Calculus II - Real numbers proofing

In summary: Then we have:|(a - y) + (y - x + x - b)| \leq |a - y| + |y - x + x - b|.Can you finish off the proof from here?In summary, we can prove that |a-b|<= |y-a|+|x-y|+|x-b| for all x,y in ℝ by using the triangle inequality and breaking down the expression into smaller parts, as shown in the conversation above. By applying the triangle inequality to these smaller parts and rearranging them, we can see that the left-hand side is always less than or equal to the right-hand side, thus proving the statement.
  • #1
FinalStand
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Homework Statement


Show that |a-b|<= |y-a|+|x-y|+|x-b|, for all x,y in ℝ


Homework Equations





The Attempt at a Solution


|a-b| <= |y-a+x-y+x-b| (correct? Not sure about this one...is it not part of the triangle rule?)
|a-b| <= |2x-b-1|
|a-b+2x-2x| <= |2x-b-1|
|a-2x| + |2x-b| <= |2x-b-1| really have clue what I am doing here, no idea if I am on the right track...doubt it is right. Please help. Thanks!
 
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  • #2
No, that's not correct at all.

Start from the expression |a-b| and use the triangle inequality (I assume you've proved this already)

Note that |a-b|*is the distance between points a and b. Therefore |a-b| ≤ |a-x| + |x-b|
 
  • #3
Oh sorry you are not supposed to start on the right hand side. But would it start out like this then:

|a-b+x-x+y-y| <= |y-a| + |x-y| + |x-b|

|a-b+x-x+y-y| <= |a-y| + |y-x| + |x-b|
|a-y+y-x+x-b|<= |a-y| + |y-x| + |x-b|

According to the triangle inequality thing this is true?
 
  • #4
FinalStand said:
Oh sorry you are not supposed to start on the right hand side.
Well, |a-b| is the left side but you can start on whichever side you prefer.

FinalStand said:
According to the triangle inequality thing this is true?

Clearly it is true as you are asked to prove it. However you need to also give the proof.
 
  • #5
Wait so is the proof I shown above correct? That's what I meant.
 
  • #6
FinalStand said:
Wait so is the proof I shown above correct? That's what I meant.

It certainly looks like you're on the right track, but you are still missing the actual proof. Why is LHS≤RHS? You should be able to take a form of triangle inequality (which I again assume has been proven in your lecture notes or somewhere) and use that explicitly to show that this holds.
 
  • #7
So I have to state that since |a+b| <= |a| + |b| therefore this is true?
 
  • #8
FinalStand said:
|a-b| <= |y-a+x-y+x-b| (correct? Not sure about this one...is it not part of the triangle rule?)
|a-b| <= |2x-b-1|
|a-b+2x-2x| <= |2x-b-1|
|a-2x| + |2x-b| <= |2x-b-1| really have clue what I am doing here, no idea if I am on the right track...doubt it is right. Please help. Thanks!
Your first line is already incorrect, as can be verified by choosing a = 2, b = 0, y = 0, x = 1. In that case the LHS is |a - b| = |2 - 0| = 2, and the RHS is |y - a + x - y + x - b| = |0 - 2 + 1 - 0 + 1 - 0| = 0.

I suggest starting as follows: [itex]|a - b| = |a - y + y - x + x - b|[/itex]. Now think about how to apply the triangle inequality to the right hand side.
 
  • #9
Didnt I already say that? Please read the full post before posting please
 
  • #10
OK, it wasn't clear what you were doing in your second post. So you have:
[tex]|a - b| = |a - y + y - x + x - b|[/tex]
What if I suggestively add parentheses as follows:
[tex]|a - y + y - x + x - b| = |(a - y) + (y - x + x - b)|[/tex]
Now apply the triangle inequality to that.
 
  • #11
what do you mean by applying the trinagle inequality to that? what does the parentheses do? Sorry never seen parenthesis in these questions before.
 
  • #12
FinalStand said:
what do you mean by applying the trinagle inequality to that? what does the parentheses do? Sorry never seen parenthesis in these questions before.
The parentheses were a hint to suggest the following: the triangle inequality says [itex]|u + v| \leq |u| + |v|[/itex]. So let [itex]u = a - y[/itex] and [itex]v = y - x + x - b[/itex].
 

Related to Calculus II - Real numbers proofing

1. What is the purpose of "Calculus II - Real numbers proofing"?

The purpose of "Calculus II - Real numbers proofing" is to develop a deeper understanding of the foundational concepts of calculus and their applications to real numbers. This includes the study of limits, continuity, derivatives, and integrals.

2. How is "Calculus II - Real numbers proofing" different from Calculus I?

"Calculus II - Real numbers proofing" builds upon the concepts introduced in Calculus I, such as limits and derivatives, and applies them to real numbers. It also introduces new topics, such as continuity and integrals, which are essential for understanding advanced calculus and its applications.

3. What are some common techniques used in "Calculus II - Real numbers proofing"?

Some common techniques used in "Calculus II - Real numbers proofing" include mathematical induction, epsilon-delta proofs, and proof by contradiction. These techniques are used to rigorously prove theorems and properties of real numbers and their functions.

4. How can "Calculus II - Real numbers proofing" be applied in real-life situations?

The concepts learned in "Calculus II - Real numbers proofing" have many real-life applications, such as in physics, engineering, economics, and statistics. For example, the study of derivatives and integrals can be used to model and analyze the motion of objects, optimize functions, and calculate areas and volumes.

5. What are some tips for success in "Calculus II - Real numbers proofing"?

To succeed in "Calculus II - Real numbers proofing", it is important to have a solid understanding of the concepts covered in Calculus I. It is also crucial to practice and master the various proof techniques, as they are essential for understanding and applying the concepts in this course. Additionally, seeking help from a tutor or professor when needed and staying organized with notes and assignments can also contribute to success in this course.

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