Simple technical question.

[ughaibu]

If I have some fraction, a/b, and an inequality stating that this is bigger than x and smaller than 1/x, and if in my base case a/b=1 and both x and 1/x approach 1 as cases become higher from the base case, do I have sufficient to state that by induction a/b=1 for all cases? The complication being that I arrived at this inequality by assuming that a/b<1, in a word, do I have enough here to say that my assumption that a/b<1 is false?

[lawcat]

Interesting. I dont think you do.

 

Solve the general equation first.

 

a/b > x

a/b < 1/x

__

 

a/b > x

-a/b > -1/x

---

sum it up

0 > x - 1/x

or

0 > (x^2 -1)/x

Since the ratio on the right side is less then zero, then it is negative.

So either

x^2 -1>0 (positive) and x < 0 (negative) in which case there is no solution since x can not be both greater than 1 and less than 0. (Draw a line diagram with arrows pointing in the positive direction from 1 and negative from 0, x can not be in both ranges at same time)

OR

x^2 -1<0 (negative) and x > 0 (positive) in which case x e (0,1).

 

So x is always in the range from 0 to 1, but never 0 or 1. (the constraint is the sign <)

 

Now, since x can be in the range from 0 to 1,

then, as x approaches 0, 1/x approaches infinity, and a/b being in between can be any value between x and infinity including less than 1.

Similarly, if x approaches 1, 1/x approaches 1, and a/b being in between can be any value between x and 1/x including just below and above 1, because x can never be 1 from the general solution x e (0,1). Therefore, being that a/b > x, and x can never be 1, then there is posibility of a/b being less than 1 and satisfying the equation.

 

You have enough information to say: a/b < 0 is false. because x is 0-1, and a/b > x, then a/b > 0. So a/b < 0 is false.

[ughaibu]

I see. Thank you for that, I'll think about it some more.

[lawcat]

Yes. think about it. imagine X as a red dot moving between 0 and 1 but never touching 0 or 1. Imagine 1/X as green dot moving from 1 to infinity but never touching 1. a/b is anything between the red and the green dot. but because the red dot never touches 1, then a/b can always be less then 1 as long as it is greater than x. But a/b can never be less then zero because the red dot never touches the zero. (You need to draw a line diagram. Picture is worth a thousand words)

 

EDIT:

 

For a special case: lim x as x approaches 1,

 

Since by definition lim x as x approaches 1 = 1

 

and since a/b > x, then a/b > lim x; a/b > 1; For this special case, by definition, a/b < 1 is false.

[Ben]

If I have some fraction, a/b, and an inequality stating that this is bigger than x and smaller than 1/x, and if in my base case a/b=1 and both x and 1/x approach 1 as cases become higher from the base case, do I have sufficient to state that by induction a/b=1 for all cases? The complication being that I arrived at this inequality by assuming that a/b<1, in a word, do I have enough here to say that my assumption that a/b<1 is false?

Why do you think induction is the right approach here? I don't, in fact I think it is likely to end in tears!

 

Is this homework? If so, as seems likely, let me just point you towards the Squeeze Theorem for sequences. Of course you may need to prove this theorem (not hard, I think), but then your own required proof follows easily.

 

If you have difficulty understanding/proving the theorem, get back to us. But, if it is homework, you should show your own working first

[ughaibu]

Why do you think induction is the right approach here?

Dont worry about it, I suspected that I didn't have enough and Lawcat convinced me. I think I've found a different way to do it.

[CraigD]

If a, b, and x are integers, I believe you can prove

[math]x < \frac{a}{b} < \frac1{x}[/math]

has no solution.

 

If x is rational, it’s trivial to prove by example it does for cases where [imath]\frac{a}{b} \not= 1[/imath]:

 

let [imath]x=\frac12, a=3, b=2[/imath]

 

[math]\frac12 < \frac32 < \frac{2}{1}[/math]

 

This seems so obvious, I suspect I may be misunderstanding the question – or else, lots of people are making a more complicated problem of it than it is. :confused: :shrug:

[ughaibu]

This seems so obvious, I suspect I may be misunderstanding the question

Sure, as a general statement it's obviously false. The situation here is that a/b=1 by examination for the base case, and both the larger and smaller terms tend to 1 in higher cases. However, the inequality was generated by assuming that a/b isn't equal to 1.

[Ben]

If a, b, and x are integers, I believe you can prove

[math]x < \frac{a}{b} < \frac1{x}[/math] has no solution.

Are you sure? What if [math]x,\,\,a < 0[/math]?

This seems so obvious, I suspect I may be misunderstanding the

question

Or maybe I am. ughaibu: You really should try to clarify what you are trying to do: what do you mean by saying that "the inequality was generated by assuming that a/b < 1"?

 

For the record, I read the OP as "if, for some [math]x \in \mathbb{R}[/math] with [imath]\frac{a}{b} \in (x,\frac{1}{x})[/imath], prove that, as [math]x \to 1[/math], there always exists some [imath]\frac{a}{b} \in (x, \frac{1}{x})\Rightarrow \frac{a}{b} = 1[/imath] ".

 

Suppose that [math]x > 0[/math]. By the implication of the OP, we will have that [imath]x < \frac{1}{x} \Rightarrow x < 1[/imath].

 

By the OP we also have that [imath]\frac{a}{b} \in (x,\frac{1} {x})[/imath]. Now every real subset has a greatest lower bound and a least upper bound. So suppose that, for the set [imath](x,\frac{1}{x})[/imath] these are [imath]x,\,\,\frac{1}{x}[/imath] respectively.

 

Now by the postulates for the real numbers, for any real [imath]x,\,\,\frac{1} {x}[/imath] there is a rational number between them; this implies that, for any rational number, there is a real number that is less, and a real number that is greater.

 

So I have that [imath]x < 1< \frac{1}{x}[/imath] and [imath]x<\frac{a}{b} < \frac{1}{x}[/imath] for any [imath]\frac{a}{b} \in \mathbb{R}[/imath] and some [math]x \in \mathbb{R}[/math], so that by the property of greatest and least bounds, this implies that [imath]1 \in (x,\frac{1}{x})[/imath], so in the limit that [math]x \to 1[/math] obviously [imath]\frac{1}{x} \to 1 \Rightarrow (x,\frac{1}{x}) \to \{1\}[/imath].

 

But [imath]\frac{a}{b} \in \{1\}[/imath] as [imath](x,\frac{1}{x}) \to \{1\}[/imath], so we must have that [imath]\frac{a}{b} = 1[/imath] at this limit.

 

Of course, for full generality, we need to say what we mean by [math]x \to 1[/math], that is, define a sequence [math]x_n[/math] such that, for ANY [math]x \in \mathbb{R}[/math] we will have that [math]\lim_{n \to \infty} x_n = 1[/math] AND [math]\lim_{n \to \infty}\frac{1}{x_n} =1[/math], but you get the general idea, I trust

[ughaibu]

ughaibu: You really should try to clarify what you are trying to do

Yes, I know, and thanks for your patient and thoughtful input.

what do you mean by saying that "the inequality was generated by assuming that a/b < 1"?

Without worrying about this, consider the following alternative scenario:

if I have some formula which generates x for all n such that n>1 and is an integer, if I can prove that for the base case of n=2 then x=1, is the establishment of (n/n-1)>x>(n-1/n) sufficient to conclude that x=1 for all cases?

[Ben]

if I have some formula which generates x for all n such that n>1 and is an integer, if I can prove that for the base case of n=2 then x=1, is the establishment of (n/n-1)>x>(n-1/n) sufficient to conclude that x=1 for all cases?

Sorry, I have no idea what you are talking about.

 

What is your "formula"? What do you mean by " base case" in this context? What does "establishment" mean in this context?

 

Look, my friend, while walking the dogs, I spent an hour thinking out my last post in this thread. It wasn't totally wasted time, but then I spent a further 20 min formatting my thoughts.

 

And the above is your response to my thinking? Did you actually read and THINK about my post? It seems not....

 

THAT convinces me I have wasted my time

[ughaibu]

THAT convinces me I have wasted my time

Well, as you apparently didn't understand anything I wrote, my time too has been wasted. Kindly stay out of this thread.