For every $epsilongt 0$, $|a-b|<epsilon $ ,then b=a . The Next CEO of Stack OverflowI need to prove the continuity of $f(x)=log x$ using a $epsilon-delta$ proofIf a is an element of R, such that $0leq a<epsilon$ for every $epsilon>0$,then a = 0If $P(A_n) ge epsilon>0$ for large $n$, then $P(A_n i.o.) ge epsilon$Proving: if $|a|<epsilon forall epsilon>0$ then $a=0$ using a direct proofHelp with composing an $epsilon-N$ proof of mineEpsilon-Delta Proof StructureProve every convergent sequence is Cauchy.Show that for every $epsilon > 0$ there exists an $M>0$ such that $|f(x)−f(y)|<epsilon$Quick epsilon-delta questionReal Analysis: Proving Continuity ($epsilon$ $delta$ argument)
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For every $epsilongt 0$, $|a-b|
The Next CEO of Stack OverflowI need to prove the continuity of $f(x)=log x$ using a $epsilon-delta$ proofIf a is an element of R, such that $0leq a<epsilon$ for every $epsilon>0$,then a = 0If $P(A_n) ge epsilon>0$ for large $n$, then $P(A_n i.o.) ge epsilon$Proving: if $|a|<epsilon forall epsilon>0$ then $a=0$ using a direct proofHelp with composing an $epsilon-N$ proof of mineEpsilon-Delta Proof StructureProve every convergent sequence is Cauchy.Show that for every $epsilon > 0$ there exists an $M>0$ such that $|f(x)−f(y)|<epsilon$Quick epsilon-delta questionReal Analysis: Proving Continuity ($epsilon$ $delta$ argument)
$begingroup$
I have done a proof by myself but not sure about it
proof: $|b-a|<epsilon $
=$a-epsilon $
real-analysis calculus measure-theory proof-writing
edited Mar 18 at 15:06
Somos
14.7k11337
asked Mar 18 at 12:43
nancynancy
155
$endgroup$
add a comment |
$begingroup$
I have done a proof by myself but not sure about it
proof: $|b-a|<epsilon $
=$a-epsilon $
real-analysis calculus measure-theory proof-writing
edited Mar 18 at 15:06
Somos
14.7k11337
asked Mar 18 at 12:43
nancynancy
155
$endgroup$
1
$begingroup$
Tryepsilonorvarepsilonto get $epsilon$ or $varepsilon$ respectively.
$endgroup$
– Randall
Mar 18 at 12:44
1
$begingroup$
Use dollar signs to wrap up math, like$epsilongt 0$for $epsilongt 0$. Read this for an introduction on how to format your content properly on this website. You can also view how others type math on this site by right clicking on the math -> show math as -> TeX commands.
$endgroup$
– learner
Mar 18 at 12:52
$begingroup$
Here's a hint to get you started: suppose $|a-b|=kgt 0$, derive a contradiction by choosing suitable $epsilon$; hence conclude that $|a-b|=0iff a-b=0iff a=b$
$endgroup$
– learner
Mar 18 at 13:01
$begingroup$
Thanks for telling how to write epsilon .it worked
$endgroup$
– nancy
Mar 18 at 13:09
$begingroup$
learner will you please elaborate it more that how to choose suitable $ epsilon $
$endgroup$
– nancy
Mar 18 at 13:10
add a comment |
$begingroup$
I have done a proof by myself but not sure about it
proof: $|b-a|<epsilon $
=$a-epsilon $
real-analysis calculus measure-theory proof-writing
edited Mar 18 at 15:06
Somos
14.7k11337
asked Mar 18 at 12:43
nancynancy
155
$endgroup$
I have done a proof by myself but not sure about it
proof: $|b-a|<epsilon $
=$a-epsilon $
real-analysis calculus measure-theory proof-writing
real-analysis calculus measure-theory proof-writing
edited Mar 18 at 15:06
Somos
14.7k11337
asked Mar 18 at 12:43
nancynancy
155
edited Mar 18 at 15:06
Somos
14.7k11337
asked Mar 18 at 12:43
nancynancy
155
edited Mar 18 at 15:06
Somos
14.7k11337
edited Mar 18 at 15:06
Somos
14.7k11337
edited Mar 18 at 15:06
Somos
14.7k11337
14.7k11337
asked Mar 18 at 12:43
nancynancy
155
asked Mar 18 at 12:43
nancynancy
155
asked Mar 18 at 12:43
nancynancy
155
155
1
$begingroup$
Tryepsilonorvarepsilonto get $epsilon$ or $varepsilon$ respectively.
$endgroup$
– Randall
Mar 18 at 12:44
1
$begingroup$
Use dollar signs to wrap up math, like$epsilongt 0$for $epsilongt 0$. Read this for an introduction on how to format your content properly on this website. You can also view how others type math on this site by right clicking on the math -> show math as -> TeX commands.
$endgroup$
– learner
Mar 18 at 12:52
$begingroup$
Here's a hint to get you started: suppose $|a-b|=kgt 0$, derive a contradiction by choosing suitable $epsilon$; hence conclude that $|a-b|=0iff a-b=0iff a=b$
$endgroup$
– learner
Mar 18 at 13:01
$begingroup$
Thanks for telling how to write epsilon .it worked
$endgroup$
– nancy
Mar 18 at 13:09
$begingroup$
learner will you please elaborate it more that how to choose suitable $ epsilon $
$endgroup$
– nancy
Mar 18 at 13:10
add a comment |
1
$begingroup$
Tryepsilonorvarepsilonto get $epsilon$ or $varepsilon$ respectively.
$endgroup$
– Randall
Mar 18 at 12:44
1
$begingroup$
Use dollar signs to wrap up math, like$epsilongt 0$for $epsilongt 0$. Read this for an introduction on how to format your content properly on this website. You can also view how others type math on this site by right clicking on the math -> show math as -> TeX commands.
$endgroup$
– learner
Mar 18 at 12:52
$begingroup$
Here's a hint to get you started: suppose $|a-b|=kgt 0$, derive a contradiction by choosing suitable $epsilon$; hence conclude that $|a-b|=0iff a-b=0iff a=b$
$endgroup$
– learner
Mar 18 at 13:01
$begingroup$
Thanks for telling how to write epsilon .it worked
$endgroup$
– nancy
Mar 18 at 13:09
$begingroup$
learner will you please elaborate it more that how to choose suitable $ epsilon $
$endgroup$
– nancy
Mar 18 at 13:10
1
1
$begingroup$
Try
epsilon or varepsilon to get $epsilon$ or $varepsilon$ respectively.$endgroup$
– Randall
Mar 18 at 12:44
$begingroup$
Try
epsilon or varepsilon to get $epsilon$ or $varepsilon$ respectively.$endgroup$
– Randall
Mar 18 at 12:44
1
1
$begingroup$
Use dollar signs to wrap up math, like
$epsilongt 0$ for $epsilongt 0$. Read this for an introduction on how to format your content properly on this website. You can also view how others type math on this site by right clicking on the math -> show math as -> TeX commands.$endgroup$
– learner
Mar 18 at 12:52
$begingroup$
Use dollar signs to wrap up math, like
$epsilongt 0$ for $epsilongt 0$. Read this for an introduction on how to format your content properly on this website. You can also view how others type math on this site by right clicking on the math -> show math as -> TeX commands.$endgroup$
– learner
Mar 18 at 12:52
$begingroup$
Here's a hint to get you started: suppose $|a-b|=kgt 0$, derive a contradiction by choosing suitable $epsilon$; hence conclude that $|a-b|=0iff a-b=0iff a=b$
$endgroup$
– learner
Mar 18 at 13:01
$begingroup$
Here's a hint to get you started: suppose $|a-b|=kgt 0$, derive a contradiction by choosing suitable $epsilon$; hence conclude that $|a-b|=0iff a-b=0iff a=b$
$endgroup$
– learner
Mar 18 at 13:01
$begingroup$
Thanks for telling how to write epsilon .it worked
$endgroup$
– nancy
Mar 18 at 13:09
$begingroup$
Thanks for telling how to write epsilon .it worked
$endgroup$
– nancy
Mar 18 at 13:09
$begingroup$
learner will you please elaborate it more that how to choose suitable $ epsilon $
$endgroup$
– nancy
Mar 18 at 13:10
$begingroup$
learner will you please elaborate it more that how to choose suitable $ epsilon $
$endgroup$
– nancy
Mar 18 at 13:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have $forall epsilon > 0,quad |a-b| < epsilon$
If we suppose that $a neq b$ then $|a-b| neq 0$, we choose $epsilon = dfrac2 > 0$
Then $|a-b| < dfrac2 implies 1 < dfrac12$, contradiction!
Conclusion : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$
answered Mar 18 at 13:37
LAGRIDALAGRIDA
295111
$endgroup$
$begingroup$
Thanku for such quick answer @LAGRIDA
$endgroup$
– nancy
Mar 18 at 13:58
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
$begingroup$
We have $forall epsilon > 0,quad |a-b| < epsilon$
If we suppose that $a neq b$ then $|a-b| neq 0$, we choose $epsilon = dfrac2 > 0$
Then $|a-b| < dfrac2 implies 1 < dfrac12$, contradiction!
Conclusion : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$
answered Mar 18 at 13:37
LAGRIDALAGRIDA
295111
$endgroup$
$begingroup$
Thanku for such quick answer @LAGRIDA
$endgroup$
– nancy
Mar 18 at 13:58
add a comment |
$begingroup$
We have $forall epsilon > 0,quad |a-b| < epsilon$
If we suppose that $a neq b$ then $|a-b| neq 0$, we choose $epsilon = dfrac2 > 0$
Then $|a-b| < dfrac2 implies 1 < dfrac12$, contradiction!
Conclusion : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$
answered Mar 18 at 13:37
LAGRIDALAGRIDA
295111
$endgroup$
$begingroup$
Thanku for such quick answer @LAGRIDA
$endgroup$
– nancy
Mar 18 at 13:58
add a comment |
$begingroup$
We have $forall epsilon > 0,quad |a-b| < epsilon$
If we suppose that $a neq b$ then $|a-b| neq 0$, we choose $epsilon = dfrac2 > 0$
Then $|a-b| < dfrac2 implies 1 < dfrac12$, contradiction!
Conclusion : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$
answered Mar 18 at 13:37
LAGRIDALAGRIDA
295111
$endgroup$
We have $forall epsilon > 0,quad |a-b| < epsilon$
If we suppose that $a neq b$ then $|a-b| neq 0$, we choose $epsilon = dfrac2 > 0$
Then $|a-b| < dfrac2 implies 1 < dfrac12$, contradiction!
Conclusion : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$
answered Mar 18 at 13:37
LAGRIDALAGRIDA
295111
answered Mar 18 at 13:37
LAGRIDALAGRIDA
295111
answered Mar 18 at 13:37
LAGRIDALAGRIDA
295111
answered Mar 18 at 13:37
LAGRIDALAGRIDA
295111
295111
$begingroup$
Thanku for such quick answer @LAGRIDA
$endgroup$
– nancy
Mar 18 at 13:58
add a comment |
$begingroup$
Thanku for such quick answer @LAGRIDA
$endgroup$
– nancy
Mar 18 at 13:58
$begingroup$
Thanku for such quick answer @LAGRIDA
$endgroup$
– nancy
Mar 18 at 13:58
$begingroup$
Thanku for such quick answer @LAGRIDA
$endgroup$
– nancy
Mar 18 at 13:58
add a comment |
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BMVJxqXi,7jfsvYy,GXOk2L0kqHmI,K10yPu eLFbyVwg77YyF6TbPR WGiBhw4cD
1
$begingroup$
Try
epsilonorvarepsilonto get $epsilon$ or $varepsilon$ respectively.$endgroup$
– Randall
Mar 18 at 12:44
1
$begingroup$
Use dollar signs to wrap up math, like
$epsilongt 0$for $epsilongt 0$. Read this for an introduction on how to format your content properly on this website. You can also view how others type math on this site by right clicking on the math -> show math as -> TeX commands.$endgroup$
– learner
Mar 18 at 12:52
$begingroup$
Here's a hint to get you started: suppose $|a-b|=kgt 0$, derive a contradiction by choosing suitable $epsilon$; hence conclude that $|a-b|=0iff a-b=0iff a=b$
$endgroup$
– learner
Mar 18 at 13:01
$begingroup$
Thanks for telling how to write epsilon .it worked
$endgroup$
– nancy
Mar 18 at 13:09
$begingroup$
learner will you please elaborate it more that how to choose suitable $ epsilon $
$endgroup$
– nancy
Mar 18 at 13:10