Proofiness, how much and what counts?

[Four Muffins]

Hello. I'm going through a problem set, and one of the questions is "Show that if 0 < a < b, then a² < b²."

Here's what I wrote, and the solution from the textbook. I suspect that what I wrote doesn't prove anything, but I'm not sure why it doesn't count.

I also have a side question I'm not sure if it is appropriate to post in any of the forums. A random maths person I met suggested learning predicate logic to better understand maths. That seems a few levels above me, so I started going through a free textbook called Forallx. Is that a good path to predicate logic, and would you consider predicate logic useful?

 

[JeffM]

No, your proof is not even close to working.

It ASSUMES the conclusion with [imath]a^2 < b^2[/imath]. That is what you are to prove.

Furthermore, the statement

[math]\sqrt{a^2} < \sqrt{b^2} \implies a < b \text { is invalid because}\\ a = 2 \text { and } b = - 3 \implies a^2 = 4 \text { and } b^2 = 9 \implies\\ \sqrt{a^2} = |a| = 2 \text { and } \sqrt{b^2} = |b| = 3 \implies \sqrt{a^2} < \sqrt{b^2}.\\ \text {But that obviously does not entail that } 2 < - 3. [/math]
It is difficult to give help on proofs because a proof depends on the definitions and axioms that you are allowed to use as well as theorems previously proved, and we do not, indeed cannot, know that.

Here, however, I suspect you are expected to know and use the the axioms of the ordered field and some theorems based on those axioms.

Here are those axioms

1.4: Ordered Field Axioms

math.libretexts.org

But we still do not know what theorems have been previously proved in your specific case. I suspect what you need is something like this

[math]p > 0 \text { and } q < r \implies pq < pr.[/math]
The predicate calculus turns our intuitions about what is a logical argument into a symbolic algebra. In one sense, all of mathematics is dependent on it. But it is not initially intuitive, and much of mathematics was developed by mathematicians who had never heard of it Your math friend has confused strict logical order with psychological order. You do not need the predicate calculus to know that proving a proposition true cannot be based on the assumption that it is true.

I hope this does not sound dismissive. Yours was a good question, with work shown. It is just that proofs are creative rather than mechanical and depend on what you are allowed to use, which we do not know. Sorry to be so little help.

 

[Steven G]

The reasons your proof is not valid:
1) You never concluded that 0<a<b
2) You are not proving the correct theorem! You are trying to prove that if a^2<b^2, then 0<a<b. Now this is NOT true since sqrt(a^2) and a are NOT equal!
Consider when a=-5. sqrt((-5)^2) = sqrt(25)=5, NOT -5

You claim that the solution states that if a<b and c<0, then a^2 < b<2. This is NOT true at all.

Suppose -7<4 which is clearly true. Now let c=-2. Then 14<-8 is NOT true. I suspect that the solution said c>0

 

[Four Muffins]

Thank you both for the explanation. Maths always seems so obvious right after its explained.

@JeffM
I didn't know ordered field axioms were a thing. I will learn them, thank you.

@Steven G
You are correct, I wrote the solution incorrectly. I didn't take a screenshot since it was scattered over a few pages.

 

[pka]

Hello. I'm going through a problem set, and one of the questions is "Show that if 0 < a < b, then a² < b²."

[imath]\text{If }0<a<b\text{ then multiply by }a\text{ to get }a^2<ab~.[/imath]
[imath]\text{Likewise then multiply by }b\text{ to get }ab<b^2~.[/imath]
[imath]\text{By the transitive property }\text{ you get }a^2<b^2~.[/imath]

 

[Al-Layth]

[imath]\text{If }0<a<b\text{ then multiply by }a\text{ to get }a^2<ab~.[/imath]
[imath]\text{Likewise then multiply by }b\text{ to get }ab<b^2~.[/imath]
[imath]\text{By the transitive property }\text{ you get }a^2<b^2~.[/imath]

can't u just square the ineqaulity ?
a< b therefore a^2 < b^2
?

 

[JeffM]

can't u just square the ineqaulity ?
a< b therefore a^2 < b^2
?

No.

[math]- 3 < - 2 \text { does NOT imply that } 9 < 4.[/math]

 

[JeffM]

can't u just square the ineqaulity ?
a< b therefore a^2 < b^2
?

It seems to me that I may have seemed curt in my previous response.

We can prove the following theorem

[math]0 \le a < b \implies a^2 < b^2.[/math]
And we frequently use that theorem without being careful to specify the limited circumstances when it is true.

BUT we cannot use that theorem in proving it, right? Now if we had a more general theorem, we could deduce the specific from the general (assuming that had been previously proved).

[math] 0 < a < b \implies a^2 < b^2 \ \text { because }\\ a < b \implies a^2 < b^2 \text { generally without restrictions.} [/math]
That proof is valid in form, but it is not substantively valid because there is no such general theorem. There cannot be because it is false.

If I was rude in my first reply, please accept my apologies.