True or False?( Calculus)












0














If $F(x)$ is a continuous function in $[0,1]$ and $F(x) = 1$ for all rational numbers then $F(1/sqrt2) = 1$. (True/ False)



I think the statement is true because since $F$ is continuous and $1/sqrt2$ is in neighbourhood of some rational number so $F(1/sqrt2)$ should be one so that limit can exist. Is this reason valid to prove the statement true.






calculus proof-verification







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  • Yes, yes, yes, yes. (Sorry I can't only type one 'yes')
    – Seewoo Lee
    Nov 24 at 5:29










  • @SeewooLee try Yes ${}{}{}{}{}{}$
    – Chase Ryan Taylor
    Nov 24 at 5:30






  • 1




    Yes, you thinking is correct. You can improve your wording. "so F(1/2–√) should be one so that limit can exist" is badly worded and semantically ... weird. But I know what you are trying to express and it is absolutely correct.
    – fleablood
    Nov 24 at 5:30










  • Joel Pereira's answer is a very nice mathematical way to express your thinking.
    – fleablood
    Nov 24 at 5:32










  • "$1/sqrt{2}$ is in neighborhood of some rational number" is not what you want to be saying. The idea you want to convey is "every neighborhood of $1/sqrt{2}$ contains a rational number", which is what @Joel Pereira does more precisely in his answer.
    – JonathanZ
    Nov 24 at 5:40
















0














If $F(x)$ is a continuous function in $[0,1]$ and $F(x) = 1$ for all rational numbers then $F(1/sqrt2) = 1$. (True/ False)



I think the statement is true because since $F$ is continuous and $1/sqrt2$ is in neighbourhood of some rational number so $F(1/sqrt2)$ should be one so that limit can exist. Is this reason valid to prove the statement true.






calculus proof-verification







share|cite|improve this question
























  • Yes, yes, yes, yes. (Sorry I can't only type one 'yes')
    – Seewoo Lee
    Nov 24 at 5:29










  • @SeewooLee try Yes ${}{}{}{}{}{}$
    – Chase Ryan Taylor
    Nov 24 at 5:30






  • 1




    Yes, you thinking is correct. You can improve your wording. "so F(1/2–√) should be one so that limit can exist" is badly worded and semantically ... weird. But I know what you are trying to express and it is absolutely correct.
    – fleablood
    Nov 24 at 5:30










  • Joel Pereira's answer is a very nice mathematical way to express your thinking.
    – fleablood
    Nov 24 at 5:32










  • "$1/sqrt{2}$ is in neighborhood of some rational number" is not what you want to be saying. The idea you want to convey is "every neighborhood of $1/sqrt{2}$ contains a rational number", which is what @Joel Pereira does more precisely in his answer.
    – JonathanZ
    Nov 24 at 5:40














0












0








0







If $F(x)$ is a continuous function in $[0,1]$ and $F(x) = 1$ for all rational numbers then $F(1/sqrt2) = 1$. (True/ False)



I think the statement is true because since $F$ is continuous and $1/sqrt2$ is in neighbourhood of some rational number so $F(1/sqrt2)$ should be one so that limit can exist. Is this reason valid to prove the statement true.






calculus proof-verification







share|cite|improve this question















If $F(x)$ is a continuous function in $[0,1]$ and $F(x) = 1$ for all rational numbers then $F(1/sqrt2) = 1$. (True/ False)



I think the statement is true because since $F$ is continuous and $1/sqrt2$ is in neighbourhood of some rational number so $F(1/sqrt2)$ should be one so that limit can exist. Is this reason valid to prove the statement true.






calculus proof-verification




calculus proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 5:29









Tianlalu

3,12321038




3,12321038










asked Nov 24 at 5:22









Mathsaddict

1658




1658












  • Yes, yes, yes, yes. (Sorry I can't only type one 'yes')
    – Seewoo Lee
    Nov 24 at 5:29










  • @SeewooLee try Yes ${}{}{}{}{}{}$
    – Chase Ryan Taylor
    Nov 24 at 5:30






  • 1




    Yes, you thinking is correct. You can improve your wording. "so F(1/2–√) should be one so that limit can exist" is badly worded and semantically ... weird. But I know what you are trying to express and it is absolutely correct.
    – fleablood
    Nov 24 at 5:30










  • Joel Pereira's answer is a very nice mathematical way to express your thinking.
    – fleablood
    Nov 24 at 5:32










  • "$1/sqrt{2}$ is in neighborhood of some rational number" is not what you want to be saying. The idea you want to convey is "every neighborhood of $1/sqrt{2}$ contains a rational number", which is what @Joel Pereira does more precisely in his answer.
    – JonathanZ
    Nov 24 at 5:40


















  • Yes, yes, yes, yes. (Sorry I can't only type one 'yes')
    – Seewoo Lee
    Nov 24 at 5:29










  • @SeewooLee try Yes ${}{}{}{}{}{}$
    – Chase Ryan Taylor
    Nov 24 at 5:30






  • 1




    Yes, you thinking is correct. You can improve your wording. "so F(1/2–√) should be one so that limit can exist" is badly worded and semantically ... weird. But I know what you are trying to express and it is absolutely correct.
    – fleablood
    Nov 24 at 5:30










  • Joel Pereira's answer is a very nice mathematical way to express your thinking.
    – fleablood
    Nov 24 at 5:32










  • "$1/sqrt{2}$ is in neighborhood of some rational number" is not what you want to be saying. The idea you want to convey is "every neighborhood of $1/sqrt{2}$ contains a rational number", which is what @Joel Pereira does more precisely in his answer.
    – JonathanZ
    Nov 24 at 5:40
















Yes, yes, yes, yes. (Sorry I can't only type one 'yes')
– Seewoo Lee
Nov 24 at 5:29




Yes, yes, yes, yes. (Sorry I can't only type one 'yes')
– Seewoo Lee
Nov 24 at 5:29












@SeewooLee try Yes ${}{}{}{}{}{}$
– Chase Ryan Taylor
Nov 24 at 5:30




@SeewooLee try Yes ${}{}{}{}{}{}$
– Chase Ryan Taylor
Nov 24 at 5:30




1




1




Yes, you thinking is correct. You can improve your wording. "so F(1/2–√) should be one so that limit can exist" is badly worded and semantically ... weird. But I know what you are trying to express and it is absolutely correct.
– fleablood
Nov 24 at 5:30




Yes, you thinking is correct. You can improve your wording. "so F(1/2–√) should be one so that limit can exist" is badly worded and semantically ... weird. But I know what you are trying to express and it is absolutely correct.
– fleablood
Nov 24 at 5:30












Joel Pereira's answer is a very nice mathematical way to express your thinking.
– fleablood
Nov 24 at 5:32




Joel Pereira's answer is a very nice mathematical way to express your thinking.
– fleablood
Nov 24 at 5:32












"$1/sqrt{2}$ is in neighborhood of some rational number" is not what you want to be saying. The idea you want to convey is "every neighborhood of $1/sqrt{2}$ contains a rational number", which is what @Joel Pereira does more precisely in his answer.
– JonathanZ
Nov 24 at 5:40




"$1/sqrt{2}$ is in neighborhood of some rational number" is not what you want to be saying. The idea you want to convey is "every neighborhood of $1/sqrt{2}$ contains a rational number", which is what @Joel Pereira does more precisely in his answer.
– JonathanZ
Nov 24 at 5:40










2 Answers
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For every $n$, there exists a rational number $q_n$ such that $mid frac{1}{sqrt{2}}-q_nmid < frac{1}{n}$. So $displaystylelim_{nrightarrow infty}q_n$ = $frac{1}{sqrt{2}}$. Since F is continuous, 1 = $displaystylelim_{nrightarrow infty}$F($q_n$) = F($frac{1}{sqrt{2}}$) .






share|cite|improve this answer





























    1














    The statement is true, although I would express it: there is a sequence ${r_n}_{ninmathbb{N}}$ of rational numbers in the interval $[0,1]$ such that $r_n to 1/sqrt{2}$, hence by continuity
    $$
    F(1/sqrt{2}) =
    F(lim_{ntoinfty} r_n) =
    lim_{ntoinfty} F(r_n) =
    lim_{ntoinfty} 1 = 1 ,.
    $$






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      For every $n$, there exists a rational number $q_n$ such that $mid frac{1}{sqrt{2}}-q_nmid < frac{1}{n}$. So $displaystylelim_{nrightarrow infty}q_n$ = $frac{1}{sqrt{2}}$. Since F is continuous, 1 = $displaystylelim_{nrightarrow infty}$F($q_n$) = F($frac{1}{sqrt{2}}$) .






      share|cite|improve this answer


























        2














        For every $n$, there exists a rational number $q_n$ such that $mid frac{1}{sqrt{2}}-q_nmid < frac{1}{n}$. So $displaystylelim_{nrightarrow infty}q_n$ = $frac{1}{sqrt{2}}$. Since F is continuous, 1 = $displaystylelim_{nrightarrow infty}$F($q_n$) = F($frac{1}{sqrt{2}}$) .






        share|cite|improve this answer
























          2












          2








          2






          For every $n$, there exists a rational number $q_n$ such that $mid frac{1}{sqrt{2}}-q_nmid < frac{1}{n}$. So $displaystylelim_{nrightarrow infty}q_n$ = $frac{1}{sqrt{2}}$. Since F is continuous, 1 = $displaystylelim_{nrightarrow infty}$F($q_n$) = F($frac{1}{sqrt{2}}$) .






          share|cite|improve this answer












          For every $n$, there exists a rational number $q_n$ such that $mid frac{1}{sqrt{2}}-q_nmid < frac{1}{n}$. So $displaystylelim_{nrightarrow infty}q_n$ = $frac{1}{sqrt{2}}$. Since F is continuous, 1 = $displaystylelim_{nrightarrow infty}$F($q_n$) = F($frac{1}{sqrt{2}}$) .







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 5:30









          Joel Pereira

          65119




          65119























              1














              The statement is true, although I would express it: there is a sequence ${r_n}_{ninmathbb{N}}$ of rational numbers in the interval $[0,1]$ such that $r_n to 1/sqrt{2}$, hence by continuity
              $$
              F(1/sqrt{2}) =
              F(lim_{ntoinfty} r_n) =
              lim_{ntoinfty} F(r_n) =
              lim_{ntoinfty} 1 = 1 ,.
              $$






              share|cite|improve this answer


























                1














                The statement is true, although I would express it: there is a sequence ${r_n}_{ninmathbb{N}}$ of rational numbers in the interval $[0,1]$ such that $r_n to 1/sqrt{2}$, hence by continuity
                $$
                F(1/sqrt{2}) =
                F(lim_{ntoinfty} r_n) =
                lim_{ntoinfty} F(r_n) =
                lim_{ntoinfty} 1 = 1 ,.
                $$






                share|cite|improve this answer
























                  1












                  1








                  1






                  The statement is true, although I would express it: there is a sequence ${r_n}_{ninmathbb{N}}$ of rational numbers in the interval $[0,1]$ such that $r_n to 1/sqrt{2}$, hence by continuity
                  $$
                  F(1/sqrt{2}) =
                  F(lim_{ntoinfty} r_n) =
                  lim_{ntoinfty} F(r_n) =
                  lim_{ntoinfty} 1 = 1 ,.
                  $$






                  share|cite|improve this answer












                  The statement is true, although I would express it: there is a sequence ${r_n}_{ninmathbb{N}}$ of rational numbers in the interval $[0,1]$ such that $r_n to 1/sqrt{2}$, hence by continuity
                  $$
                  F(1/sqrt{2}) =
                  F(lim_{ntoinfty} r_n) =
                  lim_{ntoinfty} F(r_n) =
                  lim_{ntoinfty} 1 = 1 ,.
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 at 5:42









                  mlerma54

                  1,087138




                  1,087138






























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