Proof explanation that the square root of 2 exists [closed]
$begingroup$
Theorem.
There exists $xinmathbb {R}: $ with ${x^2=2} $.
Can someone please share a simple proof of the above theorem by defining,
$${yinmathbb R: mbox{$y^2leq {2}$}}.$$
How do we proceed to use the completeness axiom to prove the theorem? The proof that my lecturer has provided uses contradictions. He also uses the same process to define the square root function. I'm finding it difficult to grasp. Thanks.
real-analysis proof-explanation intuition
asked Dec 14 '18 at 0:03
ImranImran
1955
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closed as too broad by José Carlos Santos, Jyrki Lahtonen, Alex Provost, Brian Borchers, amWhy Dec 22 '18 at 20:27
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Theorem.
There exists $xinmathbb {R}: $ with ${x^2=2} $.
Can someone please share a simple proof of the above theorem by defining,
$${yinmathbb R: mbox{$y^2leq {2}$}}.$$
How do we proceed to use the completeness axiom to prove the theorem? The proof that my lecturer has provided uses contradictions. He also uses the same process to define the square root function. I'm finding it difficult to grasp. Thanks.
real-analysis proof-explanation intuition
asked Dec 14 '18 at 0:03
ImranImran
1955
$endgroup$
closed as too broad by José Carlos Santos, Jyrki Lahtonen, Alex Provost, Brian Borchers, amWhy Dec 22 '18 at 20:27
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
If you edit the question to post your lecturer's proof and flag the place that confuses you perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 14 '18 at 0:07
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I know, sorry. I was having difficulty writing it up so that's why I didn't. I've posted a image of it. I'm confused from that part when delta is used and onwards. I'm not sure where the values for 'min' come from and how the quadratic terms change. Hope that helps.
$endgroup$
– Imran
Dec 14 '18 at 1:05
$begingroup$
A bit off-topic, but I find it funny how my answer looks like your lecturer's paper.
$endgroup$
– Lucas Henrique
Dec 14 '18 at 1:12
add a comment |
$begingroup$
Theorem.
There exists $xinmathbb {R}: $ with ${x^2=2} $.
Can someone please share a simple proof of the above theorem by defining,
$${yinmathbb R: mbox{$y^2leq {2}$}}.$$
How do we proceed to use the completeness axiom to prove the theorem? The proof that my lecturer has provided uses contradictions. He also uses the same process to define the square root function. I'm finding it difficult to grasp. Thanks.
real-analysis proof-explanation intuition
asked Dec 14 '18 at 0:03
ImranImran
1955
$endgroup$
Theorem.
There exists $xinmathbb {R}: $ with ${x^2=2} $.
Can someone please share a simple proof of the above theorem by defining,
$${yinmathbb R: mbox{$y^2leq {2}$}}.$$
How do we proceed to use the completeness axiom to prove the theorem? The proof that my lecturer has provided uses contradictions. He also uses the same process to define the square root function. I'm finding it difficult to grasp. Thanks.
real-analysis proof-explanation intuition
real-analysis proof-explanation intuition
asked Dec 14 '18 at 0:03
ImranImran
1955
asked Dec 14 '18 at 0:03
ImranImran
1955
edited Dec 14 '18 at 0:57
Imran
asked Dec 14 '18 at 0:03
ImranImran
1955
asked Dec 14 '18 at 0:03
ImranImran
1955
asked Dec 14 '18 at 0:03
ImranImran
1955
1955
closed as too broad by José Carlos Santos, Jyrki Lahtonen, Alex Provost, Brian Borchers, amWhy Dec 22 '18 at 20:27
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by José Carlos Santos, Jyrki Lahtonen, Alex Provost, Brian Borchers, amWhy Dec 22 '18 at 20:27
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
If you edit the question to post your lecturer's proof and flag the place that confuses you perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 14 '18 at 0:07
$begingroup$
I know, sorry. I was having difficulty writing it up so that's why I didn't. I've posted a image of it. I'm confused from that part when delta is used and onwards. I'm not sure where the values for 'min' come from and how the quadratic terms change. Hope that helps.
$endgroup$
– Imran
Dec 14 '18 at 1:05
$begingroup$
A bit off-topic, but I find it funny how my answer looks like your lecturer's paper.
$endgroup$
– Lucas Henrique
Dec 14 '18 at 1:12
add a comment |
3
$begingroup$
If you edit the question to post your lecturer's proof and flag the place that confuses you perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 14 '18 at 0:07
$begingroup$
I know, sorry. I was having difficulty writing it up so that's why I didn't. I've posted a image of it. I'm confused from that part when delta is used and onwards. I'm not sure where the values for 'min' come from and how the quadratic terms change. Hope that helps.
$endgroup$
– Imran
Dec 14 '18 at 1:05
$begingroup$
A bit off-topic, but I find it funny how my answer looks like your lecturer's paper.
$endgroup$
– Lucas Henrique
Dec 14 '18 at 1:12
3
3
$begingroup$
If you edit the question to post your lecturer's proof and flag the place that confuses you perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 14 '18 at 0:07
$begingroup$
If you edit the question to post your lecturer's proof and flag the place that confuses you perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 14 '18 at 0:07
$begingroup$
I know, sorry. I was having difficulty writing it up so that's why I didn't. I've posted a image of it. I'm confused from that part when delta is used and onwards. I'm not sure where the values for 'min' come from and how the quadratic terms change. Hope that helps.
$endgroup$
– Imran
Dec 14 '18 at 1:05
$begingroup$
I know, sorry. I was having difficulty writing it up so that's why I didn't. I've posted a image of it. I'm confused from that part when delta is used and onwards. I'm not sure where the values for 'min' come from and how the quadratic terms change. Hope that helps.
$endgroup$
– Imran
Dec 14 '18 at 1:05
$begingroup$
A bit off-topic, but I find it funny how my answer looks like your lecturer's paper.
$endgroup$
– Lucas Henrique
Dec 14 '18 at 1:12
$begingroup$
A bit off-topic, but I find it funny how my answer looks like your lecturer's paper.
$endgroup$
– Lucas Henrique
Dec 14 '18 at 1:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $S = {qinmathbb R: mbox{$q^2<2$}}$. Since $S subset Bbb R$ and $forall xin S(x^2<2<4)$, $2$ is upper bound for this set. Since this is a subset of real numbers, it has some real least upper bound $L$. If $L^2 -2 > 0,$ choose a sufficiently large $m$ such that $L' = L - 1/m$ and $L'^2 = L^2 - 2L/m + 1/m^2 > 2$, which contradicts the minimality of $L$ (try it yourself using the Archimedian property of the reals). The same for $L^2 - 2 < 0$. Thus $L^2 = 2$ and $L = sqrt 2 in Bbb R$.
Edit: Given the original paper, my answer doesn't really help because I used the exact same process, so I'll explain the idea: completeness is pretty much what makes $Bbb R$ without "gaps". You'll want to construct, usually, sets that look like Dedekind cuts which will have an upper bound (possibly the least, and that's what you'll seek) as your desired number. Since the least upper bound is always real, prove it is what you want! In the given case, the choices for $delta$ are not arbitrary: sketch a $delta$ that would solve your problem (small numbers that will still keep the conditions) and, if your least upper bound isn't the number you desire, you'll have a contradiction! The process of construction is like solving a one-variable linear equation: you first assume there's a solution and then find what it would be, but, to formalize, you just show that there is a solution and then prove it's unique.
answered Dec 14 '18 at 0:20
Lucas HenriqueLucas Henrique
965414
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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
$begingroup$
Let $S = {qinmathbb R: mbox{$q^2<2$}}$. Since $S subset Bbb R$ and $forall xin S(x^2<2<4)$, $2$ is upper bound for this set. Since this is a subset of real numbers, it has some real least upper bound $L$. If $L^2 -2 > 0,$ choose a sufficiently large $m$ such that $L' = L - 1/m$ and $L'^2 = L^2 - 2L/m + 1/m^2 > 2$, which contradicts the minimality of $L$ (try it yourself using the Archimedian property of the reals). The same for $L^2 - 2 < 0$. Thus $L^2 = 2$ and $L = sqrt 2 in Bbb R$.
Edit: Given the original paper, my answer doesn't really help because I used the exact same process, so I'll explain the idea: completeness is pretty much what makes $Bbb R$ without "gaps". You'll want to construct, usually, sets that look like Dedekind cuts which will have an upper bound (possibly the least, and that's what you'll seek) as your desired number. Since the least upper bound is always real, prove it is what you want! In the given case, the choices for $delta$ are not arbitrary: sketch a $delta$ that would solve your problem (small numbers that will still keep the conditions) and, if your least upper bound isn't the number you desire, you'll have a contradiction! The process of construction is like solving a one-variable linear equation: you first assume there's a solution and then find what it would be, but, to formalize, you just show that there is a solution and then prove it's unique.
answered Dec 14 '18 at 0:20
Lucas HenriqueLucas Henrique
965414
$endgroup$
add a comment |
$begingroup$
Let $S = {qinmathbb R: mbox{$q^2<2$}}$. Since $S subset Bbb R$ and $forall xin S(x^2<2<4)$, $2$ is upper bound for this set. Since this is a subset of real numbers, it has some real least upper bound $L$. If $L^2 -2 > 0,$ choose a sufficiently large $m$ such that $L' = L - 1/m$ and $L'^2 = L^2 - 2L/m + 1/m^2 > 2$, which contradicts the minimality of $L$ (try it yourself using the Archimedian property of the reals). The same for $L^2 - 2 < 0$. Thus $L^2 = 2$ and $L = sqrt 2 in Bbb R$.
Edit: Given the original paper, my answer doesn't really help because I used the exact same process, so I'll explain the idea: completeness is pretty much what makes $Bbb R$ without "gaps". You'll want to construct, usually, sets that look like Dedekind cuts which will have an upper bound (possibly the least, and that's what you'll seek) as your desired number. Since the least upper bound is always real, prove it is what you want! In the given case, the choices for $delta$ are not arbitrary: sketch a $delta$ that would solve your problem (small numbers that will still keep the conditions) and, if your least upper bound isn't the number you desire, you'll have a contradiction! The process of construction is like solving a one-variable linear equation: you first assume there's a solution and then find what it would be, but, to formalize, you just show that there is a solution and then prove it's unique.
answered Dec 14 '18 at 0:20
Lucas HenriqueLucas Henrique
965414
$endgroup$
add a comment |
$begingroup$
Let $S = {qinmathbb R: mbox{$q^2<2$}}$. Since $S subset Bbb R$ and $forall xin S(x^2<2<4)$, $2$ is upper bound for this set. Since this is a subset of real numbers, it has some real least upper bound $L$. If $L^2 -2 > 0,$ choose a sufficiently large $m$ such that $L' = L - 1/m$ and $L'^2 = L^2 - 2L/m + 1/m^2 > 2$, which contradicts the minimality of $L$ (try it yourself using the Archimedian property of the reals). The same for $L^2 - 2 < 0$. Thus $L^2 = 2$ and $L = sqrt 2 in Bbb R$.
Edit: Given the original paper, my answer doesn't really help because I used the exact same process, so I'll explain the idea: completeness is pretty much what makes $Bbb R$ without "gaps". You'll want to construct, usually, sets that look like Dedekind cuts which will have an upper bound (possibly the least, and that's what you'll seek) as your desired number. Since the least upper bound is always real, prove it is what you want! In the given case, the choices for $delta$ are not arbitrary: sketch a $delta$ that would solve your problem (small numbers that will still keep the conditions) and, if your least upper bound isn't the number you desire, you'll have a contradiction! The process of construction is like solving a one-variable linear equation: you first assume there's a solution and then find what it would be, but, to formalize, you just show that there is a solution and then prove it's unique.
answered Dec 14 '18 at 0:20
Lucas HenriqueLucas Henrique
965414
$endgroup$
Let $S = {qinmathbb R: mbox{$q^2<2$}}$. Since $S subset Bbb R$ and $forall xin S(x^2<2<4)$, $2$ is upper bound for this set. Since this is a subset of real numbers, it has some real least upper bound $L$. If $L^2 -2 > 0,$ choose a sufficiently large $m$ such that $L' = L - 1/m$ and $L'^2 = L^2 - 2L/m + 1/m^2 > 2$, which contradicts the minimality of $L$ (try it yourself using the Archimedian property of the reals). The same for $L^2 - 2 < 0$. Thus $L^2 = 2$ and $L = sqrt 2 in Bbb R$.
Edit: Given the original paper, my answer doesn't really help because I used the exact same process, so I'll explain the idea: completeness is pretty much what makes $Bbb R$ without "gaps". You'll want to construct, usually, sets that look like Dedekind cuts which will have an upper bound (possibly the least, and that's what you'll seek) as your desired number. Since the least upper bound is always real, prove it is what you want! In the given case, the choices for $delta$ are not arbitrary: sketch a $delta$ that would solve your problem (small numbers that will still keep the conditions) and, if your least upper bound isn't the number you desire, you'll have a contradiction! The process of construction is like solving a one-variable linear equation: you first assume there's a solution and then find what it would be, but, to formalize, you just show that there is a solution and then prove it's unique.
answered Dec 14 '18 at 0:20
Lucas HenriqueLucas Henrique
965414
edited Dec 14 '18 at 1:20
answered Dec 14 '18 at 0:20
Lucas HenriqueLucas Henrique
965414
answered Dec 14 '18 at 0:20
Lucas HenriqueLucas Henrique
965414
answered Dec 14 '18 at 0:20
Lucas HenriqueLucas Henrique
965414
965414
add a comment |
add a comment |
3
$begingroup$
If you edit the question to post your lecturer's proof and flag the place that confuses you perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 14 '18 at 0:07
$begingroup$
I know, sorry. I was having difficulty writing it up so that's why I didn't. I've posted a image of it. I'm confused from that part when delta is used and onwards. I'm not sure where the values for 'min' come from and how the quadratic terms change. Hope that helps.
$endgroup$
– Imran
Dec 14 '18 at 1:05
$begingroup$
A bit off-topic, but I find it funny how my answer looks like your lecturer's paper.
$endgroup$
– Lucas Henrique
Dec 14 '18 at 1:12