Hey guys, I'm getting into doing proofs and was wondering if you could tell me if my logic is correct for this one:

Given that 0 ≤ a ≤ b
Show that a ≤ sqrt((a)(b)) ≤ ((a+b)/2) ≤ b

So there should be 4 cases if I use Math. Induction:
Case 1) 0 = a = b
Case 2) 0 = a < b
Case 3) 0 < a = b
Case 4) 0 < a < b

Case 1) 0 = a = b
Sub in values of a & b into the eq'n (a, b = 0):
=> 0 ≤ sqrt((0)(0)) ≤ ((0+0)/2) ≤ 0
=> 0 ≤ 0 ≤ 0 ≤ 0
IS True

Case 2) 0 = a < b
Sub in values of a & b into the eq'n (a = 0, b = 1):
=> 0 ≤ sqrt((0)(1)) ≤ ((0+1)/2) ≤ 1
=> 0 ≤ 0 ≤ 1/2 ≤ 1
IS True

Case 3) 0 < a = b
Sub in values of a & b into the eq'n (a = 1, b = 1):
=> 1 ≤ sqrt((1)(1)) ≤ ((1+2)/2) ≤ 1
=> 1 ≤ 1 ≤ 1 ≤ 1
IS True

Case 4) 0 < a < b
Sub in values of a & b into the eq'n (a = 1, b = 2):
=> 1 ≤ sqrt((1)(2)) ≤ ((1+2)/2) ≤ 2
=> 1 ≤ sqrt(2) ≤ 3/2 ≤ 2
IS True

So that's basically it, I've proven each case to be true so I'll add in some more explanation of each step and add in a conclusion and that would be it, right?

--Thanks.

That's not at all correct. Firstly, induction is the wrong method for proving this: induction only works when the arguments of the statement are integers, and here we're dealing with real numbers greater than or equal to zero. Thirdly, using induction to prove these inequalities (for natural numbers) is labourious: giving the smallest pair of a, b such that the inequalities are true is the first step in an inductive argument to demonstrating their validity, but you then need to proceed to the general case by assuming them true for, say, a=n, b=m and proving true for the cases a=n+1, b=m, a=n, b=m+1 etc. And even then you're nowhere near showing they're true for real numbers!

I suggest the following:

Proposition 1: If 0≤a≤b then

a≤sqrt(ab)

Proof of 1:

If 0≤a≤b then, multiplying all sides of the inequality by a, which is positive,

0≤a^2≤ab

Thus, by taking square roots, we get the result.

Proposition 2: For all real, non-negative numbers a, b

sqrt((a)(b)) ≤ ((a+b)/2)

Proof of 2:

Observe that for all such a, b

0≤(sqrt(a)-sqrt(b))^2

Thus, by expanding the brackets,

0≤a+b-2sqrt(ab)

And thus we have the result.

I leave the last inequality (a+b)/2≤b to you.

Haha, great thanks.

I still don't quite understand how this proves our theorem. All you've done is express the inequalities in different ways; ie moving the square around and such. How does this actually prove as in "Proof 1" that a is less than or equal to the square root of ab?

The whole point of a proof is that you begin with a set of assumptions that define a set of objects and then proceed directly from these to establish a result describing them. In this case the assumption is 0≤a≤b, and this all we have defining a and b, therefore we only have this information to proceed with. The proof is essentially manipulation, as is often the case with algebraic results, but it demonstrates what we're required to show.

Another example of a proof would be in showing that the equation x^2=4 has a real solution: all you need do is find a real number that squares to 4. It's not a complicated piece of deduction, but it demonstrates that the assertion is true.

What do you think comprises a proof? What inclined you to use induction, and how is this preferable to you as a method of proof to what I've used?

The whole point of a proof is that you begin with a set of assumptions that define a set of objects and then proceed directly from these to establish a result describing them. In this case the assumption is 0≤a≤b, and this all we have defining a and b, therefore we only have this information to proceed with. The proof is essentially manipulation, as is often the case with algebraic results, but it demonstrates what we're required to show.

Another example of a proof would be in showing that the equation x^2=4 has a real solution: all you need do is find a real number that squares to 4. It's not a complicated piece of deduction, but it demonstrates that the assertion is true.

What do you think comprises a proof? What inclined you to use induction, and how is this preferable to you as a method of proof to what I've used?

I like induction because I actually get to sub in some numbers i.e. 1 for my base case and then n+1 for my induction hypothesis which outputs a statement that usually give me an expression that makes more sense to me. The method above that you used is exhaustion (divided it into different cases and then proved each separately), correct?

And I guess it depends on what kind of proof I'm going to be using whether it's contrapositive, natural deduction, exhaustion, induction or whatever. I read a ton of stuff last night of the different kinds of proofs so I have a pretty good idea of what comprises each kind of proof. Whether it's a "If A then B", or "If A then not B" and so forth.

I just need to keep doing problems and getting more experience because right now I have problems with proving statement a is equal/less/greater than/whatever than statement b.

So for finishing the last inequality that you left me. So for the last inequality that you left me would it be:
If 0 ≤ a ≤ b then (a+b)/2 ≤ b
=> a+b ≤ 2b
=> a ≤ b

That makes sense to me because I got the result or the true statement that we began with. So would that be right?

Your observation is correct in this case, however in general it will not be. Generally, if A implies B (here we say B if A) it does not follow that B implies A (here we say A if B). (If both A implies B and B implies A, we say A if and only if, or A iff B. Here, the first and final inequalites are iff statements, but the second isn't, because it's true for any positive a and b.) For instance, to be a man implies one is mortal, from the definition of the premise "man", but the converse is not necessarily true: if one is mortal one is not necessarily a man, because one could be a giraffe. You must show it the other way around: start from a is less than or equal to b and proove that their mean is less than or equal to b.

And no, the method I used isn't exhaustion, because there are no separate cases that a and b may fall into that exhaust the variables of the statement and should be individually examined. The triple inequality that we're asked to prove is really just the conjunction of three separate inequalities, so in order to proove the whole, larger inequality I can simply prove the smaller ones. [Or at least I wouldn't call it exhaustion.]

Exhaustion, contradiction and induction are particular methods of proof for particular classes of statements, whereas deduction is just a general logical term for the argument which takes one from a set of premises to a conclusion and provides no method at all. Often the case with an assertion and the problem of demonstrating its validity, i.e. a proof, is that the former is intuitively clear and thus the latter becomes a problem of translating this intuition into a mathematical argument that is transparent and conclusive.

In the case of these problems the final inequality says that the mean of a and b is smaller than b, which is intuitively obvious because a is less than (or equal to) b and so their mean would be somewhere between a and b, i.e. less than b. Similarly with the first inequality: ab is more than a*a=a^2 (just replace an a with a b), because a is less than b, and thus sqrt(ab) would be more than sqrt(a^2)=a. This latter intuition translates directly into the argument I gave. The proof of the second inequality involves more of what mathematicians would be inclined to call a trick, namely using some expression or device apparently quite unrelated to the problem at hand in order to solve this very problem.

So we see that there is no set of methods that facilitates the proof of any mathematical statement. I.e. the mathematician has no utility belt with which to hammer any problem with until the answer falls out, but only his intuition and the depth of his understanding of the concepts involved.