What is the Solution to This Inequality Problem?

[Jin314159]

Hi all,

I'm a math major at Berkeley with a 3.3 GPA average. I've taken calculus, linear and abstract algebra, probability and real analysis. The other day I was applying for a job as a pre-Calculus teaching assistant and my employer gave me this problem with a 2 minute time limit.

Simplify this expression:

$$\frac{\frac{1}{x}+1}{\frac{1}{x}-1}<2$$


I couldn't get it.

 

[matt grime]

The LHS is the equal to (1+x)/(1-x) so x is not equal to 1, for x>1, or x<1, clearing denominators correctly will give a simple solution, when such is possbible.

 

[HallsofIvy]

First, it's not, strictly speaking, an "expression", it's an inequality. The left hand side,
$\frac{\frac{1}{x}+1}{\frac{1}{x}-1}$ is an expression. As matt grime said, the obvious thing to do is multiply both numerator and denominator by x, resulting in $\frac{1+x}{1-x}$. If the problem was really to "simplify this expression", that or $\frac{1+x}{1-x}< 2$ is the best you can do (and should take about 10 seconds).

If the problem were, instead, to solve the inequality, then:
a) if 1-x>= 0 (that is, x<= 1), this is the same as 1+x< 2(1-x)= 2- 2x so 3x< 1 or x< 1/3. Since any number < 1/3 IS < 1, one answer is x< 1/3.
b) if 1-x< 0 (that is, x> 1), this is the same as 1+x> 2(1-x)= 2- 2x so 3x> 1 or x> 1/3.
But that only works for x> 1 so another answer is x> 1. If x< 1/3 OR x> 1, the inequality is true.

If you really are "a math major at Berkeley with a 3.3 GPA average" and could not do this problem even after taking a great deal of time, that's should be very embarassing for you (and Berkeley).

 

[Jin314159]

If you really are "a math major at Berkeley with a 3.3 GPA average" and could not do this problem even after taking a great deal of time, that's should be very embarassing for you (and Berkeley).

Much thanks for the gracious words.

Actually, I got the same solution as you did. But we're both wrong.

You assumed the expressions $$\frac{\frac{1}{x}+1}{\frac{1}{x}-1}$$ and $$\frac{x+1}{x-1}$$ are equivalent. But they're not. The former is not defined when x=0 (which is part of your solution) while the later is.

By the way, we all have our strong and weak points in math. I thought your derision was of poor taste, especially considering you are a "mentor." Please practice a bit more tact in your posts. Thank you.

 

[Jin314159]

Maybe I should have been more direct in my original posts. I might have given the impression that I had no idea what was going on, hence giving HallsOfIvy reason to doubt my claim that I was a Berkeley math student.

All I want is to confirm that the set of values which satisfies the inequality $$\frac{\frac{1}{x}+1}{\frac{1}{x}-1}<2$$

is

$$(-\infty,0) \cup(0,1/3)\cup(1,\infty)$$

 

[uart]

Hey I got x<1/3 and x>1 as the regions as well. I would likely have also overlooked the problem at x=0 if I was trying to solve it quickly, so don't feel bad. :)

Actually I'm hopeless at doing anything quickly. :)

 

[JonF]

I may not be a math major from Berkley with a 3.3 gpa but as far as I know $$\frac{\frac{1}{x}+1}{\frac{1}{x}-1}$$ and $$\frac{x+1}{1-x}$$
are equivalent expressions.

 

[fourier jr]

hey in Canada 3.3 isn't really good.
F=0, D=1, C=2, ... , A-=7, A=8, A+=9.
3.3 would be C+/B-. Maybe that's the system Berkeley uses.

& given that Berkeley only requires someone to solve problems in a given length of time IS really embarrassing. What does that have to do with being a pre-calc TA?

 

[AKG]

JonF, it is clearly explained in post 4 that the expressions are not equivalent, and why. Whenever you do something like divide top and bottom by x, you should always note the exception when x=0.

For x = 0, you cannot make the second expression from the first, and in that case, the original expression is undefined. For x != 0, you can change first to second.

 

[AKG]

hey in Canada 3.3 isn't really good.
F=0, D=1, C=2, ... , A-=7, A=8, A+=9.
3.3 would be C+/B-. Maybe that's the system Berkeley uses.

The system used in almost all places makes a 3.3 something like 75-80 (correct me if I'm wrong). I think 3.7 is something like 85, 4.0 is 90+, so a 3.3. is not great, but it's not bad. I've never heard a system like yours before, and I'm in Canada (Toronto). I think your system might be unique to your university. Do you know of ther places that use that system? And are you sure you're talking about GPA (grade point average)?

 

[e(ho0n3]

I'm a math major at Berkeley with a 3.3 GPA average.

I wish I was in Berkeley.

All I want is to confirm that the set of values which satisfies the inequality $$\frac{\frac{1}{x}+1}{\frac{1}{x}-1}<2$$

is

$$(-\infty,0) \cup(0,1/3)\cup(1,\infty)$$

Looks right to me, although your wording of the problem (see first post) is confusing.

I've been put on the spot several times before by people who have high expectations of me, and every time it happens I draw a blank. It's so wierd. A colleague of mine (who has a maths. Ph.d. from MIT by the way) also had a similar experience. Anyways, it happens.

 

[eJavier]

Much thanks for the gracious words.

Actually, I got the same solution as you did. But we're both wrong.

You assumed the expressions $$\frac{\frac{1}{x}+1}{\frac{1}{x}-1}$$ and $$\frac{x+1}{x-1}$$ are equivalent. But they're not. The former is not defined when x=0 (which is part of your solution) while the later is.

By the way, we all have our strong and weak points in math. I thought your derision was of poor taste, especially considering you are a "mentor." Please practice a bit more tact in your posts. Thank you.

But you can always define $$\frac{\frac{1}{x}+1}{\frac{1}{x}-1} = 1$$ at 0 by means of continuity.

 

[AKG]

But you can always define $$\frac{\frac{1}{x}+1}{\frac{1}{x}-1} = 1$$ at 0 by means of continuity.

First of all, that doesn't make any sense. Second of all, regardless of what you might mean by "by means of continuity," you can't define the expression at x=0, because that would mean that 1/0 would be defined, which it's not.

 

[eJavier]

First of all, that doesn't make any sense. Second of all, regardless of what you might mean by "by means of continuity," you can't define the expression at x=0, because that would mean that 1/0 would be defined, which it's not.

Nonsense.
It's customary to extend functions by means of continuity to points where it isn't defined.
For instance, $$x \sin \frac{1}{x}}$$ is defined as equals to 0 at x=0 by continuity. That doesn't mean that I'm defining 1/x in the set of Real numbers, I'm defining the whole function.

 

[futb0l]

so, what's the final answer??

i do think that the answer u got is right.
$$(-\infty,0) \cup(0,1/3)\cup(1,\infty)$$

 

[matt grime]

AKG, there are such things as removable singularities.

 

[fourier jr]

The system used in almost all places makes a 3.3 something like 75-80 (correct me if I'm wrong). I think 3.7 is something like 85, 4.0 is 90+, so a 3.3. is not great, but it's not bad. I've never heard a system like yours before, and I'm in Canada (Toronto). I think your system might be unique to your university. Do you know of ther places that use that system? And are you sure you're talking about GPA (grade point average)?

I think U of T is the only place I've heard of that uses the 4-point system. Every other school that I know of uses the 9-point one.
http://web.uvic.ca/reco/legend/legend.html

I wish I was in Berkeley.

bah Berkeley is a dump. I want to go to a real math school, U of Waterloo. They've got over 5000 math students (1/4 of all at the school), and over 6000 computers. It's the world's largest educational centre for math & comp sci. They've even got a whole faculty for math rather than one dept in the science faculty or whatever.

 

[AKG]

I think U of T is the only place I've heard of that uses the 4-point system. Every other school that I know of uses the 9-point one.

Really. I think the States use the 4-point system everywhere.

AKG, there are such things as removable singularities.

I don't understand that concept. sinx/x is not defined at x, but we can "define" it to be 1 and still have continuity and differentiability. But then, isn't it a different function? I mean, aren't:

f(x) = sinx/x (where the domain is implied to be the R\{0})

and

g(x) = sinx/x for x != 0, 1 for x = 0

two different functions, with two different domains? I mean, by default, isn't the expression that this thread about undefined at x=0 (and 1)? Therefore, how does it make sense to say what the expression is less than or greater than at x=0? We can make sense of it with a similar expression, but we're not asked about a similar expression, we're asked about the given expression, are we not? Or is it that by default removable singularities are always removed (and this is just something they don't tell you till your 3rd or 4th calculus course)?

 

[Jin314159]

A 3.3 GPA for a Berkeley math student is somewhat impressive (considering the average GPA for math is 2.7). My point was this was a tricky problem, even for a somewhat competent college math student such as myself. In fact, HallsofIvy made the same mistake and I think he's a college professor.

 

[Jin314159]

given that Berkeley only requires someone to solve problems in a given length of time IS really embarrassing. What does that have to do with being a pre-calc TA?

The time factor is important. When a TA is asked a problem by a student, he can't afford to stand there thinking about it all day. He has to be quick.

bah Berkeley is a dump. I want to go to a real math school, U of Waterloo. They've got over 5000 math students (1/4 of all at the school), and over 6000 computers. It's the world's largest educational centre for math & comp sci. They've even got a whole faculty for math rather than one dept in the science faculty or whatever.

Heh, well the last time I checked the US News rankings, Berkeley's math department is third in the nation (only Stanford and MIT beat it). But of course, we can argue over the legitimacy of these rankings.

As for the U of Waterloo, it's rather impressive that they manage to accommodate for so many math students but that doesn't necessarily mean it's a good school, does it?

 

[tomkeus]

There is one general rule which I consider very covenient: do not multuply inequalites with variable sign expressions (like x, or x-1, or sinx etc..), because multiplying with expression which is less than 0 will result in sign change. Rather move all expressions to one side of inequality and leave only zero on the right side. Then you can freely transform expression on the left side without having to vorry of signs.

 

[fourier jr]

Really. I think the States use the 4-point system everywhere.

um, I meant in Canada, & I thought U of T just changed back to the 9-point system

As for the U of Waterloo, it's rather impressive that they manage to accommodate for so many math students but that doesn't necessarily mean it's a good school, does it?

Well like you said, everything revolves around math there since such a huge number of students are in a math program. So yeah, someone can learn pretty much anything @ Waterloo since they're so much more serious than other schools. Maybe you refer to the practice of poaching faculty & students from elsewhere that elitist US schools (esp. Harvard) are notorious for, and makes them look good. That doesn't say anything about how good a school is. When you do to Harvard, Princeton, etc you get a piece of shytte "education" but you can sure get some good connections at schools like those. In math here, if someone tries to brag that they went to school at Harvard, etc they could get the (rude) response "why only Harvard, couldn't get into Waterloo?", and with good reason too. lol.

 

[master_coda]

We don't need an argument about which universities suck and which ones are great. Particularly if it's just going to be based on tired stereotypes and rankings spewed out by the media.

 

[e(ho0n3]

bah Berkeley is a dump. I want to go to a real math school, U of Waterloo. They've got over 5000 math students (1/4 of all at the school), and over 6000 computers. It's the world's largest educational centre for math & comp sci. They've even got a whole faculty for math rather than one dept in the science faculty or whatever.

Unfortunately, Berkeley is one of the few places where I can do graduate research in quantum computing. It's not like I have tons of options here.

 

[master_coda]

Unfortunately, Berkeley is one of the few places where I can do graduate research in quantum computing. It's not like I have tons of options here.

Actually, quantum computing is one of the areas you can do graduate research in here at Waterloo. It's rather funny to me that you would mention it, since it's one of the programs whose star is on the rise.

 

[KLscilevothma]

so, what's the final answer??

i do think that the answer u got is right.
$$(-\infty,0) \cup(0,1/3)\cup(1,\infty)$$

But the question asks us to "simplify" the expression, not to solve the inequality.

So I think the answer should be

$$\frac{1}{x^2} - \frac{4}{x} + 3 > 0$$
or
$$( \frac{1}{x} - 3)( \frac{1}{x} - 1) > 0$$

 

[AKG]

KLscilevothma

It's not even an expression, so "simplifying" the left hand side is answering the expressed question just as much as solving the inequality is. On top of that, I think the guy who posted this also clarified that he was meant to solve the equality, not simplify the expression to the left of the ">" sign.

 

[e(ho0n3]

Actually, quantum computing is one of the areas you can do graduate research in here at Waterloo. It's rather funny to me that you would mention it, since it's one of the programs whose star is on the rise.

I checked out the UW website and you're right. I'll add this place to my list. Thanks.

 

[ExecNight]

ok i am an high school graduate tho i got my solution :rofl:


The question equation equals to 1+x/1-x<2,


which means;

x can not be 1
x can not be -1

and everything else is ok that's all have fun guys..

Oh and x can not be 0

 

[e(ho0n3]

ok i am an high school graduate tho i got my solution :rofl:


The question equation equals to 1+x/1-x<2,


which means;

x can not be 1
x can not be -1

and everything else is ok that's all have fun guys..

Oh and x can not be 0

According to your simplification, it is perfectly valid for x to be -1 and 0. This is bogus. This question has already been answered (see previous posts).

 

[ExecNight]

can't be 1/3 can be -1 yup..

Still can't be 0 :rofl:


Try using 0 in the equation without simplifying if you wonder why..