Help with a discrete math homework problem

I'm sorry, I don't understand what you're asking in your first statement.The second statement, I believe you're asking me what I can add to the left side (r2 + s2) to make it equivalent to (r + s)2.Is that correct?If so, can I add 2rs to both sides?r < s=> r2 + r < r2 + s (by adding r2 to both sides)=> r2 + r < s2 + s (by adding s2 to both sides)Next, notice that 2rs < r2 + s2. Can you see why?yes, because if r is less than s then rs must be less than the
  • #1
aporter1
32
0

Homework Statement


show that for positive r and s, with r<s, we always have:

r<(r+s)/2<s and 2/[(1/r)+(1/s)]^2< 2rs< (r+s)^2


Homework Equations





The Attempt at a Solution


i have shown that r<s because r+r<s+r, 2r<s+r, r<(s+r)/2 and 2r< (2s+2r)/4
(r+s)^2= r^2+2rs+s^2 where are<r^2+2rs+s^2

not sure if I'm even on the right track or if i have even finished the problem. I'm kinda confused.
 
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  • #2
aporter1 said:

Homework Statement


show that for positive r and s, with r<s, we always have:

r<(r+s)/2<s and 2/[(1/r)+(1/s)]^2< 2rs< (r+s)^2


Homework Equations





The Attempt at a Solution


i have shown that r<s because r+r<s+r, 2r<s+r, r<(s+r)/2 and 2r< (2s+2r)/4
(r+s)^2= r^2+2rs+s^2 where are<r^2+2rs+s^2
You've wandered off the right track. You don't need to (and shouldn't) show that r < s. You are given that r < s.

Some of what you wrote makes sense, but a lot doesn't (e.g., "(r+s)^2= r^2+2rs+s^2 where are<r^2+2rs+s^2"), and it's not well organized.


aporter1 said:
not sure if I'm even on the right track or if i have even finished the problem. I'm kinda confused.

For each part, you should start with r < s. For the first part, the last line should be r < (r + s)/2 < s. For the second part, the last line should be 2/[(1/r)+(1/s)]^2< 2rs< (r+s)^2. Each line between the start and end should make sense logically and mathematically.
 
  • #3
okay so if i start with r<s, i have r<s.

then, how do i show r<s for r+s/2 ? Would I start with r+r<r+s, if so, how do i incorporate the 2?
 
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  • #4
aporter1 said:

Homework Statement


show that for positive r and s, with r<s, we always have:

r<(r+s)/2<s and 2/[(1/r)+(1/s)]^2< 2rs< (r+s)^2


Homework Equations





The Attempt at a Solution


i have shown that r<s because r+r<s+r, 2r<s+r, r<(s+r)/2 and 2r< (2s+2r)/4
(r+s)^2= r^2+2rs+s^2 where are<r^2+2rs+s^2

not sure if I'm even on the right track or if i have even finished the problem. I'm kinda confused.

Just break up the inequality into 2 parts, that is it.
 
  • #5
aporter1 said:
okay so if i start with r<s, i have r<s.

then, how do i show r<s for r+s/2 ? Would I start with r+r<r+s, if so, how do i incorporate the 2?
You don't show r < s for anything. You are to assume this. You need to show that r < (r + s)/2 < s.

What you started with was OK, but you need to take it a bit further.

r < s
=> r + r < s + r and s + r < s + s -- Do you understand why this follows from r < s?
=> ? What can you do with the two inequalities above?
 
  • #6


okay so,

r + r < s + r < s + s, this is true because if r < s then, r < s +r and because r < s then r+r<s, so we have what follows,

r + r < s + r < s + s , now i believe if we add 2 ,

2r < 2(s+r) < 2s then,

r< (s+r)/2 < s
 
  • #7


aporter1 said:
okay so,

r + r < s + r < s + s, this is true because if r < s then, r < s +r and because r < s then r+r<s, so we have what follows,
Here's a better way to say it.
r < s (given)
r + r < s + r (by adding r to both sides of the given inequality)

r + s < s + s (by adding s to both sides of the given inequality)

Combining these, we get
r + r < r +s < s + s
aporter1 said:
r + r < s + r < s + s , now i believe if we add 2 ,
? You're not adding 2.
aporter1 said:
2r < 2(s+r) < 2s
No. The expressions at either end are right, since r + r = 2r and s + s = 2s, but you can't just stick a factor of 2 in the middle.
aporter1 said:
then,

r< (s+r)/2 < s
This doesn't follow from your previous inequality.
 
  • #8
how to i prove the middle part then? I understand the other part now.

would i just divide out by two to get,

r<(r+s)/2<s
 
  • #9
Yes.
2r < r + s < 2s ==> r < (r + s)/2 < s

What this is saying is that if r < s, the average of r and s will be between r and s.
 
  • #10
okay now where do i start with the second part?

2/[(1/r)+(1/s)]^2< 2rs< (r+s)^2

Would I start by supposing r<s again? and then,

would i start by expanding out (r+s)^2?
 
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  • #11
so I've tried to figure it out and this is what i have come up with,

2rs< r2+2rs+s2, (by expansion of (r+s)2)

2/(2/rs) < 2rs, (by expansion of (1/r + 1/s)2) and finding a common denominator
 
  • #12
aporter1 said:
so I've tried to figure it out and this is what i have come up with,

2rs< r2+2rs+s2, (by expansion of (r+s)2)
Starting from r < s, how do you get to this?

When you expand (r + s)2, you get r2 +2rs + s2. IOW, (r + s)2 = r2 +2rs + s2.

How did you get an inequality out of that?


aporter1 said:
2/(2/rs) < 2rs, (by expansion of (1/r + 1/s)2) and finding a common denominator
This doesn't make any sense to me.
 
  • #13
well I was just comparing it to the other part in the problem
 
  • #14
You have to be able to justify each step. Each inequality you write has to be a direct consequence of the one before it.
 
  • #15
so how do i go about doing that then? I have to start with:

2/[(1/r)+(1/s)]2
 
  • #16
aporter1 said:
so how do i go about doing that then? I have to start with:

2/[(1/r)+(1/s)]2

The second part is quite a bit easier; namely, showing that 2rs < (r + s)2

As before, start with the given statement: r < s

Notice that r2 + s2 > 0 -- can you explain why?
What can you add to the left side (and the right side) to make the left side (r + s)2?
 
  • #17
its greater than zero because there are fractions
 
  • #18
would i add r2+s2 to both sides?
 
  • #19
aporter1 said:
its greater than zero because there are fractions
No, that's not why. If r < s, it does not necessarily follow that 1/r + 1/s > 0. Consider r = -1 and s = 2. We have r < s, but 1/r + 1/s = 1/(-1) + 1/2 = -1/2 < 0.
aporter1 said:
would i add r2+s2 to both sides?
Yes.
 

What is discrete math?

Discrete math is a branch of mathematics that deals with mathematical structures and objects that are countable or can be distinct, as opposed to continuous objects like real numbers or curves.

Why do I need to study discrete math?

Discrete math is used in many fields, including computer science, engineering, and cryptography. It provides the foundation for understanding and solving problems in these areas.

What is a discrete math homework problem?

A discrete math homework problem is a mathematical problem that involves discrete objects or structures, such as sets, functions, graphs, and combinatorics. It may require applying concepts and techniques from discrete math to solve.

How do I approach a discrete math homework problem?

The best approach is to understand the problem and identify the concepts and techniques that are relevant. Then, apply these concepts and techniques to solve the problem step by step. It may also be helpful to work through similar examples or seek help from a classmate or instructor.

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