Is there a way to prove numbers irrational in general?
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I'm familiar with the typical proof that $sqrt2notinmathbb{Q}$, where we assume it is equivalent to $frac ab$ for some integers $a,b$, then prove that both $a$ and $b$ are divisible by $2$, repeat infinitely, proof by contradiction, QED. I'm also familiar with the fact that if you repeat this procedure for any radical, you can similarly prove that each $a^n,b^n$ are divisible by the radicand, where $n$ is the root of the radical (though things get tricky if you don't reduce first). In other words, the proof that $sqrt2notinmathbb{Q}$ generalizes to prove that any number of the form $sqrt[m]{n}notinmathbb{Q}$.
Further, since adding or multiplying a rational and an irrational yields an irrational, this is a proof for all algebraic irrationals. (Ex. I can prove $phi=frac{1+sqrt5}{2}$ irrational just by demonstrating that the $sqrt5$ term is irrational.)
Because this relies on the algebra of radicals, this won't help for transcendental numbers. Is there a proof that generalizes to all irrational numbers, or must each transcendental number be proven irrational independently?
rationality-testing
asked Dec 25 '18 at 19:29
DonielFDonielF
515515
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show 1 more comment
$begingroup$
I'm familiar with the typical proof that $sqrt2notinmathbb{Q}$, where we assume it is equivalent to $frac ab$ for some integers $a,b$, then prove that both $a$ and $b$ are divisible by $2$, repeat infinitely, proof by contradiction, QED. I'm also familiar with the fact that if you repeat this procedure for any radical, you can similarly prove that each $a^n,b^n$ are divisible by the radicand, where $n$ is the root of the radical (though things get tricky if you don't reduce first). In other words, the proof that $sqrt2notinmathbb{Q}$ generalizes to prove that any number of the form $sqrt[m]{n}notinmathbb{Q}$.
Further, since adding or multiplying a rational and an irrational yields an irrational, this is a proof for all algebraic irrationals. (Ex. I can prove $phi=frac{1+sqrt5}{2}$ irrational just by demonstrating that the $sqrt5$ term is irrational.)
Because this relies on the algebra of radicals, this won't help for transcendental numbers. Is there a proof that generalizes to all irrational numbers, or must each transcendental number be proven irrational independently?
rationality-testing
asked Dec 25 '18 at 19:29
DonielFDonielF
515515
$endgroup$
3
$begingroup$
This may sound like silly pedantry, but what is your definition of 'number'? For instance, what (if anything) prevents 'the number which is 1 if Goldbach's conjecture is true and $sqrt{2}$ if it's false' from being a valid number under your definition?
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– Steven Stadnicki
Dec 25 '18 at 19:32
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@Steven Imagine a line which stretches infinitely in both directions, and we pick an arbitrary point on it to label 0. Let's define a number as an unmovable point on this line, and the number's value is determined by its distance from the point we labeled 0. The reason "the number which is 1 if Goldbach's conjecture is true and $sqrt2$ if it's false" is not a number is because it's not a fixed point; it's in a superposition of "conjecture is true" and "conjecture is false" (for lack of a better term) and therefore occupies multiple points on the line.
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– DonielF
Dec 25 '18 at 19:38
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@DonielF No, it's a completely fixed point - it's either $1$ or $sqrt{2}$, we just don't currently (and may never) now which it is.
$endgroup$
– Noah Schweber
Dec 25 '18 at 19:38
3
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Not all algebraic numbers admit the radical form you assume them to in your attempted proof all algebraic numbers' rationality is decidable this way. For example, quintics' roots don't in general take such a form.
$endgroup$
– J.G.
Dec 25 '18 at 19:39
$begingroup$
@DonielF For another perspective, 'its distance from the point labeled 0' is just a circular definition; you're defining the number in terms of itself. Indeed, the overwhelming majority of real numbers have no finite definition (this is a nice little exercise in infinite counting) and so the very notion of an irrationality proof for them needs a little unwinding. (Of course, all of the numbers that don't have a finite definition are immediately irrational by contradiction, but).
$endgroup$
– Steven Stadnicki
Dec 26 '18 at 3:29
|
show 1 more comment
$begingroup$
I'm familiar with the typical proof that $sqrt2notinmathbb{Q}$, where we assume it is equivalent to $frac ab$ for some integers $a,b$, then prove that both $a$ and $b$ are divisible by $2$, repeat infinitely, proof by contradiction, QED. I'm also familiar with the fact that if you repeat this procedure for any radical, you can similarly prove that each $a^n,b^n$ are divisible by the radicand, where $n$ is the root of the radical (though things get tricky if you don't reduce first). In other words, the proof that $sqrt2notinmathbb{Q}$ generalizes to prove that any number of the form $sqrt[m]{n}notinmathbb{Q}$.
Further, since adding or multiplying a rational and an irrational yields an irrational, this is a proof for all algebraic irrationals. (Ex. I can prove $phi=frac{1+sqrt5}{2}$ irrational just by demonstrating that the $sqrt5$ term is irrational.)
Because this relies on the algebra of radicals, this won't help for transcendental numbers. Is there a proof that generalizes to all irrational numbers, or must each transcendental number be proven irrational independently?
rationality-testing
asked Dec 25 '18 at 19:29
DonielFDonielF
515515
$endgroup$
I'm familiar with the typical proof that $sqrt2notinmathbb{Q}$, where we assume it is equivalent to $frac ab$ for some integers $a,b$, then prove that both $a$ and $b$ are divisible by $2$, repeat infinitely, proof by contradiction, QED. I'm also familiar with the fact that if you repeat this procedure for any radical, you can similarly prove that each $a^n,b^n$ are divisible by the radicand, where $n$ is the root of the radical (though things get tricky if you don't reduce first). In other words, the proof that $sqrt2notinmathbb{Q}$ generalizes to prove that any number of the form $sqrt[m]{n}notinmathbb{Q}$.
Further, since adding or multiplying a rational and an irrational yields an irrational, this is a proof for all algebraic irrationals. (Ex. I can prove $phi=frac{1+sqrt5}{2}$ irrational just by demonstrating that the $sqrt5$ term is irrational.)
Because this relies on the algebra of radicals, this won't help for transcendental numbers. Is there a proof that generalizes to all irrational numbers, or must each transcendental number be proven irrational independently?
rationality-testing
rationality-testing
asked Dec 25 '18 at 19:29
DonielFDonielF
515515
asked Dec 25 '18 at 19:29
DonielFDonielF
515515
asked Dec 25 '18 at 19:29
DonielFDonielF
515515
asked Dec 25 '18 at 19:29
DonielFDonielF
515515
asked Dec 25 '18 at 19:29
DonielFDonielF
515515
515515
3
$begingroup$
This may sound like silly pedantry, but what is your definition of 'number'? For instance, what (if anything) prevents 'the number which is 1 if Goldbach's conjecture is true and $sqrt{2}$ if it's false' from being a valid number under your definition?
$endgroup$
– Steven Stadnicki
Dec 25 '18 at 19:32
$begingroup$
@Steven Imagine a line which stretches infinitely in both directions, and we pick an arbitrary point on it to label 0. Let's define a number as an unmovable point on this line, and the number's value is determined by its distance from the point we labeled 0. The reason "the number which is 1 if Goldbach's conjecture is true and $sqrt2$ if it's false" is not a number is because it's not a fixed point; it's in a superposition of "conjecture is true" and "conjecture is false" (for lack of a better term) and therefore occupies multiple points on the line.
$endgroup$
– DonielF
Dec 25 '18 at 19:38
$begingroup$
@DonielF No, it's a completely fixed point - it's either $1$ or $sqrt{2}$, we just don't currently (and may never) now which it is.
$endgroup$
– Noah Schweber
Dec 25 '18 at 19:38
3
$begingroup$
Not all algebraic numbers admit the radical form you assume them to in your attempted proof all algebraic numbers' rationality is decidable this way. For example, quintics' roots don't in general take such a form.
$endgroup$
– J.G.
Dec 25 '18 at 19:39
$begingroup$
@DonielF For another perspective, 'its distance from the point labeled 0' is just a circular definition; you're defining the number in terms of itself. Indeed, the overwhelming majority of real numbers have no finite definition (this is a nice little exercise in infinite counting) and so the very notion of an irrationality proof for them needs a little unwinding. (Of course, all of the numbers that don't have a finite definition are immediately irrational by contradiction, but).
$endgroup$
– Steven Stadnicki
Dec 26 '18 at 3:29
|
show 1 more comment
3
$begingroup$
This may sound like silly pedantry, but what is your definition of 'number'? For instance, what (if anything) prevents 'the number which is 1 if Goldbach's conjecture is true and $sqrt{2}$ if it's false' from being a valid number under your definition?
$endgroup$
– Steven Stadnicki
Dec 25 '18 at 19:32
$begingroup$
@Steven Imagine a line which stretches infinitely in both directions, and we pick an arbitrary point on it to label 0. Let's define a number as an unmovable point on this line, and the number's value is determined by its distance from the point we labeled 0. The reason "the number which is 1 if Goldbach's conjecture is true and $sqrt2$ if it's false" is not a number is because it's not a fixed point; it's in a superposition of "conjecture is true" and "conjecture is false" (for lack of a better term) and therefore occupies multiple points on the line.
$endgroup$
– DonielF
Dec 25 '18 at 19:38
$begingroup$
@DonielF No, it's a completely fixed point - it's either $1$ or $sqrt{2}$, we just don't currently (and may never) now which it is.
$endgroup$
– Noah Schweber
Dec 25 '18 at 19:38
3
$begingroup$
Not all algebraic numbers admit the radical form you assume them to in your attempted proof all algebraic numbers' rationality is decidable this way. For example, quintics' roots don't in general take such a form.
$endgroup$
– J.G.
Dec 25 '18 at 19:39
$begingroup$
@DonielF For another perspective, 'its distance from the point labeled 0' is just a circular definition; you're defining the number in terms of itself. Indeed, the overwhelming majority of real numbers have no finite definition (this is a nice little exercise in infinite counting) and so the very notion of an irrationality proof for them needs a little unwinding. (Of course, all of the numbers that don't have a finite definition are immediately irrational by contradiction, but).
$endgroup$
– Steven Stadnicki
Dec 26 '18 at 3:29
3
3
$begingroup$
This may sound like silly pedantry, but what is your definition of 'number'? For instance, what (if anything) prevents 'the number which is 1 if Goldbach's conjecture is true and $sqrt{2}$ if it's false' from being a valid number under your definition?
$endgroup$
– Steven Stadnicki
Dec 25 '18 at 19:32
$begingroup$
This may sound like silly pedantry, but what is your definition of 'number'? For instance, what (if anything) prevents 'the number which is 1 if Goldbach's conjecture is true and $sqrt{2}$ if it's false' from being a valid number under your definition?
$endgroup$
– Steven Stadnicki
Dec 25 '18 at 19:32
$begingroup$
@Steven Imagine a line which stretches infinitely in both directions, and we pick an arbitrary point on it to label 0. Let's define a number as an unmovable point on this line, and the number's value is determined by its distance from the point we labeled 0. The reason "the number which is 1 if Goldbach's conjecture is true and $sqrt2$ if it's false" is not a number is because it's not a fixed point; it's in a superposition of "conjecture is true" and "conjecture is false" (for lack of a better term) and therefore occupies multiple points on the line.
$endgroup$
– DonielF
Dec 25 '18 at 19:38
$begingroup$
@Steven Imagine a line which stretches infinitely in both directions, and we pick an arbitrary point on it to label 0. Let's define a number as an unmovable point on this line, and the number's value is determined by its distance from the point we labeled 0. The reason "the number which is 1 if Goldbach's conjecture is true and $sqrt2$ if it's false" is not a number is because it's not a fixed point; it's in a superposition of "conjecture is true" and "conjecture is false" (for lack of a better term) and therefore occupies multiple points on the line.
$endgroup$
– DonielF
Dec 25 '18 at 19:38
$begingroup$
@DonielF No, it's a completely fixed point - it's either $1$ or $sqrt{2}$, we just don't currently (and may never) now which it is.
$endgroup$
– Noah Schweber
Dec 25 '18 at 19:38
$begingroup$
@DonielF No, it's a completely fixed point - it's either $1$ or $sqrt{2}$, we just don't currently (and may never) now which it is.
$endgroup$
– Noah Schweber
Dec 25 '18 at 19:38
3
3
$begingroup$
Not all algebraic numbers admit the radical form you assume them to in your attempted proof all algebraic numbers' rationality is decidable this way. For example, quintics' roots don't in general take such a form.
$endgroup$
– J.G.
Dec 25 '18 at 19:39
$begingroup$
Not all algebraic numbers admit the radical form you assume them to in your attempted proof all algebraic numbers' rationality is decidable this way. For example, quintics' roots don't in general take such a form.
$endgroup$
– J.G.
Dec 25 '18 at 19:39
$begingroup$
@DonielF For another perspective, 'its distance from the point labeled 0' is just a circular definition; you're defining the number in terms of itself. Indeed, the overwhelming majority of real numbers have no finite definition (this is a nice little exercise in infinite counting) and so the very notion of an irrationality proof for them needs a little unwinding. (Of course, all of the numbers that don't have a finite definition are immediately irrational by contradiction, but).
$endgroup$
– Steven Stadnicki
Dec 26 '18 at 3:29
$begingroup$
@DonielF For another perspective, 'its distance from the point labeled 0' is just a circular definition; you're defining the number in terms of itself. Indeed, the overwhelming majority of real numbers have no finite definition (this is a nice little exercise in infinite counting) and so the very notion of an irrationality proof for them needs a little unwinding. (Of course, all of the numbers that don't have a finite definition are immediately irrational by contradiction, but).
$endgroup$
– Steven Stadnicki
Dec 26 '18 at 3:29
|
show 1 more comment
1 Answer
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There is no known general method of dealing with this problem. Otherwise, we would already know whether or not the Euler–Mascheroni constant is rational.
answered Dec 25 '18 at 19:38
José Carlos SantosJosé Carlos Santos
167k22132235
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There is no known general method of dealing with this problem. Otherwise, we would already know whether or not the Euler–Mascheroni constant is rational.
answered Dec 25 '18 at 19:38
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
add a comment |
$begingroup$
There is no known general method of dealing with this problem. Otherwise, we would already know whether or not the Euler–Mascheroni constant is rational.
answered Dec 25 '18 at 19:38
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
add a comment |
$begingroup$
There is no known general method of dealing with this problem. Otherwise, we would already know whether or not the Euler–Mascheroni constant is rational.
answered Dec 25 '18 at 19:38
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
There is no known general method of dealing with this problem. Otherwise, we would already know whether or not the Euler–Mascheroni constant is rational.
answered Dec 25 '18 at 19:38
José Carlos SantosJosé Carlos Santos
167k22132235
answered Dec 25 '18 at 19:38
José Carlos SantosJosé Carlos Santos
167k22132235
answered Dec 25 '18 at 19:38
José Carlos SantosJosé Carlos Santos
167k22132235
answered Dec 25 '18 at 19:38
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
add a comment |
add a comment |
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$begingroup$
This may sound like silly pedantry, but what is your definition of 'number'? For instance, what (if anything) prevents 'the number which is 1 if Goldbach's conjecture is true and $sqrt{2}$ if it's false' from being a valid number under your definition?
$endgroup$
– Steven Stadnicki
Dec 25 '18 at 19:32
$begingroup$
@Steven Imagine a line which stretches infinitely in both directions, and we pick an arbitrary point on it to label 0. Let's define a number as an unmovable point on this line, and the number's value is determined by its distance from the point we labeled 0. The reason "the number which is 1 if Goldbach's conjecture is true and $sqrt2$ if it's false" is not a number is because it's not a fixed point; it's in a superposition of "conjecture is true" and "conjecture is false" (for lack of a better term) and therefore occupies multiple points on the line.
$endgroup$
– DonielF
Dec 25 '18 at 19:38
$begingroup$
@DonielF No, it's a completely fixed point - it's either $1$ or $sqrt{2}$, we just don't currently (and may never) now which it is.
$endgroup$
– Noah Schweber
Dec 25 '18 at 19:38
3
$begingroup$
Not all algebraic numbers admit the radical form you assume them to in your attempted proof all algebraic numbers' rationality is decidable this way. For example, quintics' roots don't in general take such a form.
$endgroup$
– J.G.
Dec 25 '18 at 19:39
$begingroup$
@DonielF For another perspective, 'its distance from the point labeled 0' is just a circular definition; you're defining the number in terms of itself. Indeed, the overwhelming majority of real numbers have no finite definition (this is a nice little exercise in infinite counting) and so the very notion of an irrationality proof for them needs a little unwinding. (Of course, all of the numbers that don't have a finite definition are immediately irrational by contradiction, but).
$endgroup$
– Steven Stadnicki
Dec 26 '18 at 3:29