Determine the set of values of ##n## that satisfy the inequality

[chwala]

Homework Statement

See attached

Relevant Equations

Inequalities

Find solution here;

Ok i just want clarity for part (a),
My approach is as follows, since we want positive integer values that satisfy the problem then,
$\dfrac {n^2-1}{2}≥1$ I had earlier thought of $\dfrac {n^2-1}{2}≥0$ but realized that $0$ is an integer yes but its not a positive integer.
Therefore,
$\dfrac {n^2-1}{2}≥1, n^2-1≥2, n^2≥3 ⇒n≥\sqrt 3$

For,
$\dfrac {n^2+1}{2}≥, n^2≥1⇒n≥1$. The inequality satisfying the two is

$n≥\sqrt 3$

...Any insight on this is appreciated.

Part (b) is easy...no problem there...

 

[Office_Shredder]

The typed up image has a lot of "or"s that probably are supposed to be "and"s. Is that an official grading rubric? I would give it half points at most.

It's kind of lame to not just name the smallest integer that you computed, which is 2, not $\sqrt{3}$. But just to make sure, $(2^2-1)/2=1.5$ is not an integer, so there is more work to be done.

 

[chwala]

The typed up image has a lot of "or"s that probably are supposed to be "and"s. Is that an official grading rubric? I would give it half points at most.

It's kind of lame to not just name the smallest integer that you computed, which is 2, not $\sqrt{3}$. But just to make sure, $(2^2-1)/2=1.5$ is not an integer, so there is more work to be done.

Yes, its an official ms guide from a past paper...

 

[chwala]

The typed up image has a lot of "or"s that probably are supposed to be "and"s. Is that an official grading rubric? I would give it half points at most.

It's kind of lame to not just name the smallest integer that you computed, which is 2, not $\sqrt{3}$. But just to make sure, $(2^2-1)/2=1.5$ is not an integer, so there is more work to be done.

True, i had $2$ in my original working and thought i had done something wrong! Since $n≥\sqrt 3⇒n≥2,nε\mathbb{z}^{+}$

 

[Office_Shredder]

Great. What about the part where not all choices of $n\geq 2$ satisfy this? You have the positive part down, what about the integer part?

 

[chwala]

Great. What about the part where not all choices of $n\get 2$ satisfy this? You have the positive part down, what about the integer part?

I hope i am getting you right, we have $\sqrt 3=±1.73205...$ We will not consider the negative values of $n$ as the question only requires us to find positive integer values of $n$ . In that case the value of $n=-1$ will not be a solution.

 

[SammyS]

Homework Statement:: See attached
Relevant Equations:: Inequalities

View attachment 302150

Find solution here;

View attachment 302151

Ok i just want clarity for part (a),
My approach is as follows, since we want positive integer values that satisfy the problem then,
$\dfrac {n^2-1}{2}≥1$ I had earlier thought of $\dfrac {n^2-1}{2}≥0$ but realized that $0$ is an integer yes but its not a positive integer.
Therefore,
$\dfrac {n^2-1}{2}≥1, n^2-1≥2, n^2≥3 ⇒n≥\sqrt 3$

At this point, you have given the requirement for $\dfrac {n^2-1}{2}$ to be positive.

It should be very easy to show that $\dfrac {n^2+1}{2} \gt \dfrac {n^2-1}{2}$ for all $n$, so the requirement on $n$ remains the same. I suggest using subtraction.

Seems like it's still left to show that to get integer results, we need that $n$ must be odd.

 

[Office_Shredder]

Yeah. I'm referring to the part where only odd n give actual integers. You haven't mentioned it at all in your posts.

 

[chwala]

Yeah. I'm referring to the part where only odd n give actual integers. You haven't mentioned it at all in your posts.

True, i did not mention that...i guess the ms is clear on that part...

 

[SammyS]

True, i did not mention that...i guess the ms is clear on that part...

Yes, it is true that the ms (marking scheme - or whatever) is clear in stating that $n$ must be an odd integer. What's not so clear in the ms, is how to show that.

What's even less clear is your approach to showing that $n$ must be odd.

 

[chwala]

Yes, it is true that the ms (marking scheme - or whatever) is clear in stating that $n$ must be an odd integer. What's not so clear in the ms, is how to show that.

What's even less clear is your approach to showing that $n$ must be odd.

Ok, I'll look at that later...cheers.

 

[SammyS]

Ok, I'll look at that later...cheers.

Maybe ... start with :

Suppose $\displaystyle \dfrac{n^2-1}{2}=k, \text{ for }k\text{ a positive integer.}$ Of course, you have already determined the condition for the positive part of the requirement.

 

[chwala]

Maybe ... start with :

Suppose $\displaystyle \dfrac{n^2-1}{2}=k, \text{ for }k\text{ a positive integer.}$ Of course, you have already determined the condition for the positive part of the requirement.

@SammyS let me look at this on Saturday... cheers