To prove special case when a <0Diophantine number has full measure but is meagerFor every $xinmathbb R$ and $varepsilon$ > 0 , there exist $,q,q'inmathbb Q$, such that $q<x<q'$ and $left |q-q' right |< varepsilon$Follow-up question on mathematical induction with arbitrary base caseProve that the real root of $x^3 + x + 1$ is irrationalCan this special case happen when working with L'Hopitals rule?Why are there more Irrationals than Rationals given the density of $Q$ in $R$?Proving/disproving statements with a given context of natural numbers.Proof that all real numbers have a rational Cauchy sequence?Can you prove the power rule for irrational exponents without invoking $e$?Which n for base case if number defined on $mathbb R$
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To prove special case when a
Diophantine number has full measure but is meagerFor every $xinmathbb R$ and $varepsilon$ > 0 , there exist $,q,q'inmathbb Q$, such that $q<x<q'$ and $left |q-q' right |< varepsilon$Follow-up question on mathematical induction with arbitrary base caseProve that the real root of $x^3 + x + 1$ is irrationalCan this special case happen when working with L'Hopitals rule?Why are there more Irrationals than Rationals given the density of $Q$ in $R$?Proving/disproving statements with a given context of natural numbers.Proof that all real numbers have a rational Cauchy sequence?Can you prove the power rule for irrational exponents without invoking $e$?Which n for base case if number defined on $mathbb R$
$begingroup$
Original Theorem : Given two real numbers $a$ and $b$ with $a < b$, where $a geq 0$ there exists a rational number $r$ satisfying $a < r < b$.
To be proven from above theorem :
Without doing too much work, show how to prove Theorem above in the case where a < 0 by converting into the case already proven (which is when $a geq 0$ in the above theorem)
Case when $a < 0$ (To be Proven )
Now Case1
Consider $b > 0$. Since $a < 0$ and $b > 0$, so we have $a < b$ and we have reduced to original theorem
Case 2
When $b < 0$. How do i do this ?
Thanks
real-analysis proof-writing self-learning
asked Mar 12 at 10:40
J. DeffJ. Deff
608416
$endgroup$
add a comment |
$begingroup$
Original Theorem : Given two real numbers $a$ and $b$ with $a < b$, where $a geq 0$ there exists a rational number $r$ satisfying $a < r < b$.
To be proven from above theorem :
Without doing too much work, show how to prove Theorem above in the case where a < 0 by converting into the case already proven (which is when $a geq 0$ in the above theorem)
Case when $a < 0$ (To be Proven )
Now Case1
Consider $b > 0$. Since $a < 0$ and $b > 0$, so we have $a < b$ and we have reduced to original theorem
Case 2
When $b < 0$. How do i do this ?
Thanks
real-analysis proof-writing self-learning
asked Mar 12 at 10:40
J. DeffJ. Deff
608416
$endgroup$
$begingroup$
You are done. Case 2 does not exist since $age0$ so $b>0$.
$endgroup$
– Parcly Taxel
Mar 12 at 10:41
$begingroup$
You begin by saying $ageq 0$, and then in case 1, you say $a<0$. Which is it?
$endgroup$
– 5xum
Mar 12 at 10:42
$begingroup$
@5xum please see the edit. Hope it makes clear
$endgroup$
– J. Deff
Mar 12 at 10:45
add a comment |
$begingroup$
Original Theorem : Given two real numbers $a$ and $b$ with $a < b$, where $a geq 0$ there exists a rational number $r$ satisfying $a < r < b$.
To be proven from above theorem :
Without doing too much work, show how to prove Theorem above in the case where a < 0 by converting into the case already proven (which is when $a geq 0$ in the above theorem)
Case when $a < 0$ (To be Proven )
Now Case1
Consider $b > 0$. Since $a < 0$ and $b > 0$, so we have $a < b$ and we have reduced to original theorem
Case 2
When $b < 0$. How do i do this ?
Thanks
real-analysis proof-writing self-learning
asked Mar 12 at 10:40
J. DeffJ. Deff
608416
$endgroup$
Original Theorem : Given two real numbers $a$ and $b$ with $a < b$, where $a geq 0$ there exists a rational number $r$ satisfying $a < r < b$.
To be proven from above theorem :
Without doing too much work, show how to prove Theorem above in the case where a < 0 by converting into the case already proven (which is when $a geq 0$ in the above theorem)
Case when $a < 0$ (To be Proven )
Now Case1
Consider $b > 0$. Since $a < 0$ and $b > 0$, so we have $a < b$ and we have reduced to original theorem
Case 2
When $b < 0$. How do i do this ?
Thanks
real-analysis proof-writing self-learning
real-analysis proof-writing self-learning
asked Mar 12 at 10:40
J. DeffJ. Deff
608416
asked Mar 12 at 10:40
J. DeffJ. Deff
608416
edited Mar 12 at 10:44
J. Deff
asked Mar 12 at 10:40
J. DeffJ. Deff
608416
asked Mar 12 at 10:40
J. DeffJ. Deff
608416
asked Mar 12 at 10:40
J. DeffJ. Deff
608416
608416
$begingroup$
You are done. Case 2 does not exist since $age0$ so $b>0$.
$endgroup$
– Parcly Taxel
Mar 12 at 10:41
$begingroup$
You begin by saying $ageq 0$, and then in case 1, you say $a<0$. Which is it?
$endgroup$
– 5xum
Mar 12 at 10:42
$begingroup$
@5xum please see the edit. Hope it makes clear
$endgroup$
– J. Deff
Mar 12 at 10:45
add a comment |
$begingroup$
You are done. Case 2 does not exist since $age0$ so $b>0$.
$endgroup$
– Parcly Taxel
Mar 12 at 10:41
$begingroup$
You begin by saying $ageq 0$, and then in case 1, you say $a<0$. Which is it?
$endgroup$
– 5xum
Mar 12 at 10:42
$begingroup$
@5xum please see the edit. Hope it makes clear
$endgroup$
– J. Deff
Mar 12 at 10:45
$begingroup$
You are done. Case 2 does not exist since $age0$ so $b>0$.
$endgroup$
– Parcly Taxel
Mar 12 at 10:41
$begingroup$
You are done. Case 2 does not exist since $age0$ so $b>0$.
$endgroup$
– Parcly Taxel
Mar 12 at 10:41
$begingroup$
You begin by saying $ageq 0$, and then in case 1, you say $a<0$. Which is it?
$endgroup$
– 5xum
Mar 12 at 10:42
$begingroup$
You begin by saying $ageq 0$, and then in case 1, you say $a<0$. Which is it?
$endgroup$
– 5xum
Mar 12 at 10:42
$begingroup$
@5xum please see the edit. Hope it makes clear
$endgroup$
– J. Deff
Mar 12 at 10:45
$begingroup$
@5xum please see the edit. Hope it makes clear
$endgroup$
– J. Deff
Mar 12 at 10:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In Case $1$, you did not "reduce to the original theorem". The original theorem only tells you something if $a<b$ and $ageq 0$. You only proved $a<b$, while $ageq 0$ is still not true, and therefore, the original theorem tells you nothing in this case. Your argument is incorrect.
However, in case $1$, it should be very easy to find a rational number satisfying $a<r<b$. Think about it. $a$ is smaller than $0$. $b$ is larger than $0$. That is, $a$ is on the left of $0$. $b$ is to the right of $0$. Can you think of any number that is in between $a$ and $b$? Is that number rational?
For case $2$, you can actually reduce to the original theorem. Think about what happens if you define $a'=-b$ and $b'=-a$. Do $a', b'$ satisfy the conditions of the original theorem?
answered Mar 12 at 10:48
5xum5xum
91.4k394161
$endgroup$
$begingroup$
i see. For 2nd case, i was thinking on the same lines, i was multiplying the equality with negative and reversing them but couldnot get to it,. Thanks
$endgroup$
– J. Deff
Mar 12 at 10:57
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In Case $1$, you did not "reduce to the original theorem". The original theorem only tells you something if $a<b$ and $ageq 0$. You only proved $a<b$, while $ageq 0$ is still not true, and therefore, the original theorem tells you nothing in this case. Your argument is incorrect.
However, in case $1$, it should be very easy to find a rational number satisfying $a<r<b$. Think about it. $a$ is smaller than $0$. $b$ is larger than $0$. That is, $a$ is on the left of $0$. $b$ is to the right of $0$. Can you think of any number that is in between $a$ and $b$? Is that number rational?
For case $2$, you can actually reduce to the original theorem. Think about what happens if you define $a'=-b$ and $b'=-a$. Do $a', b'$ satisfy the conditions of the original theorem?
answered Mar 12 at 10:48
5xum5xum
91.4k394161
$endgroup$
$begingroup$
i see. For 2nd case, i was thinking on the same lines, i was multiplying the equality with negative and reversing them but couldnot get to it,. Thanks
$endgroup$
– J. Deff
Mar 12 at 10:57
add a comment |
$begingroup$
In Case $1$, you did not "reduce to the original theorem". The original theorem only tells you something if $a<b$ and $ageq 0$. You only proved $a<b$, while $ageq 0$ is still not true, and therefore, the original theorem tells you nothing in this case. Your argument is incorrect.
However, in case $1$, it should be very easy to find a rational number satisfying $a<r<b$. Think about it. $a$ is smaller than $0$. $b$ is larger than $0$. That is, $a$ is on the left of $0$. $b$ is to the right of $0$. Can you think of any number that is in between $a$ and $b$? Is that number rational?
For case $2$, you can actually reduce to the original theorem. Think about what happens if you define $a'=-b$ and $b'=-a$. Do $a', b'$ satisfy the conditions of the original theorem?
answered Mar 12 at 10:48
5xum5xum
91.4k394161
$endgroup$
$begingroup$
i see. For 2nd case, i was thinking on the same lines, i was multiplying the equality with negative and reversing them but couldnot get to it,. Thanks
$endgroup$
– J. Deff
Mar 12 at 10:57
add a comment |
$begingroup$
In Case $1$, you did not "reduce to the original theorem". The original theorem only tells you something if $a<b$ and $ageq 0$. You only proved $a<b$, while $ageq 0$ is still not true, and therefore, the original theorem tells you nothing in this case. Your argument is incorrect.
However, in case $1$, it should be very easy to find a rational number satisfying $a<r<b$. Think about it. $a$ is smaller than $0$. $b$ is larger than $0$. That is, $a$ is on the left of $0$. $b$ is to the right of $0$. Can you think of any number that is in between $a$ and $b$? Is that number rational?
For case $2$, you can actually reduce to the original theorem. Think about what happens if you define $a'=-b$ and $b'=-a$. Do $a', b'$ satisfy the conditions of the original theorem?
answered Mar 12 at 10:48
5xum5xum
91.4k394161
$endgroup$
In Case $1$, you did not "reduce to the original theorem". The original theorem only tells you something if $a<b$ and $ageq 0$. You only proved $a<b$, while $ageq 0$ is still not true, and therefore, the original theorem tells you nothing in this case. Your argument is incorrect.
However, in case $1$, it should be very easy to find a rational number satisfying $a<r<b$. Think about it. $a$ is smaller than $0$. $b$ is larger than $0$. That is, $a$ is on the left of $0$. $b$ is to the right of $0$. Can you think of any number that is in between $a$ and $b$? Is that number rational?
For case $2$, you can actually reduce to the original theorem. Think about what happens if you define $a'=-b$ and $b'=-a$. Do $a', b'$ satisfy the conditions of the original theorem?
answered Mar 12 at 10:48
5xum5xum
91.4k394161
answered Mar 12 at 10:48
5xum5xum
91.4k394161
answered Mar 12 at 10:48
5xum5xum
91.4k394161
answered Mar 12 at 10:48
5xum5xum
91.4k394161
91.4k394161
$begingroup$
i see. For 2nd case, i was thinking on the same lines, i was multiplying the equality with negative and reversing them but couldnot get to it,. Thanks
$endgroup$
– J. Deff
Mar 12 at 10:57
add a comment |
$begingroup$
i see. For 2nd case, i was thinking on the same lines, i was multiplying the equality with negative and reversing them but couldnot get to it,. Thanks
$endgroup$
– J. Deff
Mar 12 at 10:57
$begingroup$
i see. For 2nd case, i was thinking on the same lines, i was multiplying the equality with negative and reversing them but couldnot get to it,. Thanks
$endgroup$
– J. Deff
Mar 12 at 10:57
$begingroup$
i see. For 2nd case, i was thinking on the same lines, i was multiplying the equality with negative and reversing them but couldnot get to it,. Thanks
$endgroup$
– J. Deff
Mar 12 at 10:57
add a comment |
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$begingroup$
You are done. Case 2 does not exist since $age0$ so $b>0$.
$endgroup$
– Parcly Taxel
Mar 12 at 10:41
$begingroup$
You begin by saying $ageq 0$, and then in case 1, you say $a<0$. Which is it?
$endgroup$
– 5xum
Mar 12 at 10:42
$begingroup$
@5xum please see the edit. Hope it makes clear
$endgroup$
– J. Deff
Mar 12 at 10:45