Mixing
Well-Ordering Irrationality
$begingroup$
Let $D$ be a positive integer and the let the square root of $D$ be a real number. Assuming that the square root of D is not an integer (i.e. $D$ is not a perfect square), use Well-Ordering to prove that there exists some integer $a$ such that $a < sqrt(D) < a+1.$
My proof attempt begins with assuming that the square root of D is not an integer and letting $a$ be the least integer that satisfies the inequality $a < $sqrt(D)$ < a + 1$, then I'm not sure what to do. My intuition tells me to square both sides, but then how would I proceed with the proof.
irrational-numbers rational-numbers well-orders
asked Jan 31 at 7:15
Sanjoy The ManjoySanjoy The Manjoy
502315
$endgroup$
add a comment |
$begingroup$
Let $D$ be a positive integer and the let the square root of $D$ be a real number. Assuming that the square root of D is not an integer (i.e. $D$ is not a perfect square), use Well-Ordering to prove that there exists some integer $a$ such that $a < sqrt(D) < a+1.$
My proof attempt begins with assuming that the square root of D is not an integer and letting $a$ be the least integer that satisfies the inequality $a < $sqrt(D)$ < a + 1$, then I'm not sure what to do. My intuition tells me to square both sides, but then how would I proceed with the proof.
irrational-numbers rational-numbers well-orders
asked Jan 31 at 7:15
Sanjoy The ManjoySanjoy The Manjoy
502315
$endgroup$
$begingroup$
It is not clear from your definition of $a$ (the least integer satisfying $a<mathit{sqrt}(D)<a+1$) that such $a$ really exists. In fact, that is what you have to prove. Try to find another definition of $a$.
$endgroup$
– Peter Elias
Jan 31 at 7:40
add a comment |
$begingroup$
Let $D$ be a positive integer and the let the square root of $D$ be a real number. Assuming that the square root of D is not an integer (i.e. $D$ is not a perfect square), use Well-Ordering to prove that there exists some integer $a$ such that $a < sqrt(D) < a+1.$
My proof attempt begins with assuming that the square root of D is not an integer and letting $a$ be the least integer that satisfies the inequality $a < $sqrt(D)$ < a + 1$, then I'm not sure what to do. My intuition tells me to square both sides, but then how would I proceed with the proof.
irrational-numbers rational-numbers well-orders
asked Jan 31 at 7:15
Sanjoy The ManjoySanjoy The Manjoy
502315
$endgroup$
Let $D$ be a positive integer and the let the square root of $D$ be a real number. Assuming that the square root of D is not an integer (i.e. $D$ is not a perfect square), use Well-Ordering to prove that there exists some integer $a$ such that $a < sqrt(D) < a+1.$
My proof attempt begins with assuming that the square root of D is not an integer and letting $a$ be the least integer that satisfies the inequality $a < $sqrt(D)$ < a + 1$, then I'm not sure what to do. My intuition tells me to square both sides, but then how would I proceed with the proof.
irrational-numbers rational-numbers well-orders
irrational-numbers rational-numbers well-orders
asked Jan 31 at 7:15
Sanjoy The ManjoySanjoy The Manjoy
502315
asked Jan 31 at 7:15
Sanjoy The ManjoySanjoy The Manjoy
502315
asked Jan 31 at 7:15
Sanjoy The ManjoySanjoy The Manjoy
502315
asked Jan 31 at 7:15
Sanjoy The ManjoySanjoy The Manjoy
502315
asked Jan 31 at 7:15
Sanjoy The ManjoySanjoy The Manjoy
502315
502315
$begingroup$
It is not clear from your definition of $a$ (the least integer satisfying $a<mathit{sqrt}(D)<a+1$) that such $a$ really exists. In fact, that is what you have to prove. Try to find another definition of $a$.
$endgroup$
– Peter Elias
Jan 31 at 7:40
add a comment |
$begingroup$
It is not clear from your definition of $a$ (the least integer satisfying $a<mathit{sqrt}(D)<a+1$) that such $a$ really exists. In fact, that is what you have to prove. Try to find another definition of $a$.
$endgroup$
– Peter Elias
Jan 31 at 7:40
$begingroup$
It is not clear from your definition of $a$ (the least integer satisfying $a<mathit{sqrt}(D)<a+1$) that such $a$ really exists. In fact, that is what you have to prove. Try to find another definition of $a$.
$endgroup$
– Peter Elias
Jan 31 at 7:40
$begingroup$
It is not clear from your definition of $a$ (the least integer satisfying $a<mathit{sqrt}(D)<a+1$) that such $a$ really exists. In fact, that is what you have to prove. Try to find another definition of $a$.
$endgroup$
– Peter Elias
Jan 31 at 7:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $A = { n in mathbb{N} ~| ~ sqrt{D} < n }$.
By the Archimedean property, the set $A$ is not the empty set,
and $A$ is a subset of $mathbb{N}$.
Thus by the well ordering principle, $A$ has the least element, say $alpha$.
The assumption for $D$ implies that $D in mathbb{N}$ and $D neq m^2$ for every $min mathbb{N}$.
Thus $D in mathbb{N}$ and $D neq 1^2$, i.e., $D neq 1$.
Hence $1<D$, so $1<sqrt{D}$.
Thus $1 notin A$.
(If $1 in A$, then $sqrt{D}<1$. Contradiction.)
Hence $alpha$ cannot be $1$, so $alpha in mathbb{N} -{1} = {2,3,4,ldots}$.
Thus $alpha -1 in mathbb{N}$.
Because $alpha$ is the least element of $A$, we know that $alpha -1 notin A$,
i.e. , $alpha-1 leq sqrt{D}$.
If $alpha -1 = sqrt{D}$, then $D = (alpha-1)^2$, and D becomes a perfect square. contradiction.
Hence $alpha -1 < sqrt{D}$. And with $sqrt{D}< alpha$, we can conclude that
$alpha -1 < sqrt{D} < alpha$.
answered Jan 31 at 7:54
Doyun NamDoyun Nam
66619
$endgroup$
$begingroup$
Why is it obvious that D is greater than 1? Since D belongs to the positive integers, D can equal 1 and then the proof fails
$endgroup$
– Sanjoy The Manjoy
Jan 31 at 18:07
$begingroup$
@SanjoyTheManjoy I edit my answer.
$endgroup$
– Doyun Nam
Feb 1 at 5:20
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $A = { n in mathbb{N} ~| ~ sqrt{D} < n }$.
By the Archimedean property, the set $A$ is not the empty set,
and $A$ is a subset of $mathbb{N}$.
Thus by the well ordering principle, $A$ has the least element, say $alpha$.
The assumption for $D$ implies that $D in mathbb{N}$ and $D neq m^2$ for every $min mathbb{N}$.
Thus $D in mathbb{N}$ and $D neq 1^2$, i.e., $D neq 1$.
Hence $1<D$, so $1<sqrt{D}$.
Thus $1 notin A$.
(If $1 in A$, then $sqrt{D}<1$. Contradiction.)
Hence $alpha$ cannot be $1$, so $alpha in mathbb{N} -{1} = {2,3,4,ldots}$.
Thus $alpha -1 in mathbb{N}$.
Because $alpha$ is the least element of $A$, we know that $alpha -1 notin A$,
i.e. , $alpha-1 leq sqrt{D}$.
If $alpha -1 = sqrt{D}$, then $D = (alpha-1)^2$, and D becomes a perfect square. contradiction.
Hence $alpha -1 < sqrt{D}$. And with $sqrt{D}< alpha$, we can conclude that
$alpha -1 < sqrt{D} < alpha$.
answered Jan 31 at 7:54
Doyun NamDoyun Nam
66619
$endgroup$
$begingroup$
Why is it obvious that D is greater than 1? Since D belongs to the positive integers, D can equal 1 and then the proof fails
$endgroup$
– Sanjoy The Manjoy
Jan 31 at 18:07
$begingroup$
@SanjoyTheManjoy I edit my answer.
$endgroup$
– Doyun Nam
Feb 1 at 5:20
add a comment |
$begingroup$
Let $A = { n in mathbb{N} ~| ~ sqrt{D} < n }$.
By the Archimedean property, the set $A$ is not the empty set,
and $A$ is a subset of $mathbb{N}$.
Thus by the well ordering principle, $A$ has the least element, say $alpha$.
The assumption for $D$ implies that $D in mathbb{N}$ and $D neq m^2$ for every $min mathbb{N}$.
Thus $D in mathbb{N}$ and $D neq 1^2$, i.e., $D neq 1$.
Hence $1<D$, so $1<sqrt{D}$.
Thus $1 notin A$.
(If $1 in A$, then $sqrt{D}<1$. Contradiction.)
Hence $alpha$ cannot be $1$, so $alpha in mathbb{N} -{1} = {2,3,4,ldots}$.
Thus $alpha -1 in mathbb{N}$.
Because $alpha$ is the least element of $A$, we know that $alpha -1 notin A$,
i.e. , $alpha-1 leq sqrt{D}$.
If $alpha -1 = sqrt{D}$, then $D = (alpha-1)^2$, and D becomes a perfect square. contradiction.
Hence $alpha -1 < sqrt{D}$. And with $sqrt{D}< alpha$, we can conclude that
$alpha -1 < sqrt{D} < alpha$.
answered Jan 31 at 7:54
Doyun NamDoyun Nam
66619
$endgroup$
$begingroup$
Why is it obvious that D is greater than 1? Since D belongs to the positive integers, D can equal 1 and then the proof fails
$endgroup$
– Sanjoy The Manjoy
Jan 31 at 18:07
$begingroup$
@SanjoyTheManjoy I edit my answer.
$endgroup$
– Doyun Nam
Feb 1 at 5:20
add a comment |
$begingroup$
Let $A = { n in mathbb{N} ~| ~ sqrt{D} < n }$.
By the Archimedean property, the set $A$ is not the empty set,
and $A$ is a subset of $mathbb{N}$.
Thus by the well ordering principle, $A$ has the least element, say $alpha$.
The assumption for $D$ implies that $D in mathbb{N}$ and $D neq m^2$ for every $min mathbb{N}$.
Thus $D in mathbb{N}$ and $D neq 1^2$, i.e., $D neq 1$.
Hence $1<D$, so $1<sqrt{D}$.
Thus $1 notin A$.
(If $1 in A$, then $sqrt{D}<1$. Contradiction.)
Hence $alpha$ cannot be $1$, so $alpha in mathbb{N} -{1} = {2,3,4,ldots}$.
Thus $alpha -1 in mathbb{N}$.
Because $alpha$ is the least element of $A$, we know that $alpha -1 notin A$,
i.e. , $alpha-1 leq sqrt{D}$.
If $alpha -1 = sqrt{D}$, then $D = (alpha-1)^2$, and D becomes a perfect square. contradiction.
Hence $alpha -1 < sqrt{D}$. And with $sqrt{D}< alpha$, we can conclude that
$alpha -1 < sqrt{D} < alpha$.
answered Jan 31 at 7:54
Doyun NamDoyun Nam
66619
$endgroup$
Let $A = { n in mathbb{N} ~| ~ sqrt{D} < n }$.
By the Archimedean property, the set $A$ is not the empty set,
and $A$ is a subset of $mathbb{N}$.
Thus by the well ordering principle, $A$ has the least element, say $alpha$.
The assumption for $D$ implies that $D in mathbb{N}$ and $D neq m^2$ for every $min mathbb{N}$.
Thus $D in mathbb{N}$ and $D neq 1^2$, i.e., $D neq 1$.
Hence $1<D$, so $1<sqrt{D}$.
Thus $1 notin A$.
(If $1 in A$, then $sqrt{D}<1$. Contradiction.)
Hence $alpha$ cannot be $1$, so $alpha in mathbb{N} -{1} = {2,3,4,ldots}$.
Thus $alpha -1 in mathbb{N}$.
Because $alpha$ is the least element of $A$, we know that $alpha -1 notin A$,
i.e. , $alpha-1 leq sqrt{D}$.
If $alpha -1 = sqrt{D}$, then $D = (alpha-1)^2$, and D becomes a perfect square. contradiction.
Hence $alpha -1 < sqrt{D}$. And with $sqrt{D}< alpha$, we can conclude that
$alpha -1 < sqrt{D} < alpha$.
answered Jan 31 at 7:54
Doyun NamDoyun Nam
66619
edited Feb 1 at 5:19
answered Jan 31 at 7:54
Doyun NamDoyun Nam
66619
answered Jan 31 at 7:54
Doyun NamDoyun Nam
66619
answered Jan 31 at 7:54
Doyun NamDoyun Nam
66619
66619
$begingroup$
Why is it obvious that D is greater than 1? Since D belongs to the positive integers, D can equal 1 and then the proof fails
$endgroup$
– Sanjoy The Manjoy
Jan 31 at 18:07
$begingroup$
@SanjoyTheManjoy I edit my answer.
$endgroup$
– Doyun Nam
Feb 1 at 5:20
add a comment |
$begingroup$
Why is it obvious that D is greater than 1? Since D belongs to the positive integers, D can equal 1 and then the proof fails
$endgroup$
– Sanjoy The Manjoy
Jan 31 at 18:07
$begingroup$
@SanjoyTheManjoy I edit my answer.
$endgroup$
– Doyun Nam
Feb 1 at 5:20
$begingroup$
Why is it obvious that D is greater than 1? Since D belongs to the positive integers, D can equal 1 and then the proof fails
$endgroup$
– Sanjoy The Manjoy
Jan 31 at 18:07
$begingroup$
Why is it obvious that D is greater than 1? Since D belongs to the positive integers, D can equal 1 and then the proof fails
$endgroup$
– Sanjoy The Manjoy
Jan 31 at 18:07
$begingroup$
@SanjoyTheManjoy I edit my answer.
$endgroup$
– Doyun Nam
Feb 1 at 5:20
$begingroup$
@SanjoyTheManjoy I edit my answer.
$endgroup$
– Doyun Nam
Feb 1 at 5:20
add a comment |
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$begingroup$
It is not clear from your definition of $a$ (the least integer satisfying $a<mathit{sqrt}(D)<a+1$) that such $a$ really exists. In fact, that is what you have to prove. Try to find another definition of $a$.
$endgroup$
– Peter Elias
Jan 31 at 7:40