If 0≤a<b, then a^2<b^2

1. Jun 11, 2013

wifi

Proof:

Two Cases:

i.) For a=0:

If a=0 and b>a $\Rightarrow$ a^2=0^2=0. Thus, a^2 < b^2.

ii.) For a>0:

If a>0 and b>a $\Rightarrow$ b-a>0 and b+a>0. By closure under addition, (b-a)(b+a)>0.

Or, b^2-a^2>0.

Or, b^2>a^2.

Or, a^2<b^2.

My friend said that this proof is wrong. I think he's just telling me that to get me mad because I don't see what I did wrong. I feel stupid because it's seems so obvious. Maybe I just don't have what it takes to work through Spivak .

2. Jun 11, 2013

MarneMath

Just doing a quick read, I don't see anything wrong that sticks out to me. However, I do know that in Spivak the intent of this problem is to use a previous problem to find the solution.

Edit: "b-a>0 and b+a>0. By closure under addition, (b-a)(b+a)>0." For some reason, i'm not seeing this as necessarily implied.

3. Jun 11, 2013

gerben

is it not much easier to say that since b > a we must have b = a + c for some c larger than 0, so b^2= a^2 + ...

4. Jun 11, 2013

Aimless

I don't see anything particularly wrong with the proof, although I agree with Marne that "by closure under addition" may not be the best way to justify that step.

However, there is an easier way (in my mind, at least) to demonstrate the proof, and it doesn't require the use of cases. Try starting from
0 $\le$ a < b $\Rightarrow \exists \epsilon$ > 0 s.t. b = a + $\epsilon$,
and see what happens.

5. Jun 11, 2013

wifi

Well, if b>a then this implies that b-a>0. So if subtracting a from b is a positive quantity, wouldn't adding a to b result in a positive quantity as well?

Oh! What I meant to say is closure under multiplication . Sorry!

6. Jun 11, 2013

MarneMath

There we go :). The simplest way suggested by the book would be to note that you already proved (or should have proved) that if 0 ≤ a < b and 0 ≤ c < d, then ac < bd. Make a 'clever' substitution and the results follows naturally.

7. Jun 11, 2013

wifi

MarneMath! How could I not see that?

If 0 ≤ a < b and 0 ≤ c < d, then ac < bd. ∴ if we let c→a and d→b $\Rightarrow$ a^2 < b^2.

Thank you!

I really enjoy finding all these different solutions. Aimless, can you elaborate more on your method of solution?

8. Jun 11, 2013

wifi

Also, I'm finding these proofs very unnatural and awkward, surely because this is my first encounter with "real" (ie. proof-based) mathematics. Is this a precursor to what I'll experience through the rest of Spivak? I'm considering giving up altogether at the moment.

Last edited: Jun 11, 2013
9. Jun 11, 2013

MarneMath

I don't know many (or anyone) who naturally grasp proofs and have the ability to do a lot of clever tricks with them initially. Proof writing is a skill that requires a good bit of training to become good at, and even more training to become reasonably proficient. I took a calculus course based on Spivak and it was extremely difficult at the time, and for better or worse, the problems get much harder soon. After the section regarding inequalities, you'll find that you'll be in some very rough terrain. HOWEVER! I don't suggest you give up on Spivak. I highly encourage you to chew on each problem for quite a bit of time before truly giving up. Some problems took me days to solve! Reread the book, look at previous problems (Spivak likes to reference previous problems a lot) and attempt to put ideas together.

Nevertheless, if the book is truly to complicated for you, then you can always start with a more gentle introduction to proofs that'll help you learn some common techniques, but in my personal opinion, the best way to learn proofs is by doing them and learning where your logic falls. By consistent work, you'll make progress and come back to this forum and help others who are stuck where you were once stuck. But before that day, you must also take the journey and endure those first difficult steps until you learn to love it. :)

10. Jun 12, 2013

wifi

MarneMath, your response comforts me & renews my confidence tenfold. I will continue to try my best. Mark my words, I will conquer Spivak!

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