True or False?( Calculus)
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
edited Nov 24 at 5:29
Tianlalu
3,12321038
asked Nov 24 at 5:22
Mathsaddict
1658
add a comment |
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
edited Nov 24 at 5:29
Tianlalu
3,12321038
asked Nov 24 at 5:22
Mathsaddict
1658
Yes, yes, yes, yes. (Sorry I can't only type one 'yes')
– Seewoo Lee
Nov 24 at 5:29
@SeewooLee tryYes ${}{}{}{}{}{}$
– 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
add a comment |
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
edited Nov 24 at 5:29
Tianlalu
3,12321038
asked Nov 24 at 5:22
Mathsaddict
1658
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
edited Nov 24 at 5:29
Tianlalu
3,12321038
asked Nov 24 at 5:22
Mathsaddict
1658
edited Nov 24 at 5:29
Tianlalu
3,12321038
asked Nov 24 at 5:22
Mathsaddict
1658
edited Nov 24 at 5:29
Tianlalu
3,12321038
edited Nov 24 at 5:29
Tianlalu
3,12321038
edited Nov 24 at 5:29
Tianlalu
3,12321038
3,12321038
asked Nov 24 at 5:22
Mathsaddict
1658
asked Nov 24 at 5:22
Mathsaddict
1658
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 tryYes ${}{}{}{}{}{}$
– 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
add a comment |
Yes, yes, yes, yes. (Sorry I can't only type one 'yes')
– Seewoo Lee
Nov 24 at 5:29
@SeewooLee tryYes ${}{}{}{}{}{}$
– 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
add a comment |
2 Answers
2
active
oldest
votes
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($frac{1}{sqrt{2}}$) .
answered Nov 24 at 5:30
Joel Pereira
65119
add a comment |
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
$$
F(1/sqrt{2}) =
F(lim_{ntoinfty} r_n) =
lim_{ntoinfty} F(r_n) =
lim_{ntoinfty} 1 = 1 ,.
$$
answered Nov 24 at 5:42
mlerma54
1,087138
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
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($frac{1}{sqrt{2}}$) .
answered Nov 24 at 5:30
Joel Pereira
65119
add a comment |
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($frac{1}{sqrt{2}}$) .
answered Nov 24 at 5:30
Joel Pereira
65119
add a comment |
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($frac{1}{sqrt{2}}$) .
answered Nov 24 at 5:30
Joel Pereira
65119
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($frac{1}{sqrt{2}}$) .
answered Nov 24 at 5:30
Joel Pereira
65119
answered Nov 24 at 5:30
Joel Pereira
65119
answered Nov 24 at 5:30
Joel Pereira
65119
answered Nov 24 at 5:30
Joel Pereira
65119
65119
add a comment |
add a comment |
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
$$
F(1/sqrt{2}) =
F(lim_{ntoinfty} r_n) =
lim_{ntoinfty} F(r_n) =
lim_{ntoinfty} 1 = 1 ,.
$$
answered Nov 24 at 5:42
mlerma54
1,087138
add a comment |
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
$$
F(1/sqrt{2}) =
F(lim_{ntoinfty} r_n) =
lim_{ntoinfty} F(r_n) =
lim_{ntoinfty} 1 = 1 ,.
$$
answered Nov 24 at 5:42
mlerma54
1,087138
add a comment |
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
$$
F(1/sqrt{2}) =
F(lim_{ntoinfty} r_n) =
lim_{ntoinfty} F(r_n) =
lim_{ntoinfty} 1 = 1 ,.
$$
answered Nov 24 at 5:42
mlerma54
1,087138
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
$$
F(1/sqrt{2}) =
F(lim_{ntoinfty} r_n) =
lim_{ntoinfty} F(r_n) =
lim_{ntoinfty} 1 = 1 ,.
$$
answered Nov 24 at 5:42
mlerma54
1,087138
answered Nov 24 at 5:42
mlerma54
1,087138
answered Nov 24 at 5:42
mlerma54
1,087138
answered Nov 24 at 5:42
mlerma54
1,087138
1,087138
add a comment |
add a comment |
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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