Please help me solve this ("easy") inequality!

[Gavriell]

I have the following inequality [MATH]\frac{1}{n^{2}}< \epsilon[/MATH], I need to isolate n.
According to the professor's answer I should get [MATH]\frac{1}{\sqrt{\epsilon}}< n[/MATH]I tried different ways of solving, but I don't even understand how can you move [MATH]n^{2}[/MATH] to the other side without even being negative.

 

[Steven G]

I have the following inequality [MATH]\frac{1}{n^{2}}< \epsilon[/MATH], I need to isolate n.
According to the professor's answer I should get [MATH]\frac{1}{\sqrt{\epsilon}}< n[/MATH]I tried different ways of solving, but I don't even understand how can you move [MATH]n^{2}[/MATH] to the other side without even being negative.

I don't even understand how can you move [MATH]n^{2}[/MATH] to the other side without even being negative. Did you think if you multiplied both sides by n^2 that n^2 will turn negative?

Since both quantities are positive (why?), you can take the reciprocal of both sides and change the direction of the inequality(why?). Show us that step then you'll receive more help.

 

[pka]

I have the following inequality [MATH]\frac{1}{n^{2}}< \epsilon[/MATH], I need to isolate n.
According to the professor's answer I should get [MATH]\frac{1}{\sqrt{\epsilon}}< n[/MATH]I tried different ways of solving, but I don't even understand how can you move [MATH]n^{2}[/MATH] to the other side without even being negative.

You did not say so, but surely \(\displaystyle \varepsilon>0 \) & you must know that \(\displaystyle n\in\mathbb{Z}^+\)
\(\displaystyle \frac{1}{n^2}<\varepsilon\\1<n^2\varepsilon\\\dfrac{1}{\varepsilon}<n^2\\\dfrac{1}{\sqrt{\varepsilon}}<n \)

 

[Gavriell]

Thank you, guys I really need to review my algebra. Now I can prove that my sequence converges to zero!

 

[HallsofIvy]

Some basic algebra students learn that to solve an equation like "x+ 4= 6" you "move the 4 to the other side of the equation and it becomes negative": x= 6- 4= 2. I hate that wording because then they solve "2x= 4" saying "you move the 2 to the other side of the equation and it becomes negative": x= 4/(-2)= -2. Of course, that is incorrect because 2(-2)= -4, not 4.

What they should learn is "the opposite of adding is subtracting and the opposite of multiplying is dividing so you undo an addition by subtracting and you undo a multiplication by dividing".

"I tried different ways of solving, but I don't even understand how can you move \(\displaystyle n^2\) to the other side without even being negative."
You don't "move \(\displaystyle x^2\) to the other side" and there is no "negative" involved here! To solve
\(\displaystyle \frac{1}{n^2}< \epsilon\), since 1 is divided by \(\displaystyle n^2\) and the opposite of dividing is multiplying, you need to multiply both sides by the positive number \(\displaystyle n^2\) to get \(\displaystyle 1< n^2\epsilon\). And then, since \(\displaystyle n^2\) is multiplied by \(\displaystyle \epsilon\) you need to divide by it: \(\displaystyle \frac{1}{\epsilon}< n^2\).

Frankly, if you want to learn Calculus it looks like you will have relearn a lot of algebra! Some of what you have "learned", "I tried different ways of solving, but I don't even understand how can you move n2n2 to the other side without even being negative" , simply isn't true!

 

[Gavriell]

Some basic algebra students learn that to solve an equation like "x+ 4= 6" you "move the 4 to the other side of the equation and it becomes negative": x= 6- 4= 2. I hate that wording because then they solve "2x= 4" saying "you move the 2 to the other side of the equation and it becomes negative": x= 4/(-2)= -2. Of course, that is incorrect because 2(-2)= -4, not 4.

What they should learn is "the opposite of adding is subtracting and the opposite of multiplying is dividing so you undo an addition by subtracting and you undo a multiplication by dividing".

"I tried different ways of solving, but I don't even understand how can you move \(\displaystyle n^2\) to the other side without even being negative."
You don't "move \(\displaystyle x^2\) to the other side" and there is no "negative" involved here! To solve
\(\displaystyle \frac{1}{n^2}< \epsilon\), since 1 is divided by \(\displaystyle n^2\) and the opposite of dividing is multiplying, you need to multiply both sides by the positive number \(\displaystyle n^2\) to get \(\displaystyle 1< n^2\epsilon\). And then, since \(\displaystyle n^2\) is multiplied by \(\displaystyle \epsilon\) you need to divide by it: \(\displaystyle \frac{1}{\epsilon}< n^2\).

Frankly, if you want to learn Calculus it looks like you will have relearn a lot of algebra! Some of what you have "learned", "I tried different ways of solving, but I don't even understand how can you move n2n2 to the other side without even being negative" , simply isn't true!

thanks for the tip.

 

[Steven G]

Thank you, guys I really need to review my algebra. Now I can prove that my sequence converges to zero!View attachment 14749

Not that hard after all!
I would have done it this way:
n>0 and \(\displaystyle \epsilon >0\)
\(\displaystyle \dfrac{1}{n^2}< \epsilon \)
So \(\displaystyle n^2>\dfrac{1}{ \epsilon}\)
So \(\displaystyle n>\dfrac{1}{\sqrt{\epsilon}}\)