Squaring infinity? Help

[Loki123]

The two pictures are how I solved the problem (which is written on the first picture right under 39 3.). I got the correct answer, however, I got very confused when I had to square infinity and even more confused when i had to square negative infinity and solve logarithms for it. I would have gotten the correct answer anyway, since the condition is =>1, but I want to know whether I made a mistake or not. The confusing parts are outlined blue. Thank you in advance. (p.s. Please don't recommend not using t because that's how we need to solve it)

 

[Dr.Peterson]

Infinity is not a number, so you can't square it. For that matter, infinity is not in the domains you are talking about, so you never really have to. If you were in calculus, or are in some pre-calculus courses, we could talk about limits, instead.

But what you can do is just to think about what happens if t is a very large number (or a very large negative number). You probably did that, and did it correctly.

I'd like to see an example of how your book deals with such problems. I suspect they don't use endpoints of intervals as you are doing. Since you say you have to do everything the way you are taught, we can't help effectively without seeing what you are taught.

 

[JeffM]

With utmost respect for Dr. Peterson, whether infinity is a "number" depends on how you define "number. Mathematicians as distinguished as Hilbert and Cantor would not agree with Dr. Peterson's definition: mathematicians like Hilbert were happy to contemplate transfinite numbers. Other mathematicians, called finitists, do agree with Dr. Peterson: infinity is not a number.

One way to define infinity as a number is through the one-point compactification of the real numbers, which is an extension of the real numbers just as real numbers are an extension of the rational numbers, the rational numbers are an extension of the integers, and the integers are an extension of the natural numbers. In the case of most extensions of the definition of number, you must learn a new arithmetic to make the extension useful and consistent. In the case of the one-point compactification, you have to have rules that specify the arithmetic applicable to infinity. See

Projectively extended real line - Wikipedia

en.wikipedia.org

Under that particular set of rules, your squaring of negative infinity is not a defined operation. Even if we define infinity in a sensible way, you must have erred in some fashion.

Now where I agree completely with Dr. Peterson is this. Infinity is not a real number. Moreover, you do not need a compactification of the real numbers to solve any problem that I have ever encountered. The way we normally teach mathematics formally never treats infinity as a number. Here the only reason that you got embroiled in squaring negative infinity is that you introduced an unnecessary complication.

[math]log_3(x) - 5\sqrt{log_3(x)} + 3 >0 \text { and } t = \sqrt{log_3(x)} \implies t^2 - 5t + 3 > 0 = t^2 - 5|t| + 3.[/math]
I suspect that you were thinking along the lines that the square root function is never negative. But all of part II is simply unnecessary nonsense. t is necessarily non-negative. So part II is dealing with an impossibility that you introduced by inserting absolute value notation into your work. Not an error exactly, but an an unnecessary complication. You came up with negative infinity because you were preceding from a false hypothesis, namely that t was negative.

 

[Loki123]

Infinity is not a number, so you can't square it. For that matter, infinity is not in the domains you are talking about, so you never really have to. If you were in calculus, or are in some pre-calculus courses, we could talk about limits, instead.

But what you can do is just to think about what happens if t is a very large number (or a very large negative number). You probably did that, and did it correctly.

I'd like to see an example of how your book deals with such problems. I suspect they don't use endpoints of intervals as you are doing. Since you say you have to do everything the way you are taught, we can't help effectively without seeing what you are taught.

Well I actually came up with using end points on my own... Because I thought it made the most sense. How am I supposed to solve it?

 

[Loki123]

With utmost respect for Dr. Peterson, whether infinity is a "number" depends on how you define "number. Mathematicians as distinguished as Hilbert and Cantor would not agree with Dr. Peterson's definition: mathematicians like Hilbert were happy to contemplate transfinite numbers. Other mathematicians, called finitists, do agree with Dr. Peterson: infinity is not a number.

One way to define infinity as a number is through the one-point compactification of the real numbers, which is an extension of the real numbers just as real numbers are an extension of the rational numbers, the rational numbers are an extension of the integers, and the integers are an extension of the natural numbers. In the case of most extensions of the definition of number, you must learn a new arithmetic to make the extension useful and consistent. In the case of the one-point compactification, you have to have rules that specify the arithmetic applicable to infinity. See

Projectively extended real line - Wikipedia

en.wikipedia.org

Under that particular set of rules, your squaring of negative infinity is not a defined operation. Even if we define infinity in a sensible way, you must have erred in some fashion.

Now where I agree completely with Dr. Peterson is this. Infinity is not a real number. Moreover, you do not need a compactification of the real numbers to solve any problem that I have ever encountered. The way we normally teach mathematics formally never treats infinity as a number. Here the only reason that you got embroiled in squaring negative infinity is that you introduced an unnecessary complication.

[math]log_3(x) - 5\sqrt{log_3(x)} + 3 >0 \text { and } t = \sqrt{log_3(x)} \implies t^2 - 5t + 3 > 0 = t^2 - 5|t| + 3.[/math]
I suspect that you were thinking along the lines that the square root function is never negative. But all of part II is simply unnecessary nonsense. t is necessarily non-negative. So part II is dealing with an impossibility that you introduced by inserting absolute value notation into your work. Not an error exactly, but an an unnecessary complication. You came up with negative infinity because you were preceding from a false hypothesis, namely that t was negative.

In the solution it says to write log3(x) as t^2, however, for some reason the solution stops there and doesn't show the answer at the end. I got it from an app. I am confused as to why t can't be less than 0, is that not how solving equations with absolute value goes?
Should I not square negative infinity? but just write it as it is and solve for endpoints that are real numbers?

 

[JeffM]

In the solution it says to write log3(x) as t^2, however, for some reason the solution stops there and doesn't show the answer at the end. I got it from an app. I am confused as to why t can't be less than 0, is that not how solving equations with absolute value goes?
Should I not square negative infinity? but just write it as it is and solve for endpoints that are real numbers?

You defined [imath]t = \sqrt{log_3(x)}[/imath].

The square root function is non-negative, right? [imath]\therefore t \ge 0.[/imath]

Thus, the whole part of your analysis under the hypothesis [imath]t < 0[/imath] makes no sense.

You are of course correct in your substitution [imath]t = |t|[/imath] although that would not be true if t were negative. In general, [imath]a \ne |a|.[/imath] That is a valid step only if a is guaranteed to be non-negative. So here is your logic

[math]t \ge 0 \text { because } t = \sqrt{log_3(x)}.[/math]
[math]t = |t| \text { because } t \ge 0.[/math]
[math]\text {Given that } t < 0 \ ….[/math]
Everything from the supposition that t < 0 is founded on a falsity.

Now, it is true generally that when you are dealing with an absolute value |p|, you must take into account both a negative and a non-negative case. But you are allowed to equate a variable with its absolute value only if there is no negative case.

When I first wrote my original answer, I said you made an error when you equated t with its absolute value. I then realized and corrected my original answer because t could not be negative. So your equating t and its absolute value was not technically an error. It was an unnecessary complication that, luckily for you, wasted your time but did not ultimately lead to an incorrect answer.

To summarize, it is not generally true that [imath]t = |t|.[/imath] If it is true, [imath]t \not < 0.[/imath]

 

[JeffM]

In the solution it says to write log3(x) as t^2, however, for some reason the solution stops there and doesn't show the answer at the end. I got it from an app. I am confused as to why t can't be less than 0, is that not how solving equations with absolute value goes?
Should I not square negative infinity? but just write it as it is and solve for endpoints that are real numbers?

I recommend that you review your work if you ever find yourself having to treat infinity as a number.

First, mathematicians who are finitists will say that you have made an error.

Second, mathematicians who accept infinity as a number (but one that is not a real number) require a non-real arithmetic for non-real numbers. If you do not know the rules for that non-real arithmetic, you may make errors.

Finally, the math that you are being taught will, if followed carefully and logically, mean that you never have to treat infinity as a number.

 

[Dr.Peterson]

In the solution it says to write log3(x) as t^2, however, for some reason the solution stops there and doesn't show the answer at the end. I got it from an app.

This confuses me. You said at the start,

Please don't recommend not using t because that's how we need to solve it

which implies that you are taking a class and were taught some specific method for this sort of problem, and are required to use it. But then you say you got your solution from some app, presumably not connected to your class. Why would you think that would help you do things the way you were taught? And you also say,

Well I actually came up with using end points on my own... Because I thought it made the most sense. How am I supposed to solve it?

What are you actually being taught? It would be really helpful if we could see that. (I don't think I've ever seen problems this complicated actually taught with a detailed method.)

As to JeffM's comments about finitism, I hope you recognize that he is talking about a different kind of math than you are doing, and my statement about infinity not being a number was made within your context, doing algebra on real numbers. Since you have to do ordinary real arithmetic, you can't do arithmetic on infinity. What we need to see is how you have been taught to do any of this, which will presumably be an appropriate way to avoid infinity. If you do what you are taught, as you initially said you wanted to do, these issues won't come up.

 

[JeffM]

I suspect Dr. Peterson and I agree 100% on the following.

Infinity is not a real number.

You cannot apply real arithmetic or real algebra to infinity.

The kind of math that you are almost certainly being taught will NEVER involve you with infinity if you apply it carefully and logically. It is designed in part to avoid infinity.

It is very hard to help you learn what your course is trying to teach you in the context of one problem and no other information. Maybe you are being taught about substitution of variables. How do we know if you do not tell us?

 

[Loki123]

You defined [imath]t = \sqrt{log_3(x)}[/imath].

The square root function is non-negative, right? [imath]\therefore t \ge 0.[/imath]

Thus, the whole part of your analysis under the hypothesis [imath]t < 0[/imath] makes no sense.

You are of course correct in your substitution [imath]t = |t|[/imath] although that would not be true if t were negative. In general, [imath]a \ne |a|.[/imath] That is a valid step only if a is guaranteed to be non-negative. So here is your logic

[math]t \ge 0 \text { because } t = \sqrt{log_3(x)}.[/math]
[math]t = |t| \text { because } t \ge 0.[/math]
[math]\text {Given that } t < 0 \ ….[/math]
Everything from the supposition that t < 0 is founded on a falsity.

Now, it is true generally that when you are dealing with an absolute value |p|, you must take into account both a negative and a non-negative case. But you are allowed to equate a variable with its absolute value only if there is no negative case.

When I first wrote my original answer, I said you made an error when you equated t with its absolute value. I then realized and corrected my original answer because t could not be negative. So your equating t and its absolute value was not technically an error. It was an unnecessary complication that, luckily for you, wasted your time but did not ultimately lead to an incorrect answer.

To summarize, it is not generally true that [imath]t = |t|.[/imath] If it is true, [imath]t \not < 0.[/imath]

Oh, I didn't think of that. Thank you. I'll be more aware of such things from now on.

 

[JeffM]

Oh, I didn't think of that. Thank you. I'll be more aware of such things from now on.

There was a lot of nuance in that problem.

I am still wondering if it was trying to teach about substitution of variables.