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Say I have a number, "a". 0 < a < 5. Let's say I have another number, "b". 0 < b < 10

Is ab < 50?

If you were to write this as a * b < 5 * b, you would find that the upper bound is < than 50, since b is strictly less than 10. 5 * <10 = < 50

If the upper bound is less than 50, doesn't a have to be less than a number that's less than 50, and not 50 itself?

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What do you mean by "less than less than"? –  Michael Albanese Dec 22 '12 at 2:40
    
Basically, in the example, shouldn't ab < 49.999999999999... (the largest number preceding 50) –  Tim Dec 22 '12 at 2:49
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There is no largest number less than $50$, the number you wrote down is equal to $50$. What is true is that if $ab < 50$, then $ab < k$ for some (in fact infinitely many) $k < 50$. –  Michael Albanese Dec 22 '12 at 2:53
    
Yes, I know that 0.9999999999999999999... = 1. But if there is no largest number < 50, then what about 49.5? 49.9? Since b is not exactly 10 (it has to be less), shouldn't the actual upper bound is less than 50? And therefore it would not be 50, but some number < 50? –  Tim Dec 22 '12 at 2:54
    
As MJD put in his answer, the upper bound can't be any smaller than $50$ because given a potential upper bound $k$ with $k < 50$, you can find $a$ and $b$ with $0 < a < 5$ and $0 < b < 10$ such that $ab \geq k$. –  Michael Albanese Dec 22 '12 at 3:01
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3 Answers 3

up vote 4 down vote accepted

You use only one "less than" sign.

Consider this as an analogy: Just as we would not say: $a = 5$, $b = 5$, so therefore $a + b$ "equals equals" $10$, but rather, we'd conclude that $a + b = 10$. In your case, we would not say $ab$ "is less than less than" $50$, but rather, $ab < 50$.

For any two real numbers, $x, y$, one and only one of the following is true:

  • $x < y\quad$ OR (x less than y)
  • $x = y\quad$ OR (x equals y)
  • $x > y\quad$ (x is greater than y)

EDIT following question edit:

To answer the question in the comment below: There is no largest number preceding 50: For every $k < 50$, there is a $j$ such that $k < j < 50$, and there is an $m$ such that $k < j < m < 50...$ and on and on and on...for every proposed "largest number $n$ preceding 50", there is a larger number than $n$ preceding 50.

In this case, for example, take any possible choice of $ab < 50$. Then there exists $x = \large \frac {50 + ab}{2}$ $> ab$ but nonetheless, $x < 50$. If we then let $ab = x$, there exists $y = \large\frac {50 + x}{2}$ $> x,$ with $y< 50$...and so on.

The least upper bound for $ab$, in this case, is not in the set of possible values for $ab$. Think of it like this: $$ 0 < a < 5,\;\;0 < b < 10, \;\; \implies \;\; ab \in (0, 50),$$ where $(0, 50)$ is the open interval of reals greater than $0$ but less than $50$ (the interval of all real numbers from $0$ to $50$, excluding $0$ and $50$:

$$ab \in \{x \in \mathbb{R}: 0 < x < 50\}$$

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By "less than less than", I meant, "shouldn't a < the largest number preceding 50, instead of 50 itself?" –  Tim Dec 22 '12 at 2:53
    
But if 10b < 50, and ab < 10b, then isn't "50" an inexact upper bound? Isn't the actual upper bound less than 50? –  Tim Dec 22 '12 at 3:00
    
$50$ is the least upper bound in this case, even given the fact that $ab < 50$. That is not to say $ab = 50$. –  amWhy Dec 22 '12 at 3:01
    
but wouldn't 50 not satisfy the inequality x < 50? –  Tim Dec 22 '12 at 3:03
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Tim: Yes, you're correct. $50$ is an upper bound because there is no largest (real) number less than 50. But $50$ is more than an upper bound of $ab$: it is the least upper bound of $ab$. –  amWhy Dec 22 '12 at 3:16
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The product $ab$ is less than 50, but it's not "less than less than 50", because it can be as close to 50 as you want to make it: for any number $x$ at all, if $x<50$, then there is some $a$ and $b$ with $x < ab < 50$.

For example, say $x = 50 - \frac1{1,000}$. But then with $a = 5 - \frac1{1,000,000}$ and $b = 10-\frac1{1,000,000}$, you get $x < ab < 50$: $x = 49.999$, but $ab = 49.999985000001$, which is even closer to 50 than $x$ is. And you can do this for any $x$ at all, no matter how close to 50 it is.

There is no "largest number preceding 50". For any $x < 50$, the number $25 + \frac x2$ is bigger than $x$ and still smaller than 50.

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But if 10b < 50, then isn't the upper bound of ab less than 50, and not 50 itself? –  Tim Dec 22 '12 at 3:01
    
No; suppose you have some number $x$ which you think might be an upper bound for $ab$, but with $x<50$. It's always possible to find $a$ and $b$ so that $ab$ will be closer to 50 than $x$ is, with $x < ab < 50$, which shows that $x$ isn't in fact an upper bound for $ab$. –  MJD Dec 22 '12 at 16:06
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The basic definition of "<" for real numbers is that $u < v$ means that there is a positive real number $w$ such that $u + w = v$.

If $0 < a < 5$, there is a positive number $p < 5$ such that $a+p = 5$. Similarly, if $0 < b < 10$, there is a positive number $q < 10$ such that $b+q = 10$.

So $a b = (5-p)(10-q) = 50 - 10p -5q + pq $. Since $p < 5$ and $q < 10$, both $10p$ and $5q$ are greater than $pq$, so $10p+5q-pq > 0$, so $ab < 50$.

This clearly generalizes.

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