Mixing
Epsilon proof 10
0
$begingroup$
Assume for every real number $x$, there is an integer $N$ such that $N>x^3$. Prove that for every positive real number $epsilon$, there exists a positive integer $N$ such that for all $ngeq N$, $frac{1}{n}<epsilon$
Ok so I know I am suppose to show an attempt or at least explain my thought process, but honestly I don't even know how to start this proof. Any suggestions?
analysis proof-writing
edited Jan 30 at 4:37
YuiTo Cheng
2,1863937
asked Jan 30 at 2:54
ForextraderForextrader
988
$endgroup$
|
show 1 more comment
0
$begingroup$
Assume for every real number $x$, there is an integer $N$ such that $N>x^3$. Prove that for every positive real number $epsilon$, there exists a positive integer $N$ such that for all $ngeq N$, $frac{1}{n}<epsilon$
Ok so I know I am suppose to show an attempt or at least explain my thought process, but honestly I don't even know how to start this proof. Any suggestions?
analysis proof-writing
edited Jan 30 at 4:37
YuiTo Cheng
2,1863937
asked Jan 30 at 2:54
ForextraderForextrader
988
$endgroup$
$begingroup$
First step. Figure out what they are asking. Don't worry about proving, but what, in english, are the asking.
$endgroup$
– fleablood
Jan 30 at 2:58
$begingroup$
And thats probably the part I am having difficulty with is understanding what in english they are asking.
$endgroup$
– Forextrader
Jan 30 at 3:02
$begingroup$
The problem is asking you to show that for every positive real (ie number greater than 0) there is always a smaller 1/n. Even as $epsilon$ gets close to zero it should make intuitive sense that you can find a large enough n st 1/n is smaller.
$endgroup$
– T. Fo
Jan 30 at 3:28
$begingroup$
Consider if $epsilon$ is a positive real number, then $frac 1{epsilon}$ is also a positive real number. Now you are assuming that there is an $N > (frac 1{epsilon})^3> 0$. If $n > N > (frac 1{epsilon})^3 > 0$ then $frac 1n < frac 1N < epsilon^3$. ... Okay, I have to admit the statement of "assume if $x$ is real there is an $N > x^3$" is strange but... if $epsilon le 1$ then $epsilon^3 le epsilon$ and if $epsilon > 1$ then $frac 1n < 1 < epsilon$.
$endgroup$
– fleablood
Jan 30 at 6:25
$begingroup$
Are you sure you are being asked to prove that if $x$ is real there is an $N > x^3$ and not that there is an $N > x$? Both are true, but $N > x$ is more straightforward and direct.
$endgroup$
– fleablood
Jan 30 at 6:27
|
show 1 more comment
0
0
0
$begingroup$
Assume for every real number $x$, there is an integer $N$ such that $N>x^3$. Prove that for every positive real number $epsilon$, there exists a positive integer $N$ such that for all $ngeq N$, $frac{1}{n}<epsilon$
Ok so I know I am suppose to show an attempt or at least explain my thought process, but honestly I don't even know how to start this proof. Any suggestions?
analysis proof-writing
edited Jan 30 at 4:37
YuiTo Cheng
2,1863937
asked Jan 30 at 2:54
ForextraderForextrader
988
$endgroup$
Assume for every real number $x$, there is an integer $N$ such that $N>x^3$. Prove that for every positive real number $epsilon$, there exists a positive integer $N$ such that for all $ngeq N$, $frac{1}{n}<epsilon$
Ok so I know I am suppose to show an attempt or at least explain my thought process, but honestly I don't even know how to start this proof. Any suggestions?
analysis proof-writing
analysis proof-writing
edited Jan 30 at 4:37
YuiTo Cheng
2,1863937
asked Jan 30 at 2:54
ForextraderForextrader
988
edited Jan 30 at 4:37
YuiTo Cheng
2,1863937
asked Jan 30 at 2:54
ForextraderForextrader
988
edited Jan 30 at 4:37
YuiTo Cheng
2,1863937
edited Jan 30 at 4:37
YuiTo Cheng
2,1863937
edited Jan 30 at 4:37
YuiTo Cheng
2,1863937
2,1863937
asked Jan 30 at 2:54
ForextraderForextrader
988
asked Jan 30 at 2:54
ForextraderForextrader
988
asked Jan 30 at 2:54
ForextraderForextrader
988
988
$begingroup$
First step. Figure out what they are asking. Don't worry about proving, but what, in english, are the asking.
$endgroup$
– fleablood
Jan 30 at 2:58
$begingroup$
And thats probably the part I am having difficulty with is understanding what in english they are asking.
$endgroup$
– Forextrader
Jan 30 at 3:02
$begingroup$
The problem is asking you to show that for every positive real (ie number greater than 0) there is always a smaller 1/n. Even as $epsilon$ gets close to zero it should make intuitive sense that you can find a large enough n st 1/n is smaller.
$endgroup$
– T. Fo
Jan 30 at 3:28
$begingroup$
Consider if $epsilon$ is a positive real number, then $frac 1{epsilon}$ is also a positive real number. Now you are assuming that there is an $N > (frac 1{epsilon})^3> 0$. If $n > N > (frac 1{epsilon})^3 > 0$ then $frac 1n < frac 1N < epsilon^3$. ... Okay, I have to admit the statement of "assume if $x$ is real there is an $N > x^3$" is strange but... if $epsilon le 1$ then $epsilon^3 le epsilon$ and if $epsilon > 1$ then $frac 1n < 1 < epsilon$.
$endgroup$
– fleablood
Jan 30 at 6:25
$begingroup$
Are you sure you are being asked to prove that if $x$ is real there is an $N > x^3$ and not that there is an $N > x$? Both are true, but $N > x$ is more straightforward and direct.
$endgroup$
– fleablood
Jan 30 at 6:27
|
show 1 more comment
$begingroup$
First step. Figure out what they are asking. Don't worry about proving, but what, in english, are the asking.
$endgroup$
– fleablood
Jan 30 at 2:58
$begingroup$
And thats probably the part I am having difficulty with is understanding what in english they are asking.
$endgroup$
– Forextrader
Jan 30 at 3:02
$begingroup$
The problem is asking you to show that for every positive real (ie number greater than 0) there is always a smaller 1/n. Even as $epsilon$ gets close to zero it should make intuitive sense that you can find a large enough n st 1/n is smaller.
$endgroup$
– T. Fo
Jan 30 at 3:28
$begingroup$
Consider if $epsilon$ is a positive real number, then $frac 1{epsilon}$ is also a positive real number. Now you are assuming that there is an $N > (frac 1{epsilon})^3> 0$. If $n > N > (frac 1{epsilon})^3 > 0$ then $frac 1n < frac 1N < epsilon^3$. ... Okay, I have to admit the statement of "assume if $x$ is real there is an $N > x^3$" is strange but... if $epsilon le 1$ then $epsilon^3 le epsilon$ and if $epsilon > 1$ then $frac 1n < 1 < epsilon$.
$endgroup$
– fleablood
Jan 30 at 6:25
$begingroup$
Are you sure you are being asked to prove that if $x$ is real there is an $N > x^3$ and not that there is an $N > x$? Both are true, but $N > x$ is more straightforward and direct.
$endgroup$
– fleablood
Jan 30 at 6:27
$begingroup$
First step. Figure out what they are asking. Don't worry about proving, but what, in english, are the asking.
$endgroup$
– fleablood
Jan 30 at 2:58
$begingroup$
First step. Figure out what they are asking. Don't worry about proving, but what, in english, are the asking.
$endgroup$
– fleablood
Jan 30 at 2:58
$begingroup$
And thats probably the part I am having difficulty with is understanding what in english they are asking.
$endgroup$
– Forextrader
Jan 30 at 3:02
$begingroup$
And thats probably the part I am having difficulty with is understanding what in english they are asking.
$endgroup$
– Forextrader
Jan 30 at 3:02
$begingroup$
The problem is asking you to show that for every positive real (ie number greater than 0) there is always a smaller 1/n. Even as $epsilon$ gets close to zero it should make intuitive sense that you can find a large enough n st 1/n is smaller.
$endgroup$
– T. Fo
Jan 30 at 3:28
$begingroup$
The problem is asking you to show that for every positive real (ie number greater than 0) there is always a smaller 1/n. Even as $epsilon$ gets close to zero it should make intuitive sense that you can find a large enough n st 1/n is smaller.
$endgroup$
– T. Fo
Jan 30 at 3:28
$begingroup$
Consider if $epsilon$ is a positive real number, then $frac 1{epsilon}$ is also a positive real number. Now you are assuming that there is an $N > (frac 1{epsilon})^3> 0$. If $n > N > (frac 1{epsilon})^3 > 0$ then $frac 1n < frac 1N < epsilon^3$. ... Okay, I have to admit the statement of "assume if $x$ is real there is an $N > x^3$" is strange but... if $epsilon le 1$ then $epsilon^3 le epsilon$ and if $epsilon > 1$ then $frac 1n < 1 < epsilon$.
$endgroup$
– fleablood
Jan 30 at 6:25
$begingroup$
Consider if $epsilon$ is a positive real number, then $frac 1{epsilon}$ is also a positive real number. Now you are assuming that there is an $N > (frac 1{epsilon})^3> 0$. If $n > N > (frac 1{epsilon})^3 > 0$ then $frac 1n < frac 1N < epsilon^3$. ... Okay, I have to admit the statement of "assume if $x$ is real there is an $N > x^3$" is strange but... if $epsilon le 1$ then $epsilon^3 le epsilon$ and if $epsilon > 1$ then $frac 1n < 1 < epsilon$.
$endgroup$
– fleablood
Jan 30 at 6:25
$begingroup$
Are you sure you are being asked to prove that if $x$ is real there is an $N > x^3$ and not that there is an $N > x$? Both are true, but $N > x$ is more straightforward and direct.
$endgroup$
– fleablood
Jan 30 at 6:27
$begingroup$
Are you sure you are being asked to prove that if $x$ is real there is an $N > x^3$ and not that there is an $N > x$? Both are true, but $N > x$ is more straightforward and direct.
$endgroup$
– fleablood
Jan 30 at 6:27
|
show 1 more comment
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$begingroup$
First step. Figure out what they are asking. Don't worry about proving, but what, in english, are the asking.
$endgroup$
– fleablood
Jan 30 at 2:58
$begingroup$
And thats probably the part I am having difficulty with is understanding what in english they are asking.
$endgroup$
– Forextrader
Jan 30 at 3:02
$begingroup$
The problem is asking you to show that for every positive real (ie number greater than 0) there is always a smaller 1/n. Even as $epsilon$ gets close to zero it should make intuitive sense that you can find a large enough n st 1/n is smaller.
$endgroup$
– T. Fo
Jan 30 at 3:28
$begingroup$
Consider if $epsilon$ is a positive real number, then $frac 1{epsilon}$ is also a positive real number. Now you are assuming that there is an $N > (frac 1{epsilon})^3> 0$. If $n > N > (frac 1{epsilon})^3 > 0$ then $frac 1n < frac 1N < epsilon^3$. ... Okay, I have to admit the statement of "assume if $x$ is real there is an $N > x^3$" is strange but... if $epsilon le 1$ then $epsilon^3 le epsilon$ and if $epsilon > 1$ then $frac 1n < 1 < epsilon$.
$endgroup$
– fleablood
Jan 30 at 6:25
$begingroup$
Are you sure you are being asked to prove that if $x$ is real there is an $N > x^3$ and not that there is an $N > x$? Both are true, but $N > x$ is more straightforward and direct.
$endgroup$
– fleablood
Jan 30 at 6:27