Check my proof of 0 < |r - q| < epsilon. (Real # - Rational #) [duplicate]
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This question already has an answer here:
Real Analysis Inequality Proof Involving Reals and Rationals $0 < |r - q| < varepsilon$
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I am working on this exercise:
$$ forall varepsilon > 0, exists q in Q text{ where } 0 < |r - q| < varepsilon $$
To clarify, r is a real number, q is a rational number. This is what I have so far:
$$ 0 < |r - q| < varepsilon $$
$$ -varepsilon < (r-q) < varepsilon $$
$$ r - varepsilon < q < r + varepsilon $$
Then making use of a given theorem that "Between any two distinct real numbers there is a rational number and an irrational number," I conclude the proof by claiming that the inequality must hold by the theorem and given that $r - epsilon$ and $r + epsilon$ are two distinct real numbers.
Is this sufficient for the proof? Otherwise, would appreciate some help on how to proof the above inequality more rigorously.
I recall the Archimedean Property that for each positive real number r, there exists a positive integer n such that $frac{1}{n} < r$. There is a hint to use this condition, but I am unsure how to apply it into the proof.
Thanks in advance for any help!
real-analysis proof-verification inequality proof-writing rational-numbers
asked Jan 13 at 19:53
Zen'zZen'z
393
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Jan 13 at 20:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Real Analysis Inequality Proof Involving Reals and Rationals $0 < |r - q| < varepsilon$
2 answers
I am working on this exercise:
$$ forall varepsilon > 0, exists q in Q text{ where } 0 < |r - q| < varepsilon $$
To clarify, r is a real number, q is a rational number. This is what I have so far:
$$ 0 < |r - q| < varepsilon $$
$$ -varepsilon < (r-q) < varepsilon $$
$$ r - varepsilon < q < r + varepsilon $$
Then making use of a given theorem that "Between any two distinct real numbers there is a rational number and an irrational number," I conclude the proof by claiming that the inequality must hold by the theorem and given that $r - epsilon$ and $r + epsilon$ are two distinct real numbers.
Is this sufficient for the proof? Otherwise, would appreciate some help on how to proof the above inequality more rigorously.
I recall the Archimedean Property that for each positive real number r, there exists a positive integer n such that $frac{1}{n} < r$. There is a hint to use this condition, but I am unsure how to apply it into the proof.
Thanks in advance for any help!
real-analysis proof-verification inequality proof-writing rational-numbers
asked Jan 13 at 19:53
Zen'zZen'z
393
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Jan 13 at 20:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Your proof is backwards. You're starting from the thing that you want to prove, then working backwards from that. Go the other way: start from what you know, and show that such a $q$ has to exist.
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– user3482749
Jan 13 at 19:57
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You already asked a similar question yesterday
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– rtybase
Jan 13 at 20:24
add a comment |
$begingroup$
This question already has an answer here:
Real Analysis Inequality Proof Involving Reals and Rationals $0 < |r - q| < varepsilon$
2 answers
I am working on this exercise:
$$ forall varepsilon > 0, exists q in Q text{ where } 0 < |r - q| < varepsilon $$
To clarify, r is a real number, q is a rational number. This is what I have so far:
$$ 0 < |r - q| < varepsilon $$
$$ -varepsilon < (r-q) < varepsilon $$
$$ r - varepsilon < q < r + varepsilon $$
Then making use of a given theorem that "Between any two distinct real numbers there is a rational number and an irrational number," I conclude the proof by claiming that the inequality must hold by the theorem and given that $r - epsilon$ and $r + epsilon$ are two distinct real numbers.
Is this sufficient for the proof? Otherwise, would appreciate some help on how to proof the above inequality more rigorously.
I recall the Archimedean Property that for each positive real number r, there exists a positive integer n such that $frac{1}{n} < r$. There is a hint to use this condition, but I am unsure how to apply it into the proof.
Thanks in advance for any help!
real-analysis proof-verification inequality proof-writing rational-numbers
asked Jan 13 at 19:53
Zen'zZen'z
393
$endgroup$
This question already has an answer here:
Real Analysis Inequality Proof Involving Reals and Rationals $0 < |r - q| < varepsilon$
2 answers
I am working on this exercise:
$$ forall varepsilon > 0, exists q in Q text{ where } 0 < |r - q| < varepsilon $$
To clarify, r is a real number, q is a rational number. This is what I have so far:
$$ 0 < |r - q| < varepsilon $$
$$ -varepsilon < (r-q) < varepsilon $$
$$ r - varepsilon < q < r + varepsilon $$
Then making use of a given theorem that "Between any two distinct real numbers there is a rational number and an irrational number," I conclude the proof by claiming that the inequality must hold by the theorem and given that $r - epsilon$ and $r + epsilon$ are two distinct real numbers.
Is this sufficient for the proof? Otherwise, would appreciate some help on how to proof the above inequality more rigorously.
I recall the Archimedean Property that for each positive real number r, there exists a positive integer n such that $frac{1}{n} < r$. There is a hint to use this condition, but I am unsure how to apply it into the proof.
Thanks in advance for any help!
This question already has an answer here:
Real Analysis Inequality Proof Involving Reals and Rationals $0 < |r - q| < varepsilon$
2 answers
real-analysis proof-verification inequality proof-writing rational-numbers
real-analysis proof-verification inequality proof-writing rational-numbers
asked Jan 13 at 19:53
Zen'zZen'z
393
asked Jan 13 at 19:53
Zen'zZen'z
393
asked Jan 13 at 19:53
Zen'zZen'z
393
asked Jan 13 at 19:53
Zen'zZen'z
393
asked Jan 13 at 19:53
Zen'zZen'z
393
393
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Jan 13 at 20:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Jan 13 at 20:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Your proof is backwards. You're starting from the thing that you want to prove, then working backwards from that. Go the other way: start from what you know, and show that such a $q$ has to exist.
$endgroup$
– user3482749
Jan 13 at 19:57
$begingroup$
You already asked a similar question yesterday
$endgroup$
– rtybase
Jan 13 at 20:24
add a comment |
$begingroup$
Your proof is backwards. You're starting from the thing that you want to prove, then working backwards from that. Go the other way: start from what you know, and show that such a $q$ has to exist.
$endgroup$
– user3482749
Jan 13 at 19:57
$begingroup$
You already asked a similar question yesterday
$endgroup$
– rtybase
Jan 13 at 20:24
$begingroup$
Your proof is backwards. You're starting from the thing that you want to prove, then working backwards from that. Go the other way: start from what you know, and show that such a $q$ has to exist.
$endgroup$
– user3482749
Jan 13 at 19:57
$begingroup$
Your proof is backwards. You're starting from the thing that you want to prove, then working backwards from that. Go the other way: start from what you know, and show that such a $q$ has to exist.
$endgroup$
– user3482749
Jan 13 at 19:57
$begingroup$
You already asked a similar question yesterday
$endgroup$
– rtybase
Jan 13 at 20:24
$begingroup$
You already asked a similar question yesterday
$endgroup$
– rtybase
Jan 13 at 20:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you're using the density of the rationals, then your proof is upside down, i.e., to fix it, go from $r - epsilon < q < r + epsilon$ to your claim.
Nevertheless, you're using a rocket launcher against a fly; in fact, the two statements are pretty much equal, so it's almost circular reasoning. Suppose WLOG $r > 0$; since you have the Archimedian property, fix $epsilon$ and pick the least $n$ such that $epsilon geq 1/n$. If $r < epsilon$ pick $q=0$ and you're done. Otherwise, pick the biggest $m$ such that $r - m/n geq 0$; since $r - (m+1)/n < 0$ (because $m$ is maximal), $r-m/n < 1/n leq epsilon$. Pick $q = m/n$ and we're done.
answered Jan 13 at 20:13
Lucas HenriqueLucas Henrique
1,031414
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you're using the density of the rationals, then your proof is upside down, i.e., to fix it, go from $r - epsilon < q < r + epsilon$ to your claim.
Nevertheless, you're using a rocket launcher against a fly; in fact, the two statements are pretty much equal, so it's almost circular reasoning. Suppose WLOG $r > 0$; since you have the Archimedian property, fix $epsilon$ and pick the least $n$ such that $epsilon geq 1/n$. If $r < epsilon$ pick $q=0$ and you're done. Otherwise, pick the biggest $m$ such that $r - m/n geq 0$; since $r - (m+1)/n < 0$ (because $m$ is maximal), $r-m/n < 1/n leq epsilon$. Pick $q = m/n$ and we're done.
answered Jan 13 at 20:13
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
add a comment |
$begingroup$
If you're using the density of the rationals, then your proof is upside down, i.e., to fix it, go from $r - epsilon < q < r + epsilon$ to your claim.
Nevertheless, you're using a rocket launcher against a fly; in fact, the two statements are pretty much equal, so it's almost circular reasoning. Suppose WLOG $r > 0$; since you have the Archimedian property, fix $epsilon$ and pick the least $n$ such that $epsilon geq 1/n$. If $r < epsilon$ pick $q=0$ and you're done. Otherwise, pick the biggest $m$ such that $r - m/n geq 0$; since $r - (m+1)/n < 0$ (because $m$ is maximal), $r-m/n < 1/n leq epsilon$. Pick $q = m/n$ and we're done.
answered Jan 13 at 20:13
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
add a comment |
$begingroup$
If you're using the density of the rationals, then your proof is upside down, i.e., to fix it, go from $r - epsilon < q < r + epsilon$ to your claim.
Nevertheless, you're using a rocket launcher against a fly; in fact, the two statements are pretty much equal, so it's almost circular reasoning. Suppose WLOG $r > 0$; since you have the Archimedian property, fix $epsilon$ and pick the least $n$ such that $epsilon geq 1/n$. If $r < epsilon$ pick $q=0$ and you're done. Otherwise, pick the biggest $m$ such that $r - m/n geq 0$; since $r - (m+1)/n < 0$ (because $m$ is maximal), $r-m/n < 1/n leq epsilon$. Pick $q = m/n$ and we're done.
answered Jan 13 at 20:13
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
If you're using the density of the rationals, then your proof is upside down, i.e., to fix it, go from $r - epsilon < q < r + epsilon$ to your claim.
Nevertheless, you're using a rocket launcher against a fly; in fact, the two statements are pretty much equal, so it's almost circular reasoning. Suppose WLOG $r > 0$; since you have the Archimedian property, fix $epsilon$ and pick the least $n$ such that $epsilon geq 1/n$. If $r < epsilon$ pick $q=0$ and you're done. Otherwise, pick the biggest $m$ such that $r - m/n geq 0$; since $r - (m+1)/n < 0$ (because $m$ is maximal), $r-m/n < 1/n leq epsilon$. Pick $q = m/n$ and we're done.
answered Jan 13 at 20:13
Lucas HenriqueLucas Henrique
1,031414
answered Jan 13 at 20:13
Lucas HenriqueLucas Henrique
1,031414
answered Jan 13 at 20:13
Lucas HenriqueLucas Henrique
1,031414
answered Jan 13 at 20:13
Lucas HenriqueLucas Henrique
1,031414
1,031414
add a comment |
add a comment |
$begingroup$
Your proof is backwards. You're starting from the thing that you want to prove, then working backwards from that. Go the other way: start from what you know, and show that such a $q$ has to exist.
$endgroup$
– user3482749
Jan 13 at 19:57
$begingroup$
You already asked a similar question yesterday
$endgroup$
– rtybase
Jan 13 at 20:24