit or implicit.
If $sqrt{x_1}+sqrt{x_2}+sqrt{x_3}=a$ is rational, then moving $sqrt{x_3}$ to the right-hand side and squaring we get
$$ 2sqrt{x_1x_2} = a_1-2asqrt{x_3}, $$
where $a_1=a^2+x_3-x_1-x_2$ is rational. Squaring again,
$$ a_2 = -4aa_1sqrt{x_3} $$
with $a_2=4x_1x_2-4a^2x_3-a_1^2$. Since $a>0$, it follows that either $sqrt{x_3}$ is rational, or $a_1=a_2=0$. In the former case we are done, in the latter case $x_1x_2=a^2x_3$ and also $a^2+x_3-x_1-x_2=0$. Excluding $a^2$ we then get $x_3in{x_1,x_2}$. Thus, either $sqrt{4x_1}+sqrt{x_2}$, or $sqrt{x_1}+sqrt{4x_2}$ is rational, and the claim follows by induction.
In the same way it should be possible to show that if none of the products $x_ix_j$ is a square, and $alpha_1sqrt{x_1}+alpha_2sqrt{x_2}+alpha_3sqrt{x_3}$ is rational, then in fact all summands are rational. (Here $alpha_i$ and $x_ige 0$ are rational.)
In fact, I strongly suspect that one should be able to prove the general claim, with $n$ summands, using nothing but induction. The trick is, the induction should be by the rank of the group generated (multiplicatively) by $x_1,dotsc,x_n$, not by $n$. This, however, would be a little longer to describe in details.
answered Jan 15 at 15:43
W-t-PW-t-P
1,137611
$endgroup$
$begingroup$
Yes, with some (nontrivial) effort we can generalize elementary methods like that, e.g. see this elementary proof. But one pays a price for avoiding (nontrivial) field theory - namely increased induction complexity.
$endgroup$
– Bill Dubuque
Jan 15 at 16:54
add a comment |
$begingroup$
The case $n=3$ is actually very easy and requires no field theory, explicit or implicit.
If $sqrt{x_1}+sqrt{x_2}+sqrt{x_3}=a$ is rational, then moving $sqrt{x_3}$ to the right-hand side and squaring we get
$$ 2sqrt{x_1x_2} = a_1-2asqrt{x_3}, $$
where $a_1=a^2+x_3-x_1-x_2$ is rational. Squaring again,
$$ a_2 = -4aa_1sqrt{x_3} $$
with $a_2=4x_1x_2-4a^2x_3-a_1^2$. Since $a>0$, it follows that either $sqrt{x_3}$ is rational, or $a_1=a_2=0$. In the former case we are done, in the latter case $x_1x_2=a^2x_3$ and also $a^2+x_3-x_1-x_2=0$. Excluding $a^2$ we then get $x_3in{x_1,x_2}$. Thus, either $sqrt{4x_1}+sqrt{x_2}$, or $sqrt{x_1}+sqrt{4x_2}$ is rational, and the claim follows by induction.
In the same way it should be possible to show that if none of the products $x_ix_j$ is a square, and $alpha_1sqrt{x_1}+alpha_2sqrt{x_2}+alpha_3sqrt{x_3}$ is rational, then in fact all summands are rational. (Here $alpha_i$ and $x_ige 0$ are rational.)
In fact, I strongly suspect that one should be able to prove the general claim, with $n$ summands, using nothing but induction. The trick is, the induction should be by the rank of the group generated (multiplicatively) by $x_1,dotsc,x_n$, not by $n$. This, however, would be a little longer to describe in details.
answered Jan 15 at 15:43
W-t-PW-t-P
1,137611
$endgroup$
The case $n=3$ is actually very easy and requires no field theory, explicit or implicit.
If $sqrt{x_1}+sqrt{x_2}+sqrt{x_3}=a$ is rational, then moving $sqrt{x_3}$ to the right-hand side and squaring we get
$$ 2sqrt{x_1x_2} = a_1-2asqrt{x_3}, $$
where $a_1=a^2+x_3-x_1-x_2$ is rational. Squaring again,
$$ a_2 = -4aa_1sqrt{x_3} $$
with $a_2=4x_1x_2-4a^2x_3-a_1^2$. Since $a>0$, it follows that either $sqrt{x_3}$ is rational, or $a_1=a_2=0$. In the former case we are done, in the latter case $x_1x_2=a^2x_3$ and also $a^2+x_3-x_1-x_2=0$. Excluding $a^2$ we then get $x_3in{x_1,x_2}$. Thus, either $sqrt{4x_1}+sqrt{x_2}$, or $sqrt{x_1}+sqrt{4x_2}$ is rational, and the claim follows by induction.
In the same way it should be possible to show that if none of the products $x_ix_j$ is a square, and $alpha_1sqrt{x_1}+alpha_2sqrt{x_2}+alpha_3sqrt{x_3}$ is rational, then in fact all summands are rational. (Here $alpha_i$ and $x_ige 0$ are rational.)
In fact, I strongly suspect that one should be able to prove the general claim, with $n$ summands, using nothing but induction. The trick is, the induction should be by the rank of the group generated (multiplicatively) by $x_1,dotsc,x_n$, not by $n$. This, however, would be a little longer to describe in details.
answered Jan 15 at 15:43
W-t-PW-t-P
1,137611
answered Jan 15 at 15:43
W-t-PW-t-P
1,137611
answered Jan 15 at 15:43
W-t-PW-t-P
1,137611
answered Jan 15 at 15:43
W-t-PW-t-P
1,137611
1,137611
$begingroup$
Yes, with some (nontrivial) effort we can generalize elementary methods like that, e.g. see this elementary proof. But one pays a price for avoiding (nontrivial) field theory - namely increased induction complexity.
$endgroup$
– Bill Dubuque
Jan 15 at 16:54
add a comment |
$begingroup$
Yes, with some (nontrivial) effort we can generalize elementary methods like that, e.g. see this elementary proof. But one pays a price for avoiding (nontrivial) field theory - namely increased induction complexity.
$endgroup$
– Bill Dubuque
Jan 15 at 16:54
$begingroup$
Yes, with some (nontrivial) effort we can generalize elementary methods like that, e.g. see this elementary proof. But one pays a price for avoiding (nontrivial) field theory - namely increased induction complexity.
$endgroup$
– Bill Dubuque
Jan 15 at 16:54
$begingroup$
Yes, with some (nontrivial) effort we can generalize elementary methods like that, e.g. see this elementary proof. But one pays a price for avoiding (nontrivial) field theory - namely increased induction complexity.
$endgroup$
– Bill Dubuque
Jan 15 at 16:54
add a comment |
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1
$begingroup$
Could you clarify your proof for the case of $n=2$?
$endgroup$
– ShapeOfMatter
Jan 14 at 19:32
1
$begingroup$
If you can show this for $n=2$, you should be able to prove by induction. Can you include your attempt?
$endgroup$
– Morgan Rodgers
Jan 14 at 19:33
1
$begingroup$
I tried using induction the following way: for n=1 this is trivial. Our induction hypothesis looks like this: If $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational then all $sqrt{x_i}$ are rational. Now i would like to that that if $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational then so are all $sqrt{x_i}$. I cannot see how the induction hypothesis would help to be honest. If the sum $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}+sqrt{x_{n+1}}$ is rational we cannot conclude that $sqrt{x_1}+sqrt{x_2}+...+sqrt{x_n}$ is rational as well so as to use the hypothesis.
$endgroup$
– The Cat
Jan 14 at 19:49
2
$begingroup$
Related: math.stackexchange.com/q/30687/42969.
$endgroup$
– Martin R
Jan 14 at 19:51
5
$begingroup$
I would be legitimately surprised (and delighted) to see a good high school-level proof of this fact. Where did you come across the problem?
$endgroup$
– Steven Stadnicki
Jan 14 at 20:17