Mixing
Abstract Algebra Square Roots Are Irrational
$begingroup$
For part (a), I begin by trying to prove $S$ is empty implies the square root of $D$ is irrational. If we take the contrapositive of this implication, this is equivalent to proving that if the square root of $D$ is rational then $S$ is not empty. Let $D$ be a positive integer and suppose $D$ is rational. Then it follows that $D^2$ is a perfect square. Moreover, S = n*D^1/2 and since n and D are positive integers, then by definition, S is non-empty. Conversely, we wish to prove that the square root of D is not rational implies S is empty. But this trivially follows from the product of a rational and irrational number. I'm not sure if my reasoning is sound, but constructive criticism would be appreciated.
For part (b), I'm not sure how exactly to use Well-Ordering to prove the inequality. I suppose my initial thought was to square both sides of the inequality and then rearrange some terms. Because the square root of D is not rational, it follows D^2 is not a perfect square. Then we can bound it appropriately?
For part (c), I think what we should begin with is to consider the number m*(D^1/2 - a). From there, I do not know how to proceed. Any help would greatly be appreciated.
abstract-algebra irrational-numbers rational-numbers
asked Jan 29 at 6:45
Sanjoy The ManjoySanjoy The Manjoy
502315
$endgroup$
add a comment |
$begingroup$
For part (a), I begin by trying to prove $S$ is empty implies the square root of $D$ is irrational. If we take the contrapositive of this implication, this is equivalent to proving that if the square root of $D$ is rational then $S$ is not empty. Let $D$ be a positive integer and suppose $D$ is rational. Then it follows that $D^2$ is a perfect square. Moreover, S = n*D^1/2 and since n and D are positive integers, then by definition, S is non-empty. Conversely, we wish to prove that the square root of D is not rational implies S is empty. But this trivially follows from the product of a rational and irrational number. I'm not sure if my reasoning is sound, but constructive criticism would be appreciated.
For part (b), I'm not sure how exactly to use Well-Ordering to prove the inequality. I suppose my initial thought was to square both sides of the inequality and then rearrange some terms. Because the square root of D is not rational, it follows D^2 is not a perfect square. Then we can bound it appropriately?
For part (c), I think what we should begin with is to consider the number m*(D^1/2 - a). From there, I do not know how to proceed. Any help would greatly be appreciated.
abstract-algebra irrational-numbers rational-numbers
asked Jan 29 at 6:45
Sanjoy The ManjoySanjoy The Manjoy
502315
$endgroup$
$begingroup$
Can you please type your question out instead of pasting an image? Also, can you use MathJax to typeset your formulae?
$endgroup$
– Morgan Rodgers
Jan 29 at 7:08
$begingroup$
Sorry I just need help
$endgroup$
– Sanjoy The Manjoy
Jan 29 at 19:06
add a comment |
$begingroup$
For part (a), I begin by trying to prove $S$ is empty implies the square root of $D$ is irrational. If we take the contrapositive of this implication, this is equivalent to proving that if the square root of $D$ is rational then $S$ is not empty. Let $D$ be a positive integer and suppose $D$ is rational. Then it follows that $D^2$ is a perfect square. Moreover, S = n*D^1/2 and since n and D are positive integers, then by definition, S is non-empty. Conversely, we wish to prove that the square root of D is not rational implies S is empty. But this trivially follows from the product of a rational and irrational number. I'm not sure if my reasoning is sound, but constructive criticism would be appreciated.
For part (b), I'm not sure how exactly to use Well-Ordering to prove the inequality. I suppose my initial thought was to square both sides of the inequality and then rearrange some terms. Because the square root of D is not rational, it follows D^2 is not a perfect square. Then we can bound it appropriately?
For part (c), I think what we should begin with is to consider the number m*(D^1/2 - a). From there, I do not know how to proceed. Any help would greatly be appreciated.
abstract-algebra irrational-numbers rational-numbers
asked Jan 29 at 6:45
Sanjoy The ManjoySanjoy The Manjoy
502315
$endgroup$
For part (a), I begin by trying to prove $S$ is empty implies the square root of $D$ is irrational. If we take the contrapositive of this implication, this is equivalent to proving that if the square root of $D$ is rational then $S$ is not empty. Let $D$ be a positive integer and suppose $D$ is rational. Then it follows that $D^2$ is a perfect square. Moreover, S = n*D^1/2 and since n and D are positive integers, then by definition, S is non-empty. Conversely, we wish to prove that the square root of D is not rational implies S is empty. But this trivially follows from the product of a rational and irrational number. I'm not sure if my reasoning is sound, but constructive criticism would be appreciated.
For part (b), I'm not sure how exactly to use Well-Ordering to prove the inequality. I suppose my initial thought was to square both sides of the inequality and then rearrange some terms. Because the square root of D is not rational, it follows D^2 is not a perfect square. Then we can bound it appropriately?
For part (c), I think what we should begin with is to consider the number m*(D^1/2 - a). From there, I do not know how to proceed. Any help would greatly be appreciated.
abstract-algebra irrational-numbers rational-numbers
abstract-algebra irrational-numbers rational-numbers
asked Jan 29 at 6:45
Sanjoy The ManjoySanjoy The Manjoy
502315
asked Jan 29 at 6:45
Sanjoy The ManjoySanjoy The Manjoy
502315
edited Jan 29 at 23:40
Sanjoy The Manjoy
asked Jan 29 at 6:45
Sanjoy The ManjoySanjoy The Manjoy
502315
asked Jan 29 at 6:45
Sanjoy The ManjoySanjoy The Manjoy
502315
asked Jan 29 at 6:45
Sanjoy The ManjoySanjoy The Manjoy
502315
502315
$begingroup$
Can you please type your question out instead of pasting an image? Also, can you use MathJax to typeset your formulae?
$endgroup$
– Morgan Rodgers
Jan 29 at 7:08
$begingroup$
Sorry I just need help
$endgroup$
– Sanjoy The Manjoy
Jan 29 at 19:06
add a comment |
$begingroup$
Can you please type your question out instead of pasting an image? Also, can you use MathJax to typeset your formulae?
$endgroup$
– Morgan Rodgers
Jan 29 at 7:08
$begingroup$
Sorry I just need help
$endgroup$
– Sanjoy The Manjoy
Jan 29 at 19:06
$begingroup$
Can you please type your question out instead of pasting an image? Also, can you use MathJax to typeset your formulae?
$endgroup$
– Morgan Rodgers
Jan 29 at 7:08
$begingroup$
Can you please type your question out instead of pasting an image? Also, can you use MathJax to typeset your formulae?
$endgroup$
– Morgan Rodgers
Jan 29 at 7:08
$begingroup$
Sorry I just need help
$endgroup$
– Sanjoy The Manjoy
Jan 29 at 19:06
$begingroup$
Sorry I just need help
$endgroup$
– Sanjoy The Manjoy
Jan 29 at 19:06
add a comment |
1 Answer
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$begingroup$
In part (a) you are right about the contrapositive of the implication but you proved it wrong because you said that "and since $n$ and $D$ are positive integers, then by definition, S is non-empty" but I don't see that this is true because $sqrt D in mathbb Q$ doesn't imply that $Din mathbb Z$
So instead you can say that if $sqrt D in mathbb Q implies sqrt D= frac{a}{b}$ where $a,bin mathbb Z$
(Note that $a$ and $b$ are positive here since a square root is always positive so $a,bin mathbb N$)
Which gives that $bsqrt D=a in mathbb Z$ so S is not empty since $bin S$.
Now for the sufficient condition, yes it is trivial since if $sqrt D not in mathbb Q implies nsqrt D not in mathbb Q, forall nin mathbb N implies S=phi$
answered Jan 29 at 7:02
Fareed AFFareed AF
632112
$endgroup$
add a comment |
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1 Answer
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$begingroup$
In part (a) you are right about the contrapositive of the implication but you proved it wrong because you said that "and since $n$ and $D$ are positive integers, then by definition, S is non-empty" but I don't see that this is true because $sqrt D in mathbb Q$ doesn't imply that $Din mathbb Z$
So instead you can say that if $sqrt D in mathbb Q implies sqrt D= frac{a}{b}$ where $a,bin mathbb Z$
(Note that $a$ and $b$ are positive here since a square root is always positive so $a,bin mathbb N$)
Which gives that $bsqrt D=a in mathbb Z$ so S is not empty since $bin S$.
Now for the sufficient condition, yes it is trivial since if $sqrt D not in mathbb Q implies nsqrt D not in mathbb Q, forall nin mathbb N implies S=phi$
answered Jan 29 at 7:02
Fareed AFFareed AF
632112
$endgroup$
add a comment |
$begingroup$
In part (a) you are right about the contrapositive of the implication but you proved it wrong because you said that "and since $n$ and $D$ are positive integers, then by definition, S is non-empty" but I don't see that this is true because $sqrt D in mathbb Q$ doesn't imply that $Din mathbb Z$
So instead you can say that if $sqrt D in mathbb Q implies sqrt D= frac{a}{b}$ where $a,bin mathbb Z$
(Note that $a$ and $b$ are positive here since a square root is always positive so $a,bin mathbb N$)
Which gives that $bsqrt D=a in mathbb Z$ so S is not empty since $bin S$.
Now for the sufficient condition, yes it is trivial since if $sqrt D not in mathbb Q implies nsqrt D not in mathbb Q, forall nin mathbb N implies S=phi$
answered Jan 29 at 7:02
Fareed AFFareed AF
632112
$endgroup$
add a comment |
$begingroup$
In part (a) you are right about the contrapositive of the implication but you proved it wrong because you said that "and since $n$ and $D$ are positive integers, then by definition, S is non-empty" but I don't see that this is true because $sqrt D in mathbb Q$ doesn't imply that $Din mathbb Z$
So instead you can say that if $sqrt D in mathbb Q implies sqrt D= frac{a}{b}$ where $a,bin mathbb Z$
(Note that $a$ and $b$ are positive here since a square root is always positive so $a,bin mathbb N$)
Which gives that $bsqrt D=a in mathbb Z$ so S is not empty since $bin S$.
Now for the sufficient condition, yes it is trivial since if $sqrt D not in mathbb Q implies nsqrt D not in mathbb Q, forall nin mathbb N implies S=phi$
answered Jan 29 at 7:02
Fareed AFFareed AF
632112
$endgroup$
In part (a) you are right about the contrapositive of the implication but you proved it wrong because you said that "and since $n$ and $D$ are positive integers, then by definition, S is non-empty" but I don't see that this is true because $sqrt D in mathbb Q$ doesn't imply that $Din mathbb Z$
So instead you can say that if $sqrt D in mathbb Q implies sqrt D= frac{a}{b}$ where $a,bin mathbb Z$
(Note that $a$ and $b$ are positive here since a square root is always positive so $a,bin mathbb N$)
Which gives that $bsqrt D=a in mathbb Z$ so S is not empty since $bin S$.
Now for the sufficient condition, yes it is trivial since if $sqrt D not in mathbb Q implies nsqrt D not in mathbb Q, forall nin mathbb N implies S=phi$
answered Jan 29 at 7:02
Fareed AFFareed AF
632112
edited Jan 29 at 7:13
answered Jan 29 at 7:02
Fareed AFFareed AF
632112
answered Jan 29 at 7:02
Fareed AFFareed AF
632112
answered Jan 29 at 7:02
Fareed AFFareed AF
632112
632112
add a comment |
add a comment |
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$begingroup$
Can you please type your question out instead of pasting an image? Also, can you use MathJax to typeset your formulae?
$endgroup$
– Morgan Rodgers
Jan 29 at 7:08
$begingroup$
Sorry I just need help
$endgroup$
– Sanjoy The Manjoy
Jan 29 at 19:06