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)










0












$begingroup$


I have done a proof by myself but not sure about it



proof: $|b-a|<epsilon $
=$a-epsilon $










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Try epsilon or varepsilon to 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
















0












$begingroup$


I have done a proof by myself but not sure about it



proof: $|b-a|<epsilon $
=$a-epsilon $










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Try epsilon or varepsilon to 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














0












0








0





$begingroup$


I have done a proof by myself but not sure about it



proof: $|b-a|<epsilon $
=$a-epsilon $










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 18 at 15:06









Somos

14.7k11337




14.7k11337










asked Mar 18 at 12:43









nancynancy

155




155







  • 1




    $begingroup$
    Try epsilon or varepsilon to 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




    $begingroup$
    Try epsilon or varepsilon to 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











1 Answer
1






active

oldest

votes


















5












$begingroup$

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 : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thanku for such quick answer @LAGRIDA
    $endgroup$
    – nancy
    Mar 18 at 13:58











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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 : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thanku for such quick answer @LAGRIDA
    $endgroup$
    – nancy
    Mar 18 at 13:58















5












$begingroup$

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 : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thanku for such quick answer @LAGRIDA
    $endgroup$
    – nancy
    Mar 18 at 13:58













5












5








5





$begingroup$

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 : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$






share|cite|improve this answer









$endgroup$



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 : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 18 at 13:37









LAGRIDALAGRIDA

295111




295111











  • $begingroup$
    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




$begingroup$
Thanku for such quick answer @LAGRIDA
$endgroup$
– nancy
Mar 18 at 13:58

















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