Irrationality of $sqrt{15}$
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Could someone verify the correctness of this proof for the irrationality of $sqrt{15}$?
Assume $sqrt{15}inmathbf{Q}$, then $sqrt{15}=frac{p}{q}$ with $p,qinmathbf{Z}$ ($qne0$ and $gcd(p,q)=1$).
$implies 15q^2=p^2 implies 15mid p^2 implies 3mid p^2 implies 3mid p$ (Euclid's Lemma)
Now we write $p=3k$ for $kinmathbf{Z}$, then we have $15q^2=9k^2 implies 5q^2=3k^2 implies 3mid 5q^2$. Since $gcd(3,5)=1$ (this lemma: if $amid bc$ and $gcd(a,b)=1$ then $amid c$) this gives $3mid q^2 implies 3mid q$ (Euclid's Lemma).
Therefore $3mid p$ and $3mid q$. Contradiction.
number-theory proof-verification irrational-numbers
edited Oct 25 '17 at 16:34
Rodrigo de Azevedo
12.9k41856
asked Oct 25 '17 at 16:15
Heinz DoofenschmirtzHeinz Doofenschmirtz
602620
$endgroup$
add a comment |
$begingroup$
Could someone verify the correctness of this proof for the irrationality of $sqrt{15}$?
Assume $sqrt{15}inmathbf{Q}$, then $sqrt{15}=frac{p}{q}$ with $p,qinmathbf{Z}$ ($qne0$ and $gcd(p,q)=1$).
$implies 15q^2=p^2 implies 15mid p^2 implies 3mid p^2 implies 3mid p$ (Euclid's Lemma)
Now we write $p=3k$ for $kinmathbf{Z}$, then we have $15q^2=9k^2 implies 5q^2=3k^2 implies 3mid 5q^2$. Since $gcd(3,5)=1$ (this lemma: if $amid bc$ and $gcd(a,b)=1$ then $amid c$) this gives $3mid q^2 implies 3mid q$ (Euclid's Lemma).
Therefore $3mid p$ and $3mid q$. Contradiction.
number-theory proof-verification irrational-numbers
edited Oct 25 '17 at 16:34
Rodrigo de Azevedo
12.9k41856
asked Oct 25 '17 at 16:15
Heinz DoofenschmirtzHeinz Doofenschmirtz
602620
$endgroup$
7
$begingroup$
That's a great proof.
$endgroup$
– Yanko
Oct 25 '17 at 16:18
2
$begingroup$
looks correct to me well done
$endgroup$
– Isham
Oct 25 '17 at 16:34
2
$begingroup$
Agreed and well explained
$endgroup$
– imranfat
Oct 25 '17 at 16:53
7
$begingroup$
Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
$endgroup$
– rtybase
Oct 25 '17 at 18:12
add a comment |
$begingroup$
Could someone verify the correctness of this proof for the irrationality of $sqrt{15}$?
Assume $sqrt{15}inmathbf{Q}$, then $sqrt{15}=frac{p}{q}$ with $p,qinmathbf{Z}$ ($qne0$ and $gcd(p,q)=1$).
$implies 15q^2=p^2 implies 15mid p^2 implies 3mid p^2 implies 3mid p$ (Euclid's Lemma)
Now we write $p=3k$ for $kinmathbf{Z}$, then we have $15q^2=9k^2 implies 5q^2=3k^2 implies 3mid 5q^2$. Since $gcd(3,5)=1$ (this lemma: if $amid bc$ and $gcd(a,b)=1$ then $amid c$) this gives $3mid q^2 implies 3mid q$ (Euclid's Lemma).
Therefore $3mid p$ and $3mid q$. Contradiction.
number-theory proof-verification irrational-numbers
edited Oct 25 '17 at 16:34
Rodrigo de Azevedo
12.9k41856
asked Oct 25 '17 at 16:15
Heinz DoofenschmirtzHeinz Doofenschmirtz
602620
$endgroup$
Could someone verify the correctness of this proof for the irrationality of $sqrt{15}$?
Assume $sqrt{15}inmathbf{Q}$, then $sqrt{15}=frac{p}{q}$ with $p,qinmathbf{Z}$ ($qne0$ and $gcd(p,q)=1$).
$implies 15q^2=p^2 implies 15mid p^2 implies 3mid p^2 implies 3mid p$ (Euclid's Lemma)
Now we write $p=3k$ for $kinmathbf{Z}$, then we have $15q^2=9k^2 implies 5q^2=3k^2 implies 3mid 5q^2$. Since $gcd(3,5)=1$ (this lemma: if $amid bc$ and $gcd(a,b)=1$ then $amid c$) this gives $3mid q^2 implies 3mid q$ (Euclid's Lemma).
Therefore $3mid p$ and $3mid q$. Contradiction.
number-theory proof-verification irrational-numbers
number-theory proof-verification irrational-numbers
edited Oct 25 '17 at 16:34
Rodrigo de Azevedo
12.9k41856
asked Oct 25 '17 at 16:15
Heinz DoofenschmirtzHeinz Doofenschmirtz
602620
edited Oct 25 '17 at 16:34
Rodrigo de Azevedo
12.9k41856
asked Oct 25 '17 at 16:15
Heinz DoofenschmirtzHeinz Doofenschmirtz
602620
edited Oct 25 '17 at 16:34
Rodrigo de Azevedo
12.9k41856
edited Oct 25 '17 at 16:34
Rodrigo de Azevedo
12.9k41856
edited Oct 25 '17 at 16:34
Rodrigo de Azevedo
12.9k41856
12.9k41856
asked Oct 25 '17 at 16:15
Heinz DoofenschmirtzHeinz Doofenschmirtz
602620
asked Oct 25 '17 at 16:15
Heinz DoofenschmirtzHeinz Doofenschmirtz
602620
asked Oct 25 '17 at 16:15
Heinz DoofenschmirtzHeinz Doofenschmirtz
602620
602620
7
$begingroup$
That's a great proof.
$endgroup$
– Yanko
Oct 25 '17 at 16:18
2
$begingroup$
looks correct to me well done
$endgroup$
– Isham
Oct 25 '17 at 16:34
2
$begingroup$
Agreed and well explained
$endgroup$
– imranfat
Oct 25 '17 at 16:53
7
$begingroup$
Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
$endgroup$
– rtybase
Oct 25 '17 at 18:12
add a comment |
7
$begingroup$
That's a great proof.
$endgroup$
– Yanko
Oct 25 '17 at 16:18
2
$begingroup$
looks correct to me well done
$endgroup$
– Isham
Oct 25 '17 at 16:34
2
$begingroup$
Agreed and well explained
$endgroup$
– imranfat
Oct 25 '17 at 16:53
7
$begingroup$
Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
$endgroup$
– rtybase
Oct 25 '17 at 18:12
7
7
$begingroup$
That's a great proof.
$endgroup$
– Yanko
Oct 25 '17 at 16:18
$begingroup$
That's a great proof.
$endgroup$
– Yanko
Oct 25 '17 at 16:18
2
2
$begingroup$
looks correct to me well done
$endgroup$
– Isham
Oct 25 '17 at 16:34
$begingroup$
looks correct to me well done
$endgroup$
– Isham
Oct 25 '17 at 16:34
2
2
$begingroup$
Agreed and well explained
$endgroup$
– imranfat
Oct 25 '17 at 16:53
$begingroup$
Agreed and well explained
$endgroup$
– imranfat
Oct 25 '17 at 16:53
7
7
$begingroup$
Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
$endgroup$
– rtybase
Oct 25 '17 at 18:12
$begingroup$
Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
$endgroup$
– rtybase
Oct 25 '17 at 18:12
add a comment |
1 Answer
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I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $15$ with any product $p_0 r$ where $p_0$ is prime and $r$ is relatively prime to $p_0$.
So you have really proven the following:
Proposition. Let $n$ be an integer and let $p_0$ be a prime number such that $p_0$ divides into $n$ exactly once (that is, $p_0$ divides $n$ but $p_0^2$ does not divide $n$). Then $sqrt{n}$ is irrational.
answered Dec 7 '18 at 18:13
60056005
36.1k751125
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add a comment |
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$begingroup$
I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $15$ with any product $p_0 r$ where $p_0$ is prime and $r$ is relatively prime to $p_0$.
So you have really proven the following:
Proposition. Let $n$ be an integer and let $p_0$ be a prime number such that $p_0$ divides into $n$ exactly once (that is, $p_0$ divides $n$ but $p_0^2$ does not divide $n$). Then $sqrt{n}$ is irrational.
answered Dec 7 '18 at 18:13
60056005
36.1k751125
$endgroup$
add a comment |
$begingroup$
I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $15$ with any product $p_0 r$ where $p_0$ is prime and $r$ is relatively prime to $p_0$.
So you have really proven the following:
Proposition. Let $n$ be an integer and let $p_0$ be a prime number such that $p_0$ divides into $n$ exactly once (that is, $p_0$ divides $n$ but $p_0^2$ does not divide $n$). Then $sqrt{n}$ is irrational.
answered Dec 7 '18 at 18:13
60056005
36.1k751125
$endgroup$
add a comment |
$begingroup$
I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $15$ with any product $p_0 r$ where $p_0$ is prime and $r$ is relatively prime to $p_0$.
So you have really proven the following:
Proposition. Let $n$ be an integer and let $p_0$ be a prime number such that $p_0$ divides into $n$ exactly once (that is, $p_0$ divides $n$ but $p_0^2$ does not divide $n$). Then $sqrt{n}$ is irrational.
answered Dec 7 '18 at 18:13
60056005
36.1k751125
$endgroup$
I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $15$ with any product $p_0 r$ where $p_0$ is prime and $r$ is relatively prime to $p_0$.
So you have really proven the following:
Proposition. Let $n$ be an integer and let $p_0$ be a prime number such that $p_0$ divides into $n$ exactly once (that is, $p_0$ divides $n$ but $p_0^2$ does not divide $n$). Then $sqrt{n}$ is irrational.
answered Dec 7 '18 at 18:13
60056005
36.1k751125
answered Dec 7 '18 at 18:13
60056005
36.1k751125
answered Dec 7 '18 at 18:13
60056005
36.1k751125
answered Dec 7 '18 at 18:13
60056005
36.1k751125
36.1k751125
add a comment |
add a comment |
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7
$begingroup$
That's a great proof.
$endgroup$
– Yanko
Oct 25 '17 at 16:18
2
$begingroup$
looks correct to me well done
$endgroup$
– Isham
Oct 25 '17 at 16:34
2
$begingroup$
Agreed and well explained
$endgroup$
– imranfat
Oct 25 '17 at 16:53
7
$begingroup$
Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
$endgroup$
– rtybase
Oct 25 '17 at 18:12