Prove or disprove non-constructively there exist irrationals $a, b, c$ such that $a^{b^c}$ is rational.
Consider the interesting question:
Do there exist irrationals $a$ and $b$ such that $a^b$ is a rational? Alternatively, prove or disprove that there exist irrationals $a$ and $b$ such that $a^b$ is a rational.
There has been an interesting nonconstructive proof:
Let $x:=sqrt{2}^sqrt{2}$. $x$ is either a rational or an irrational. (Law of excluded middle is used here.) If assuming that $x$ is rational, than its existence gives a proof. If assuming that $x$ is irrational, then we put $a:=x:=sqrt{2}^sqrt{2}$ and $b:=sqrt{2}$ and we have $a^b=(sqrt{2}^sqrt{2})^sqrt{2}=sqrt{2}^2=2$, a rational. So, no matter we assuming x is a rational or not, there is always a proof.
A more interesting question:
Do there exist irrationals $a$ and $b$ and $c$ such that $a^{b^c}$ is a rational? Alternatively, prove or disprove that there exist irrationals $a$ and $b$ and $c$ such that $a^{b^c}$ is a rational.
Can we obtain a non-constructive proof, in which law of excluded middle is used, again?
Note: We follow the Tower rule ( https://brilliant.org/wiki/exponential-functions-properties/#the-tower-rule ) to compute $a^{b^c}=a^{(b^c)}$
irrational-numbers rational-numbers constructive-mathematics
edited Dec 1 at 11:26
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 30 at 14:15
JZH
727
add a comment |
Consider the interesting question:
Do there exist irrationals $a$ and $b$ such that $a^b$ is a rational? Alternatively, prove or disprove that there exist irrationals $a$ and $b$ such that $a^b$ is a rational.
There has been an interesting nonconstructive proof:
Let $x:=sqrt{2}^sqrt{2}$. $x$ is either a rational or an irrational. (Law of excluded middle is used here.) If assuming that $x$ is rational, than its existence gives a proof. If assuming that $x$ is irrational, then we put $a:=x:=sqrt{2}^sqrt{2}$ and $b:=sqrt{2}$ and we have $a^b=(sqrt{2}^sqrt{2})^sqrt{2}=sqrt{2}^2=2$, a rational. So, no matter we assuming x is a rational or not, there is always a proof.
A more interesting question:
Do there exist irrationals $a$ and $b$ and $c$ such that $a^{b^c}$ is a rational? Alternatively, prove or disprove that there exist irrationals $a$ and $b$ and $c$ such that $a^{b^c}$ is a rational.
Can we obtain a non-constructive proof, in which law of excluded middle is used, again?
Note: We follow the Tower rule ( https://brilliant.org/wiki/exponential-functions-properties/#the-tower-rule ) to compute $a^{b^c}=a^{(b^c)}$
irrational-numbers rational-numbers constructive-mathematics
edited Dec 1 at 11:26
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 30 at 14:15
JZH
727
probably the (positive) solution to the equation $$x^{x^{1/x}}=2$$ is not rational
– Masacroso
Nov 30 at 14:24
Every constructive proof induces a non-constructive: Assume that no such numbers exist, use the constructive proof $rightarrow$ contradiction.
– gammatester
Nov 30 at 15:09
Non-constructive prrof is not proving by contradiction. It is that: whether you assuming such a, b, c exist or not , they all induce the same conclusion. (x is either a rational or an irrational.)
– JZH
Nov 30 at 15:12
This is a weird definition of non-constructive.
– gammatester
Nov 30 at 15:15
en.wikipedia.org/wiki/Law_of_excluded_middle#Examples
– JZH
Nov 30 at 15:16
add a comment |
Consider the interesting question:
Do there exist irrationals $a$ and $b$ such that $a^b$ is a rational? Alternatively, prove or disprove that there exist irrationals $a$ and $b$ such that $a^b$ is a rational.
There has been an interesting nonconstructive proof:
Let $x:=sqrt{2}^sqrt{2}$. $x$ is either a rational or an irrational. (Law of excluded middle is used here.) If assuming that $x$ is rational, than its existence gives a proof. If assuming that $x$ is irrational, then we put $a:=x:=sqrt{2}^sqrt{2}$ and $b:=sqrt{2}$ and we have $a^b=(sqrt{2}^sqrt{2})^sqrt{2}=sqrt{2}^2=2$, a rational. So, no matter we assuming x is a rational or not, there is always a proof.
A more interesting question:
Do there exist irrationals $a$ and $b$ and $c$ such that $a^{b^c}$ is a rational? Alternatively, prove or disprove that there exist irrationals $a$ and $b$ and $c$ such that $a^{b^c}$ is a rational.
Can we obtain a non-constructive proof, in which law of excluded middle is used, again?
Note: We follow the Tower rule ( https://brilliant.org/wiki/exponential-functions-properties/#the-tower-rule ) to compute $a^{b^c}=a^{(b^c)}$
irrational-numbers rational-numbers constructive-mathematics
edited Dec 1 at 11:26
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 30 at 14:15
JZH
727
Consider the interesting question:
Do there exist irrationals $a$ and $b$ such that $a^b$ is a rational? Alternatively, prove or disprove that there exist irrationals $a$ and $b$ such that $a^b$ is a rational.
There has been an interesting nonconstructive proof:
Let $x:=sqrt{2}^sqrt{2}$. $x$ is either a rational or an irrational. (Law of excluded middle is used here.) If assuming that $x$ is rational, than its existence gives a proof. If assuming that $x$ is irrational, then we put $a:=x:=sqrt{2}^sqrt{2}$ and $b:=sqrt{2}$ and we have $a^b=(sqrt{2}^sqrt{2})^sqrt{2}=sqrt{2}^2=2$, a rational. So, no matter we assuming x is a rational or not, there is always a proof.
A more interesting question:
Do there exist irrationals $a$ and $b$ and $c$ such that $a^{b^c}$ is a rational? Alternatively, prove or disprove that there exist irrationals $a$ and $b$ and $c$ such that $a^{b^c}$ is a rational.
Can we obtain a non-constructive proof, in which law of excluded middle is used, again?
Note: We follow the Tower rule ( https://brilliant.org/wiki/exponential-functions-properties/#the-tower-rule ) to compute $a^{b^c}=a^{(b^c)}$
irrational-numbers rational-numbers constructive-mathematics
irrational-numbers rational-numbers constructive-mathematics
edited Dec 1 at 11:26
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 30 at 14:15
JZH
727
edited Dec 1 at 11:26
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 30 at 14:15
JZH
727
edited Dec 1 at 11:26
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Dec 1 at 11:26
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Dec 1 at 11:26
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Nov 30 at 14:15
JZH
727
asked Nov 30 at 14:15
JZH
727
asked Nov 30 at 14:15
JZH
727
727
probably the (positive) solution to the equation $$x^{x^{1/x}}=2$$ is not rational
– Masacroso
Nov 30 at 14:24
Every constructive proof induces a non-constructive: Assume that no such numbers exist, use the constructive proof $rightarrow$ contradiction.
– gammatester
Nov 30 at 15:09
Non-constructive prrof is not proving by contradiction. It is that: whether you assuming such a, b, c exist or not , they all induce the same conclusion. (x is either a rational or an irrational.)
– JZH
Nov 30 at 15:12
This is a weird definition of non-constructive.
– gammatester
Nov 30 at 15:15
en.wikipedia.org/wiki/Law_of_excluded_middle#Examples
– JZH
Nov 30 at 15:16
add a comment |
probably the (positive) solution to the equation $$x^{x^{1/x}}=2$$ is not rational
– Masacroso
Nov 30 at 14:24
Every constructive proof induces a non-constructive: Assume that no such numbers exist, use the constructive proof $rightarrow$ contradiction.
– gammatester
Nov 30 at 15:09
Non-constructive prrof is not proving by contradiction. It is that: whether you assuming such a, b, c exist or not , they all induce the same conclusion. (x is either a rational or an irrational.)
– JZH
Nov 30 at 15:12
This is a weird definition of non-constructive.
– gammatester
Nov 30 at 15:15
en.wikipedia.org/wiki/Law_of_excluded_middle#Examples
– JZH
Nov 30 at 15:16
probably the (positive) solution to the equation $$x^{x^{1/x}}=2$$ is not rational
– Masacroso
Nov 30 at 14:24
probably the (positive) solution to the equation $$x^{x^{1/x}}=2$$ is not rational
– Masacroso
Nov 30 at 14:24
Every constructive proof induces a non-constructive: Assume that no such numbers exist, use the constructive proof $rightarrow$ contradiction.
– gammatester
Nov 30 at 15:09
Every constructive proof induces a non-constructive: Assume that no such numbers exist, use the constructive proof $rightarrow$ contradiction.
– gammatester
Nov 30 at 15:09
Non-constructive prrof is not proving by contradiction. It is that: whether you assuming such a, b, c exist or not , they all induce the same conclusion. (x is either a rational or an irrational.)
– JZH
Nov 30 at 15:12
Non-constructive prrof is not proving by contradiction. It is that: whether you assuming such a, b, c exist or not , they all induce the same conclusion. (x is either a rational or an irrational.)
– JZH
Nov 30 at 15:12
This is a weird definition of non-constructive.
– gammatester
Nov 30 at 15:15
This is a weird definition of non-constructive.
– gammatester
Nov 30 at 15:15
en.wikipedia.org/wiki/Law_of_excluded_middle#Examples
– JZH
Nov 30 at 15:16
en.wikipedia.org/wiki/Law_of_excluded_middle#Examples
– JZH
Nov 30 at 15:16
add a comment |
2 Answers
2
active
oldest
votes
I'm not exactly sure if this is what you want, but off the top of my mind, by using $a^{log_a b} = a$, you could come up with many examples of a rational number for $a^{b^c}$ with $a$, $b$, and $c$ being irrational, such as
$$large{(sqrt 2)^{e^{ln 2}}}$$
or more generally,
$$large{(sqrt[n] x)^{a^{log_a n}}}$$
for irrational values of $sqrt[n] x$, $a$, and $log_a n$.
answered Nov 30 at 14:30
KM101
4,528418
Nice observation, which is more general than mine. $+1$
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 14:58
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:08
Well, you could just use the expressions given by either answer to conclude that there there exists irrational numbers for $a$, $b$, and $c$, such that $a^{b^c}$ becomes rational.
– KM101
Nov 30 at 15:27
@KM101 I don't quite understand we you mean. the currently 2 answers here solve my problem by giving explicit expressions, so they are constructive. I am curious about non-constructive. en.wikipedia.org/wiki/Law_of_excluded_middle#Examples "x is either a rational or an irrational".
– JZH
Nov 30 at 15:41
I'm not really into these different types of proofs, so I may have misunderstood what exactly you wanted.
– KM101
Nov 30 at 15:44
|
show 1 more comment
Take $a=c=sqrt2, b={sqrt2}^{sqrt2}$.
$$a^{b^c}={sqrt2}^{({sqrt2}^{sqrt2})^{sqrt2}}={sqrt2}^{(sqrt2)^2}=2$$
answered Nov 30 at 14:29
GNUSupporter 8964民主女神 地下教會
12.8k72445
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:11
In your current way you need to firstly prove that $sqrt{2}^sqrt{2}$ is an irrational, so you have an example of irrationals $a$ and $b$ and $c$.
– JZH
Nov 30 at 23:38
In your current (constructive) way..., I mean.
– JZH
Nov 30 at 23:45
@JZH Like the other answerer, I've no interest in finding a "non-constructive proof". Add a bounty if you like. Here's a non-constructive proof that you asked in the comments. Consider $b^2=2^{sqrt2}=r/s$, with $r,s$ integers. We have a positive irrational exponent of $2$ on the left-hand side, but rational exponents of any prime factors on the right-hand side.
– GNUSupporter 8964民主女神 地下教會
Dec 1 at 9:03
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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I'm not exactly sure if this is what you want, but off the top of my mind, by using $a^{log_a b} = a$, you could come up with many examples of a rational number for $a^{b^c}$ with $a$, $b$, and $c$ being irrational, such as
$$large{(sqrt 2)^{e^{ln 2}}}$$
or more generally,
$$large{(sqrt[n] x)^{a^{log_a n}}}$$
for irrational values of $sqrt[n] x$, $a$, and $log_a n$.
answered Nov 30 at 14:30
KM101
4,528418
Nice observation, which is more general than mine. $+1$
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 14:58
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:08
Well, you could just use the expressions given by either answer to conclude that there there exists irrational numbers for $a$, $b$, and $c$, such that $a^{b^c}$ becomes rational.
– KM101
Nov 30 at 15:27
@KM101 I don't quite understand we you mean. the currently 2 answers here solve my problem by giving explicit expressions, so they are constructive. I am curious about non-constructive. en.wikipedia.org/wiki/Law_of_excluded_middle#Examples "x is either a rational or an irrational".
– JZH
Nov 30 at 15:41
I'm not really into these different types of proofs, so I may have misunderstood what exactly you wanted.
– KM101
Nov 30 at 15:44
|
show 1 more comment
I'm not exactly sure if this is what you want, but off the top of my mind, by using $a^{log_a b} = a$, you could come up with many examples of a rational number for $a^{b^c}$ with $a$, $b$, and $c$ being irrational, such as
$$large{(sqrt 2)^{e^{ln 2}}}$$
or more generally,
$$large{(sqrt[n] x)^{a^{log_a n}}}$$
for irrational values of $sqrt[n] x$, $a$, and $log_a n$.
answered Nov 30 at 14:30
KM101
4,528418
Nice observation, which is more general than mine. $+1$
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 14:58
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:08
Well, you could just use the expressions given by either answer to conclude that there there exists irrational numbers for $a$, $b$, and $c$, such that $a^{b^c}$ becomes rational.
– KM101
Nov 30 at 15:27
@KM101 I don't quite understand we you mean. the currently 2 answers here solve my problem by giving explicit expressions, so they are constructive. I am curious about non-constructive. en.wikipedia.org/wiki/Law_of_excluded_middle#Examples "x is either a rational or an irrational".
– JZH
Nov 30 at 15:41
I'm not really into these different types of proofs, so I may have misunderstood what exactly you wanted.
– KM101
Nov 30 at 15:44
|
show 1 more comment
I'm not exactly sure if this is what you want, but off the top of my mind, by using $a^{log_a b} = a$, you could come up with many examples of a rational number for $a^{b^c}$ with $a$, $b$, and $c$ being irrational, such as
$$large{(sqrt 2)^{e^{ln 2}}}$$
or more generally,
$$large{(sqrt[n] x)^{a^{log_a n}}}$$
for irrational values of $sqrt[n] x$, $a$, and $log_a n$.
answered Nov 30 at 14:30
KM101
4,528418
I'm not exactly sure if this is what you want, but off the top of my mind, by using $a^{log_a b} = a$, you could come up with many examples of a rational number for $a^{b^c}$ with $a$, $b$, and $c$ being irrational, such as
$$large{(sqrt 2)^{e^{ln 2}}}$$
or more generally,
$$large{(sqrt[n] x)^{a^{log_a n}}}$$
for irrational values of $sqrt[n] x$, $a$, and $log_a n$.
answered Nov 30 at 14:30
KM101
4,528418
answered Nov 30 at 14:30
KM101
4,528418
answered Nov 30 at 14:30
KM101
4,528418
answered Nov 30 at 14:30
KM101
4,528418
4,528418
Nice observation, which is more general than mine. $+1$
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 14:58
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:08
Well, you could just use the expressions given by either answer to conclude that there there exists irrational numbers for $a$, $b$, and $c$, such that $a^{b^c}$ becomes rational.
– KM101
Nov 30 at 15:27
@KM101 I don't quite understand we you mean. the currently 2 answers here solve my problem by giving explicit expressions, so they are constructive. I am curious about non-constructive. en.wikipedia.org/wiki/Law_of_excluded_middle#Examples "x is either a rational or an irrational".
– JZH
Nov 30 at 15:41
I'm not really into these different types of proofs, so I may have misunderstood what exactly you wanted.
– KM101
Nov 30 at 15:44
|
show 1 more comment
Nice observation, which is more general than mine. $+1$
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 14:58
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:08
Well, you could just use the expressions given by either answer to conclude that there there exists irrational numbers for $a$, $b$, and $c$, such that $a^{b^c}$ becomes rational.
– KM101
Nov 30 at 15:27
@KM101 I don't quite understand we you mean. the currently 2 answers here solve my problem by giving explicit expressions, so they are constructive. I am curious about non-constructive. en.wikipedia.org/wiki/Law_of_excluded_middle#Examples "x is either a rational or an irrational".
– JZH
Nov 30 at 15:41
I'm not really into these different types of proofs, so I may have misunderstood what exactly you wanted.
– KM101
Nov 30 at 15:44
Nice observation, which is more general than mine. $+1$
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 14:58
Nice observation, which is more general than mine. $+1$
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 14:58
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:08
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:08
Well, you could just use the expressions given by either answer to conclude that there there exists irrational numbers for $a$, $b$, and $c$, such that $a^{b^c}$ becomes rational.
– KM101
Nov 30 at 15:27
Well, you could just use the expressions given by either answer to conclude that there there exists irrational numbers for $a$, $b$, and $c$, such that $a^{b^c}$ becomes rational.
– KM101
Nov 30 at 15:27
@KM101 I don't quite understand we you mean. the currently 2 answers here solve my problem by giving explicit expressions, so they are constructive. I am curious about non-constructive. en.wikipedia.org/wiki/Law_of_excluded_middle#Examples "x is either a rational or an irrational".
– JZH
Nov 30 at 15:41
@KM101 I don't quite understand we you mean. the currently 2 answers here solve my problem by giving explicit expressions, so they are constructive. I am curious about non-constructive. en.wikipedia.org/wiki/Law_of_excluded_middle#Examples "x is either a rational or an irrational".
– JZH
Nov 30 at 15:41
I'm not really into these different types of proofs, so I may have misunderstood what exactly you wanted.
– KM101
Nov 30 at 15:44
I'm not really into these different types of proofs, so I may have misunderstood what exactly you wanted.
– KM101
Nov 30 at 15:44
|
show 1 more comment
Take $a=c=sqrt2, b={sqrt2}^{sqrt2}$.
$$a^{b^c}={sqrt2}^{({sqrt2}^{sqrt2})^{sqrt2}}={sqrt2}^{(sqrt2)^2}=2$$
answered Nov 30 at 14:29
GNUSupporter 8964民主女神 地下教會
12.8k72445
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:11
In your current way you need to firstly prove that $sqrt{2}^sqrt{2}$ is an irrational, so you have an example of irrationals $a$ and $b$ and $c$.
– JZH
Nov 30 at 23:38
In your current (constructive) way..., I mean.
– JZH
Nov 30 at 23:45
@JZH Like the other answerer, I've no interest in finding a "non-constructive proof". Add a bounty if you like. Here's a non-constructive proof that you asked in the comments. Consider $b^2=2^{sqrt2}=r/s$, with $r,s$ integers. We have a positive irrational exponent of $2$ on the left-hand side, but rational exponents of any prime factors on the right-hand side.
– GNUSupporter 8964民主女神 地下教會
Dec 1 at 9:03
add a comment |
Take $a=c=sqrt2, b={sqrt2}^{sqrt2}$.
$$a^{b^c}={sqrt2}^{({sqrt2}^{sqrt2})^{sqrt2}}={sqrt2}^{(sqrt2)^2}=2$$
answered Nov 30 at 14:29
GNUSupporter 8964民主女神 地下教會
12.8k72445
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:11
In your current way you need to firstly prove that $sqrt{2}^sqrt{2}$ is an irrational, so you have an example of irrationals $a$ and $b$ and $c$.
– JZH
Nov 30 at 23:38
In your current (constructive) way..., I mean.
– JZH
Nov 30 at 23:45
@JZH Like the other answerer, I've no interest in finding a "non-constructive proof". Add a bounty if you like. Here's a non-constructive proof that you asked in the comments. Consider $b^2=2^{sqrt2}=r/s$, with $r,s$ integers. We have a positive irrational exponent of $2$ on the left-hand side, but rational exponents of any prime factors on the right-hand side.
– GNUSupporter 8964民主女神 地下教會
Dec 1 at 9:03
add a comment |
Take $a=c=sqrt2, b={sqrt2}^{sqrt2}$.
$$a^{b^c}={sqrt2}^{({sqrt2}^{sqrt2})^{sqrt2}}={sqrt2}^{(sqrt2)^2}=2$$
answered Nov 30 at 14:29
GNUSupporter 8964民主女神 地下教會
12.8k72445
Take $a=c=sqrt2, b={sqrt2}^{sqrt2}$.
$$a^{b^c}={sqrt2}^{({sqrt2}^{sqrt2})^{sqrt2}}={sqrt2}^{(sqrt2)^2}=2$$
answered Nov 30 at 14:29
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 30 at 14:29
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 30 at 14:29
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 30 at 14:29
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:11
In your current way you need to firstly prove that $sqrt{2}^sqrt{2}$ is an irrational, so you have an example of irrationals $a$ and $b$ and $c$.
– JZH
Nov 30 at 23:38
In your current (constructive) way..., I mean.
– JZH
Nov 30 at 23:45
@JZH Like the other answerer, I've no interest in finding a "non-constructive proof". Add a bounty if you like. Here's a non-constructive proof that you asked in the comments. Consider $b^2=2^{sqrt2}=r/s$, with $r,s$ integers. We have a positive irrational exponent of $2$ on the left-hand side, but rational exponents of any prime factors on the right-hand side.
– GNUSupporter 8964民主女神 地下教會
Dec 1 at 9:03
add a comment |
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:11
In your current way you need to firstly prove that $sqrt{2}^sqrt{2}$ is an irrational, so you have an example of irrationals $a$ and $b$ and $c$.
– JZH
Nov 30 at 23:38
In your current (constructive) way..., I mean.
– JZH
Nov 30 at 23:45
@JZH Like the other answerer, I've no interest in finding a "non-constructive proof". Add a bounty if you like. Here's a non-constructive proof that you asked in the comments. Consider $b^2=2^{sqrt2}=r/s$, with $r,s$ integers. We have a positive irrational exponent of $2$ on the left-hand side, but rational exponents of any prime factors on the right-hand side.
– GNUSupporter 8964民主女神 地下教會
Dec 1 at 9:03
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:11
Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used.
– JZH
Nov 30 at 15:11
In your current way you need to firstly prove that $sqrt{2}^sqrt{2}$ is an irrational, so you have an example of irrationals $a$ and $b$ and $c$.
– JZH
Nov 30 at 23:38
In your current way you need to firstly prove that $sqrt{2}^sqrt{2}$ is an irrational, so you have an example of irrationals $a$ and $b$ and $c$.
– JZH
Nov 30 at 23:38
In your current (constructive) way..., I mean.
– JZH
Nov 30 at 23:45
In your current (constructive) way..., I mean.
– JZH
Nov 30 at 23:45
@JZH Like the other answerer, I've no interest in finding a "non-constructive proof". Add a bounty if you like. Here's a non-constructive proof that you asked in the comments. Consider $b^2=2^{sqrt2}=r/s$, with $r,s$ integers. We have a positive irrational exponent of $2$ on the left-hand side, but rational exponents of any prime factors on the right-hand side.
– GNUSupporter 8964民主女神 地下教會
Dec 1 at 9:03
@JZH Like the other answerer, I've no interest in finding a "non-constructive proof". Add a bounty if you like. Here's a non-constructive proof that you asked in the comments. Consider $b^2=2^{sqrt2}=r/s$, with $r,s$ integers. We have a positive irrational exponent of $2$ on the left-hand side, but rational exponents of any prime factors on the right-hand side.
– GNUSupporter 8964民主女神 地下教會
Dec 1 at 9:03
add a comment |
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probably the (positive) solution to the equation $$x^{x^{1/x}}=2$$ is not rational
– Masacroso
Nov 30 at 14:24
Every constructive proof induces a non-constructive: Assume that no such numbers exist, use the constructive proof $rightarrow$ contradiction.
– gammatester
Nov 30 at 15:09
Non-constructive prrof is not proving by contradiction. It is that: whether you assuming such a, b, c exist or not , they all induce the same conclusion. (x is either a rational or an irrational.)
– JZH
Nov 30 at 15:12
This is a weird definition of non-constructive.
– gammatester
Nov 30 at 15:15
en.wikipedia.org/wiki/Law_of_excluded_middle#Examples
– JZH
Nov 30 at 15:16