# GRE physics question need explained

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1. Oct 27, 2012

### Fellowroot

1. The problem statement, all variables and given/known data

See image.

2. Relevant equations

3. The attempt at a solution

Here is the solution provided and the answer is B, but I disagree with this explanation.

I agree that g(x)=1 and g(x)=e^(0)=1 makes a continuous function, however I don't see how one would ever be able to use x=0 in the function g(x)=e^x since only x values which are irrational can only be used in g(x)=e^x and x=0 is not an irrational number.

2. Oct 27, 2012

### HallsofIvy

I don't understand what is meant by "this type of function" nor what is meant by "their intersection". The intersection of what sets? On the other hand, I don't know what you mean by "g(x)= 1 and g(x)= e^(0)= 1"!

What we can say is that, because for every real number, there exist a sequence of irrational numbers converging to it and a sequence of rational numbers converging to it, this function can be continuous only at x= a where $\lim_{i\to a}e^x= \lim_{r\to a} 1$ where i is irrational and r is rational. Then, because ex itself is continuous, that reduce to saying that ex= 1 which is only true for x= 0.

3. Oct 27, 2012

### Ray Vickson

The function g(x) is continuous at x = 0 because limx→0g(x) = g(0). Obviously, when x → 0 through rational x-values we just have g = 1 for all such x. But, because exp(x) is a continuous function on ℝ, we have exp(x) → 1 as x → 0 through irrational values (or through any values). We don't need to use exp(0) here!

RGV

4. Oct 27, 2012

### Fellowroot

Sorry, I meant to say that in the piecewise function the top function

g(x)=1, can only be used if the x-value is rational. Zero is rational therefore g(0)=1.

The bottom function in the piecewise is

g(x)=e^x, and can only be used when the x-value is irrational.

In the solution they plug in x=0 for both functions and say that they are continuous at x=0, what I don't understand is that you can't even plug in x=0 into g(x)=e^x since zero is not an irrational number.

Also how can something be continuous at a single point (0,1)? I would only think that you would need to have at least an interval of x-values for something to be continuous.

Just wondering. If you have a function which is a straight vertical line, would that function be considered to be continuous?

5. Oct 27, 2012

### Ray Vickson

Please tell me what part of the following you do not understand: (i) limx→0g(x) = 1. (ii) g(0) = 1. Nobody is computing exp(0).

RGV

6. Oct 27, 2012

### Fellowroot

Okay, so all you really do is just take the limit as x -> 0 of g(x) and since you can't use g(x)=e^x you use g(x)=1. So I understand that.

What I still don't understand is...

We know that the function g(x)=1 when x is rational is just a bunch of discontinuous points.

In advanced calculus we were shown that if you take a neighborhood at any given x-value in R it will always contain rational and irrational numbers.

So I would only think that if I were to see the graph of g(x)=1 when x is rational it would be a straight line with many points, but it could never be continuous because for every point there is an irrational number next to it so that you could never have any type of interval an order to call it continuous.

It seems like all we are really doing is taking the limit of a single point (0,1), I didn't think you could do that.

Last edited: Oct 27, 2012
7. Oct 27, 2012

### haruspex

I assume you meant when you can't use g(x)=e^x, because x is rational, you use g(x)=1, or maybe since you can't use g(0)=e^0 you use g(0)=1
Need to be careful about the domain of the function. If you had a function g(x)=1 for which the domain is only the rationals it would be continuous according to the definition. Here, g is defined on all reals; what's confusing you is that its definition depends on whether its argument is rational.
There's nothing in the definition of continuity that requires it be over an interval. It is defined for each point separately. That's what this example is showing you: a function can be continuous at isolated points.

8. Oct 27, 2012