Proving ac < bd: Inequality Proof

[Kzmo]

If $0 ≤ a < b$ and $0 ≤ c <d$, then prove that $ac < bd$

I have taken the proof approach from some previous problems in Spivak's book on Calculus (3rd edition).
This is problem 5.(viii) in chapter 1: Basic Properties of Numbers.

I did as follows:

If $a = 0$ or $c = 0$, then $ac = 0$, but since $0 < bd$, so $ac < bd$
Otherwise,

$0 < a$ & $b > a$ & $0 < c$ & $d > c$
Therefore,
$a \in \mathbb{R}$ and $(b-a) \in \mathbb{R}$ and $c\in \mathbb{R}$ and $(d-c) \in \mathbb{R}$
$a(b-a)\in \mathbb{R}$ and $c(d-c)\in \mathbb{R}$
$a(b-a)$ and $c(d-c)\in \mathbb{R}$

This is where I am stuck.
From here, How can I algebraically deduce that $ac < bd$?

Any hint, would be greatly appreciated,

Thank you very much in advance.

 

[micromass]

If $0 ≤ a < b$ and $0 ≤ c <d$, then prove that $ac < bd$

I have taken the proof approach from some previous problems in Spivak's book on Calculus (3rd edition).
This is problem 5.(viii) in chapter 1: Basic Properties of Numbers.

I did as follows:

If $a = 0$ or $c = 0$, then $ac = 0$, but since $0 < bd$, so $ac < bd$
Otherwise,

Why is $bd>0$?

$0 < a$ & $b > a$ & $0 < c$ & $d > c$
Therefore,
$a \in \mathbb{R}$ and $(b-a) \in \mathbb{R}$ and $c\in \mathbb{R}$ and $(d-c) \in \mathbb{R}$

OK, this follows, but this isn't what you want. Of course $b-a\in \mathbb{R}$, but that's not what $a<b$ means. What it means is something like $b-a\in \mathbb{R}^+\setminus\{0\}$.

Anyway, can you start by listing the relevant axioms here. That is, the axioms that deal with inequalities?

Try to prove the following first: If $0\leq a<b$ and $0\leq c$, then $0\leq ac<bc$. This is simpler and will actually help to prove the general case.

By the way, I've moved this to homework. Please post there in the future (yes, even if you're self-studying )

 

[Kzmo]

micromass,

Than you very much for a very quick response.

$0 < bd$ has to be true if $a=0$ or $c=0$,
because it is part of the question to prove that $ac < bd$, and in the special case of $a=0$ or $c=0$, it should follow that $0 < bd$.
(Only in the specific case when $a=0$ or $c=0$.)
I guess a more simpler reason would be to say that, if $0 > bd$ & $a=0$ or $c=0$, it would make no sense! since $-bd > 0$ does not make sense, nor does $0 > 0$ make sense (the case where they are both zero).
Please correct me if I am wrong!

I have a general question regarding:
$a<b$ meaning $b-a\in \mathbb{R}^+\setminus\{0\}$.
I understand that $b-a\in \mathbb{R}^+$ means that the difference must belong to only the positive numbers in $\mathbb{R}$.
What does $\setminus\{0\}$ mean? (How would you pronounce this in English?). I'm guessing the set of 1 element (zero), is not contained in $(b-a)$ ? (which would make sense, since b is strictly greater than a).

As for listing the relevant axioms regarding inequalities:

Transitive
If $a > b and b > c, then a > c.$
If $a < b and b < c, then a < c.$

Addition 
If $a < b, then a + c < b + c.$
If $a > b, then a + c > b + c.$

Subtraction
If $a < b, then a ‐ c < b ‐ c.$
If $a > b, then a ‐ c > b ‐ c.$

Multiplication
If $a < b and c > 0, then ac < bc.$
If $a < b and c < 0, then ac > bc.$
If $a > b and c > 0, then ac > bc.$
If $a > b and c < 0, then ac < bc.$

Division:
If $a < b and c > 0, then \frac{a}{c} < \frac{b}{c}.$
If $a < b and c < 0, then \frac{a}{c} > \frac{b}{c}.$
If $a > b and c > 0, then \frac{a}{c} > \frac{b}{c}.$
If $a < b and c < 0, then \frac{a}{c} < \frac{b}{c}.$

Substitution
If $a = b$, then $b$ can be substituted for $a$ in any equation or inequality.

As for the Main question of yours,
Please let me know if I have done the proof correct; Please see below.

$0\leq a<b$ and $0\leq c$, then $0\leq ac<bc$.
$0 < a$ & $b > a$ & $0 < c$.
$a\in \mathbb{R}^+$.
$b-a\in \mathbb{R}^+$.
$c\in \mathbb{R}^+$.
$ac(b-a)\in \mathbb{R}^+$.
$ac(b>a)\in \mathbb{R}^+$.
$(abc>a^2c)\in \mathbb{R}^+$.
$(bc>ac)\in \mathbb{R}^+$, $a≠0$.

However, I still haven't proved that $ac≥0$. How would I go about doing this?

Best regards.

 

[micromass]

$0 < bd$ has to be true if $a=0$ or $c=0$,
because it is part of the question to prove that $ac < bd$, and in the special case of $a=0$ or $c=0$, it should follow that $0 < bd$.
(Only in the specific case when $a=0$ or $c=0$.)
I guess a more simpler reason would be to say that, if $0 > bd$ & $a=0$ or $c=0$, it would make no sense! since $-bd > 0$ does not make sense, nor does $0 > 0$ make sense (the case where they are both zero).
Please correct me if I am wrong!

OK. But can you prove from the axioms that if $b>0$ and $d>0$, then $bd>0$? I know it is "easy to see" and the converse "doesn't make sense". But the goal of this chapter of Spivak is that you prove everything rigorously from the axioms and only the axioms.

I have a general question regarding:
$a<b$ meaning $b-a\in \mathbb{R}^+\setminus\{0\}$.
I understand that $b-a\in \mathbb{R}^+$ means that the difference must belong to only the positive numbers in $\mathbb{R}$.
What does $\setminus\{0\}$ mean? (How would you pronounce this in English?). I'm guessing the set of 1 element (zero), is not contained in $(b-a)$ ? (which would make sense, since b is strictly greater than a).

The notation $\setminus\{0\}$ means that I leave out $0$. So $\mathbb{R}^+\setminus\{0\}$ are all the positive real numbers (thus without zero). Another notation you might have seen is $\mathbb{R}^+_0$.

As for listing the relevant axioms regarding inequalities:

Transitive
If $a > b and b > c, then a > c.$
If $a < b and b < c, then a < c.$

Addition 
If $a < b, then a + c < b + c.$
If $a > b, then a + c > b + c.$

Subtraction
If $a < b, then a ‐ c < b ‐ c.$
If $a > b, then a ‐ c > b ‐ c.$

Multiplication
If $a < b and c > 0, then ac < bc.$
If $a < b and c < 0, then ac > bc.$
If $a > b and c > 0, then ac > bc.$
If $a > b and c < 0, then ac < bc.$

Division:
If $a < b and c > 0, then \frac{a}{c} < \frac{b}{c}.$
If $a < b and c < 0, then \frac{a}{c} > \frac{b}{c}.$
If $a > b and c > 0, then \frac{a}{c} > \frac{b}{c}.$
If $a < b and c < 0, then \frac{a}{c} < \frac{b}{c}.$

Substitution
If $a = b$, then $b$ can be substituted for $a$ in any equation or inequality.

Are you sure all these axioms are necessary? I didn't check Spivak, but I know for certain that he had a lot less axioms than this. Or is this list just a list of relevant properties that you've already proven.

What you can use in your proofs are only the axioms or theorems you've already proven before!

As for the Main question of yours,
Please let me know if I have done the proof correct; Please see below.

$0\leq a<b$ and $0\leq c$, then $0\leq ac<bc$.
$0 < a$ & $b > a$ & $0 < c$.
$a\in \mathbb{R}^+$.
$b-a\in \mathbb{R}^+$.
$c\in \mathbb{R}^+$.
$ac(b-a)\in \mathbb{R}^+$.
$ac(b>a)\in \mathbb{R}^+$.
$(abc>a^2c)\in \mathbb{R}^+$.
$(bc>ac)\in \mathbb{R}^+$, $a≠0$.

OK. But what I want to hear from you is that you justify every step by specifying explicitely what axioms or previous proven theorems you use. What you wrote down might be correct, but you need to specify clearly what you use.

Also, if I check your multiplication-axioms, then I see exactly the theorem I asked you to prove! So you could have just said "it follows directly from this axiom" instead of giving a proof!

However, I still haven't proved that $ac≥0$. How would I go about doing this?

By contradiction, I guess. Assume that $ac<0$. Then deduce from this a contradiction (again: by only using the axioms and proven theorems!)

 

[1MileCrash]

What can we take for granted here? May I say that a/b < 1?

If so, what if you wrote it like:

$bd = \Sigma ^{d}_{i=1} b = \Sigma ^{c}_{i=1} b + \Sigma ^{d-c}_{i=1} b$

What is that bigger than? And what's that bigger than?

Whoops, this is in R, not N!

 

[Kzmo]

micromass,

Than you again for your reply, and sorry for my delayed response.

I understood your point that I must prove everything rigorously from the axioms and only the axioms - and I agree.
With that said, I have gave it some thought, and I am not sure how to proceed with proving:
If $b>0$ and $d>0$, then $bd>0$?

Thank you for the explanation regarding the notation $\mathbb{R}^+_0$.

The list contains axioms found thought the chapter. He actually has about 12 axioms in Chapter 1.
(3rd edition).

I understand Your point regarding listing the axioms I used step by step. I will try to do this.
As of now, instead of saying "it follows directly from this axiom" I would actually like to prove it from scratch a few times myself, so I get comfortable saying "it follows directly from this axiom" in the near future.

I have also gave some thought to Assuming that $ac<0$. I do not know where to go with this,
As I am new to proofs and have spent my educational career as an engineer doing computational algebra and Calculus. But I find it very dull and am very interested in learning the rigor behind why I'm allowed to do the computation in the first place. Thus, I am asking you not to be frustrated, but understand that I am very new to this level of mathematics, and am seeking the most elementary explanations as possible. Thank you in advance.

As for my original question - I am still left clueless on how to proceed.
I was able to answer your question (besides proving how $ac≥0$).
But I am not understanding something fundamental - Which is preventing me from answering my original question; Please see my work below, and please tell me at which number I have messed up.

If $0 ≤ a < b$ and $0 ≤ c <d$, then prove that $ac < bd$
1. $0 < a$ & $b > a$ & $0 < c$ & $d > c$ which all $\in$ $\mathbb{R}^+_0$.
2. $a(b-a)$ and $c(d-c)\in \mathbb{R}^+_0$
3. $ac(b-a)(d-c)$$\in \mathbb{R}^+_0$
4. $(abc-a^2c)(d-c)$$\in \mathbb{R}^+_0$
5. $abcd-abc^2-a^2cd+a^2c^2$$\in \mathbb{R}^+_0$
6. $(abcd-a^2cd) + (a^2c^2-abc^2)$$\in \mathbb{R}^+_0$

There is no way I am able to manipulate step 6 (or 5) to give me $bd-ac$, which would result in $ac < bd$. This is my main problem I am not able to get over.

 

[Kzmo]

1MileCrash,

Thank you for your reply, but I am unable to see how this would help answer my question.
I am not saying it won't. I am just unable to see.

Thank you.

 

[slider142]

Hello Kzmo,
There is no need to go all the way back to the ordered field axioms if you have been doing the exercises that lead up to Problem 5(viii). In particular, recall that in Problem 5(iv), you prove the fact that "If a < b and c > 0, then ac < bc." This is now a proven theorem, the conclusion of which you may apply anywhere the antecedent is true. In particular, you can apply it to the antecedent of Problem 5(viii) twice to get the required conclusion.
You will find that Spivak frequently does this, especially in multi-part problems. He will start with a basic theorem for you to prove, with the latter theorems being consequences of the former ones.
As a hint, the antecedent of Problem 5(viii) is equivalent to the statement that "[(a < b and c > 0) or (a < b and c = 0)] and (c < d and b > 0)." What can we conclude from these three statements with their logical connectives if we apply Problem 5(iv) to those particular statements that satisfy its antecedent ?

 

[Ray Vickson]

1MileCrash,

Thank you for your reply, but I am unable to see how this would help answer my question.
I am not saying it won't. I am just unable to see.

Thank you.

There are two cases: (1) c = 0; and (2) c > 0.

Can you deal with case (1)?

Now look at case (2). Can you prove first that ac < bc? If so, can you then prove that bc <= bd? If so, you are done. Sometimes going a few small steps is easier than trying to go one long step.