Need help on a proof from Spivak's Calculus

In summary: No. I meant to start with the claim, and transform it until a true statement is found. That's legitimate if all the steps are reversible, but squaring both sides is not one of them. However, since it is given that both a and b are positive, squaring both sides is reversible in this case.In summary, the conversation discusses a problem from Spivak's book and the attempt to solve it using different strategies. The problem states that if 0 < a < b, then a < √(ab) < (a+b)/2 < b. The conversation includes hints and tips on how to approach the problem, including starting with the end statement and working backwards, and the idea of squaring both sides to
  • #1
CaptainAmerica17
59
10

Homework Statement


I'm currently working through Spivak independently and have reached the problems at the end of ch. 1.
The problem is:
Prove that if [itex] 0 < a < b [/itex], then [tex] a < \sqrt{ab} < \frac{a+b}{2} < b [/tex]

Homework Equations



Spivak's properties P1 - P12

The Attempt at a Solution


I was thinking that if I do it by cases, I can first try to prove [itex] a < \sqrt{ab} [/itex] and then do [itex] \frac{a+b}{2} < b[/itex]. After that, I can use [itex] 0 < a < b [/itex] to bring it all together. To start with [itex] a < \sqrt{ab} [/itex], I have: If [itex] a > 0 [/itex] and [itex] b > 0 [/itex], then [itex] ab > 0 [/itex]. I'm not sure where to go from there or if my approach is even correct.
 
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  • #2
Try and solve it from behind: start with ##\sqrt{ab} \leq \dfrac{a+b}{2}## and square this equation, do some algebra, and finally write it down in reverse order.
 
  • #3
fresh_42 said:
Try and solve it from behind: start with ##\sqrt{ab} \leq \dfrac{a+b}{2}## and square this equation, do some algebra, and finally write it down in reverse order.
Thanks for the hint. It helped a bit.
 
  • #4
CaptainAmerica17 said:
Thanks for the hint. It helped a bit.
The trick is to turn ##(a+b)^2>0## into ##(a-b)^2>0## via ##4ab##.
 
  • #5
CaptainAmerica17 said:

Homework Statement


I'm currently working through Spivak independently and have reached the problems at the end of ch. 1.
The problem is:
Prove that if [itex] 0 < a < b [/itex], then [tex] a < \sqrt{ab} < \frac{a+b}{2} < b [/tex]

Homework Equations



Spivak's properties P1 - P12

The Attempt at a Solution


I was thinking that if I do it by cases, I can first try to prove [itex] a < \sqrt{ab} [/itex] and then do [itex] \frac{a+b}{2} < b[/itex]. After that, I can use [itex] 0 < a < b [/itex] to bring it all together. To start with [itex] a < \sqrt{ab} [/itex], I have: If [itex] a > 0 [/itex] and [itex] b > 0 [/itex], then [itex] ab > 0 [/itex]. I'm not sure where to go from there or if my approach is even correct.

(1) What do you get when you expand out ##(\sqrt{b}-\sqrt{a})^2\: ?##
(2) What can you do with the fact that ##0 < a < b \; \Rightarrow 0 < a^2 < ab \: ?##
 
  • #6
fresh_42 said:
Try and solve it from behind: start with ##\sqrt{ab} \leq \dfrac{a+b}{2}## and square this equation, do some algebra, and finally write it down in reverse order.
Did you mean "start with ##\sqrt{ab} \geq \dfrac{a+b}{2}## and square this inequality?
 
  • #7
what happens if you multiply the inequality a < b by a? or if you let b = a+e where e>0? or almost anything. what did you try?
 
  • #8
Mark44 said:
Did you mean "start with ##\sqrt{ab} \geq \dfrac{a+b}{2}## and square this inequality?
No. I meant to start with the claim, and transform it until a true statement is found. Afterwards, it has to be written in the correct order, i.e. reverse to what has been done. In this case, there are no losses between the steps due to the inequalities, and as ##0< a < b## the root - function in this case - also behaves nicely.

I haven't recommended to turn a proof upside down, I only said that starting at the end sometimes gives the right hints and shows the way to proceed.
 
  • #9
fresh_42 said:
No. I meant to start with the claim, and transform it until a true statement is found.
That's legitimate if all the steps are reversible, but squaring both sides is not one of them. However, since it is given that both a and b are positive, squaring both sides is reversible in this case.
 
  • #10
fresh_42 said:
No. I meant to start with the claim, and transform it until a true statement is found. Afterwards, it has to be written in the correct order, i.e. reverse to what has been done. In this case, there are no losses between the steps due to the inequalities, and as ##0< a < b## the root - function in this case - also behaves nicely.

I haven't recommended to turn a proof upside down, I only said that starting at the end sometimes gives the right hints and shows the way to proceed.
When I did that, I ended up with the statement [itex] 2 \leq a+b [/itex], which must be true given what's stated in the problem.
 
  • #11
I think it is legitimate even if single steps aren't reversible. This - of course - won't result in a proof, as it is here the case, where we don't need an indirect procedure, but it might lead to insights where the critical points are. So to get ideas about how to prove something, it is often a good idea to play with the result. E.g. ##\varepsilon-## proofs can frequently be found this way, simply because we don't know how to set the variable at the existence quantifier at the start - it's filled in when it's obvious how, even if the formally correct version of the proof has it at the beginning.

However, I admit one has to know that such a procedure does not yield a proof - only hints. And if we're lucky, as in the example above, everything is reversible.
 
  • #12
CaptainAmerica17 said:
When I did that, I ended up with the statement [itex] 2 \leq a+b [/itex], which must be true given what's stated in the problem.
No, that's not necessarily true. For example, if a = 1/2 and b = 1, then the chain of inequalities given in post #1 is true.
 
  • #13
CaptainAmerica17 said:
When I did that, I ended up with the statement [itex] 2 \leq a+b [/itex], which must be true given what's stated in the problem.
I get ##4ab \leq (a+b)^2 \Longleftrightarrow 0 \leq (a-b)^2##. How did you arrive at ##2 \leq a+b\,?##
 
  • #14
fresh_42 said:
I think it is legitimate even if single steps aren't reversible. This - of course - won't result in a proof, as it is here the case, where we don't need an indirect procedure, but it might lead to insights where the critical points are. So to get ideas about how to prove something, it is often a good idea to play with the result. E.g. ##\varepsilon-## proofs can frequently be found this way, simply because we don't know how to set the variable at the existence quantifier at the start - it's filled in when it's obvious how, even if the formally correct version of the proof has it at the beginning.

However, I admit one has to know that such a procedure does not yield a proof - only hints. And if we're lucky, as in the example above, everything is reversible.

This is a skill I have been learning very slowly, but surely. It's been especially useful for some set theory proofs I've done recently.
 
  • #15
fresh_42 said:
I think it is legitimate even if single steps aren't reversible.
Maybe sometimes, but not in general.

For example, if we are given that a < b, by squaring both sides we might conclude that ##a^2 < b^2##. This conclusion would be false if a = -1 and b = 1/2, because we would be concluding that ##(-1)^2 = 1 < (\frac 1 2)^2 = \frac 1 4##.
 
  • #16
fresh_42 said:
I get ##4ab \leq (a+b)^2 \Longleftrightarrow 0 \leq (a-b)^2##. How did you arrive at ##2 \leq a+b\,?##
I started with [itex] \sqrt{ab} \leq \frac{a+b}{2} [/itex] and just went from there like you said in your original message.
 
  • #17
Mark44 said:
Maybe sometimes, but not in general.

For example, if we are given that a < b, by squaring both sides we might conclude that ##a^2 < b^2##. This conclusion would be false if a = -1 and b = 1/2, because we would be concluding that ##(-1)^2 = 1 < (\frac 1 2)^2 = \frac 1 4##.
Yes. One has to be cautious. I remember I had a thesis on the desk with basically the following conclusion:
$$
a < b \Longrightarrow ax < bx
$$
Of course ##x## had been an entire expression. Too bad ##x## could have been negative!

And in the case above, I agree, the crucial part is to clearly state, why squaring does no harm to the equivalence. And in this sense, my hint had been far from being a proof. Just a hint how to get an idea.
 
  • #18
CaptainAmerica17 said:
I started with [itex] \sqrt{ab} \leq \frac{a+b}{2} [/itex] and just went from there like you said in your original message.
But taking the square yields ##ab \leq \dfrac{(a+b)^2}{4}##. But read my previous post, as to why this is an allowed operation in this case!
 
  • #19
fresh_42 said:
But taking the square yields ##ab \leq \dfrac{(a+b)^2}{4}##. But read my previous post, as to why this is an allowed operation in this case!

I think I took to many steps in the wrong direction. I just kept reducing. I see how you got what you did, I'm going to reassess.
 
  • #20
Just a comment that the lh side is the so-cold AGI or Arohmetric-Geometric inequality. The leftmost is the geometric mean and the center one is the Arithmetic mean. It holds true for n>1.
 
  • #21
Make that so-called and Arithmetic.
EarlyWinter getting to me.
 

Related to Need help on a proof from Spivak's Calculus

1. How do I approach proofs in Spivak's Calculus?

The key to approaching proofs in Spivak's Calculus is to break them down into smaller, manageable steps. Begin by carefully understanding the problem and identifying what is given and what is being asked. Then, use your knowledge of theorems and definitions to come up with a plan for solving the proof. Make sure to clearly outline each step and provide justifications for your reasoning.

2. What resources can I use to help me with proofs in Spivak's Calculus?

There are several resources that can be helpful when working on proofs in Spivak's Calculus. These include study guides, online forums, and study groups. It can also be helpful to consult your textbook for additional examples and explanations.

3. How can I improve my proof-writing skills for Spivak's Calculus?

The best way to improve your proof-writing skills is to practice regularly. Start with simpler proofs and work your way up to more complex ones. Additionally, pay attention to the feedback you receive from your instructor or peers and use it to improve your approach and reasoning.

4. I'm struggling with a specific proof in Spivak's Calculus. What should I do?

If you are struggling with a specific proof, it can be helpful to break it down into smaller parts and try to solve them individually. You can also consult with your instructor or peers for assistance. Additionally, revisiting previous proofs and examples from your textbook can provide insight and help you better understand the steps involved in solving proofs.

5. Are there any tips for writing clear and concise proofs in Spivak's Calculus?

To write clear and concise proofs in Spivak's Calculus, make sure to use precise and unambiguous language. Clearly state your assumptions and justify each step in your reasoning. It can also be helpful to use diagrams or examples to illustrate your reasoning and make your proof easier to follow.

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