Math Is Fun Forum

At first let's put it clear: I totally agree with both of you "it means what we define it to mean", but the definition has already been made. However, about the suspicions which you produced in your arguments, the mathematically consistent treatments for both arguments would be as stated below:
Infinity is not a number, infinity is a special concept, even if we use the symbol "∞" to represent it we don't mean it is an ordinary number, since it has properties which ordinary numbers are lack of. For this reason mathematicians created a coherent conceptual system to deal with it. So the first argument (which anonimnystefy has presented) should, according to the mathematical practice, go as follow:

So we have to use the concept of "limit" when we deal with this situation. the other argument (which is persented by bob and which, in fact, is another aspect of the first argument) treat "∞" (the number of 9s) as an ordinary number which we can define "precisely", but infinity is not an ordinary number and can not be reached, so for any natural number n threre will be a number between 1 and

also there will be a difference between them. However, as n approaches ∞ (but could not reach it) the difference between these two numbers shrinks more and more (but will not be Zero) and since the limit of a function at "∞" is defined as the number the function approaches more and more, no matter how large the number is, as the independent variable appoaches (but not reach) ∞, therefore "by the definition" the difference between them will equal "zero" as a "limit".

Note: Someone may wonders "why we have to distinguish between these two concepts?, that is why:

The reason is simply because if we did not, we would be misleaded!! To see why, let's follow the reasoning of those who prefer to adopt the definition "

" on the ground of that the more we add digits to "

"the more we get close to "1" and since "∞" is so large then the difference will be Zero at the end. If we accept this argument we must accept also that

for the same reason!! However, if we distinguish between them and adopt the "limit" concept we will reach the correct answer because the definition of the "limit" at "∞" states that the limit exists if for any ε>0 there is N>0 such that:

Hence applying this condition to these two cases we will have respectively:

And

Which are the correct results. These results hold since the infinite seires in both cases are geometric with general ratio less then one.

Q.E.F

When a celestial object (e.g. Comet, asteroid, planet...) orbet another object, the orbet could be one of the three types of conic sections (circle, ellipse or parabola it could also be a hyperbola in some hypothetical universes where the gravity repulsing instead of attracting) in the case of comets the orbets are parabola and the longest distance is "∞" (the comet does not return again). This fact is reflected in the equation that relates time with the distance from the center of jupiter's moon which you have provided, so the distance from the moon at the turnig point is not the absolute longest distance from the center of jupiter's moon, It is a local longest distance between the successive nearest approach of the comet.

It turns out that the third possibilty could be solved just in three weighing attempts, so the test ,in its three attempts and restrictions, is in fact conclusive!!!

In the third possibility, when we weigh the subgruops of group A and the subgroups of group B (using the first two attempts). We will have, in group A, one of the scale's arm be lifted and the same for group B, but we are using a scale!!, so we can know the following results:

the difference between the lighter and the heavier arms in in the first weighing attempt (weighing the subgroups of A), call it "d1", is equal to or more or less then the same difference in the second attempt (when weighing group B), call it "d2".

Now if they are equal then the two rotten apples are equal in weight to two good apples. If they are not, we cann't know which side of the balance contains the rotten apple in both weighing attempts of each group, so we remove the lighter subgroup of B from the arm (in the second attempt) and put on it the lighter side of the subgroups of A (using the third and last attempt), If the two arms balanced then we know ,for sure, that group A contains the heavier rotten apple and group B have the lighter one, So we have:

1) d1>d2 (in this case the two rotten apples weigh more then two normal ones)
2) d1<d2 ( in this case the two rotten apples weigh less then two normal)

Finally, if in the theird attempt, the scale which contains the heavier subgroup of group B remains unlifted, then we know, for sure, that group B contains the heavier rotten apple, and we have the results:

1) d2>d1 (the two rotten apples are more in weight then two normal ones)
2) d2<d1 (the two rotten are less in weight then two normal).

Note: the third possible case (the heavier subgroup of group B in the third attempt is lifted) can not happen, since the heavier rotten apple must be either in group "A" or in group "B".

Q.E.F

You think that "because we don't know which subgroup will contain the good apples and which subgroup will contain the rotten apple" there is a possibility that the good apples will be mixed with a subgroup which contains a rotten apple, so we have to find a way to avoid this situation when cross mixing!! In fact, this is impossible to happen!! In cross mixing the subgroups to form two new collections (each consists of four apples) the rotten apples will "spontaneously" be both in one collection and the other collection will contain only good apples "in all cases" (this is the main point and the tricky one of the puzzle which you always miss).

You don't have to know which side of the scale (in both group) contains the rotten apple!!! Why? To explain this situation suppose "without loss of generality" that group A contains the heavier rotten apple in subgroup A1, then when you weigh the subgroups of this group (group A) the arm of the scale containing the subgroup which include the heavier rotten apple (it could be A1 or A2) will lift the other arm which containing the subgroup of the normal apples and when you do the same with group B which contains the lighter rotten apple the subgroup which contains that apple (it could be B1 or B2 no matter) will be lifted, so when you cross mix the subgroups of each group,  the lifting side will match the lifted side and the rotten apples will be in the same collections.

Note: it is obvious that the thing which made you miss the point is confusing the heavier and the lighter sides with the notations of the subgrouops. I think you asked yourself the following question:
"If I donate a particular subgroup of group A "A1" and the other subgroup "A2" and do the same for group B then after weighing, the cross mix will be as follow :(A1+B2) and (A2+B1) but wait a minute!! How the hell could I know that B2 contains the lighter rotten apple or the heavier one and the same for B1?!!!" This is the mistake which you have made. We did not talk about the notations we talked about the heavier and the lighter sides!! So if group A contais  the heavier side it could be A1 or A2 no matter and the same for the lighter side it could be B1 or B2 so focus on the heavier and the lighter side not on the notations. I think it is very clear now (please let me know in any case).

Divide the 8 apples into two groups each consists of four apples (call them group A and group B) and divide each goup into subgroup each contains two apples. Now put on each the scale's arm one subgroup of group A , and do the same for group B. So two of the weighing attempts will be used and you will get one of the following possibilities:

1) The subgroups of each group are balanced
2) One group has its subgroups balanced and the othe has not.
3) Both groups have their subgroups unbalanced.

In the first possibility you will know of course that the two rotten apples are equal to two good ones. If you get the second possibility you will know that the group which has its subgroups balanced are all consists of good apples, so put each of its subgroup together and each of the subgroup of the other group together and put each collection in on arm of the scale (using the third attempt) then you will have one of the following results:

1) both of them palanced (in this case the two rotten apples equal in weight to a two good apples)
2) the collection of the group of the unbalanced subgroup weighs more then the collection of the other group (in this case the  two rotten apples weigh more then the two good apples)
3) the collection of the group of the balanced subgroup weighs more then the collection of the group of the unbalanced subgroup (in this case the two rotten apples weighs less two good apples)

Finally, if you get the third possibility, put the subgroup of the group A which weighs less with the subgroup of group B which weighs more and vice versa, and put on each arm of the scale one collection (using the third attempt), then you will get the same results that we get in the second prossibility hence you know either the two rotten apples are equal to or more or less in weight then two good apple.

Q.E.F

In logic there is a concept called "the universe of discourse" .For example, if someone talks about something he doesn't have to provide all the details, those details can be infered from the cotext. If another person bring irrelevant ideas to the topic which should be rolled out by those implicit details they would claim that he diverges from the universe of discourse and consider him as a "sophist", so if the argument which I have provided gets out the universe of discourse of the problem then someone can claim that I have produced a fallacy. However, if we suppose that the implicit details imply that the bases of the numbers in the list are decimal then the problem will be "impossible", but the person who asked the question claimed that it has a solution, this clue actually is a detail which included in the universe of discourse of the problem. In other ways it says that the numbers in the list are not in decimal base because ,if they are, the problem would be "impossible". Therefore we are in the universe of discourse and did not commit a fallacy. Now there are two possible solutions:
1) To change the base of each number in the list as we would like (your proposed solution)
2) To change all the numbers in the list to the same base.
If we choose the first way, we can do that for any number in the list, so that will send us back to the first square, but if we choose the second way, there will be just one number which satisfies the conditions (which is "71"), so we could reach a solution to the problem.

Q.E.D

You can define a piecewise fuction to be consisted of as many pieces as you would like, and the number of pieces alog with their shapes will determine the domain and the range of the function. The question which you provide don't explicitly contain those information, but according to the description of the piecewise fuction in the question, it is obvious (if there aren't missing information) that the function consists of two pieces one of them is an absolute value piece with the independent variable "x" extending from -∞ to 2.5 and the other piece is a graph with vertex at height 12 from the x-axis in the first quarter covering the interval from 2.5 to 5. If this description is the correct one, then the domaim and the range are (-∞,5] and [-3.25,∞) respectevily. Otherwise, the question should include more information.

Q.E.F

They are very simlpe (straightforward applications of Algebra) the first and second problems can be solved as provided by "anna_99" and you have mentioned above that you have solved the fourth problem. About the solution of the extra problem ,which you displayed alon, your solution is correct, and you can factorize the polynomial as follow:

or

Now you made a little mistake in your reasoning about the negative value of the velocity since you can ,in fact, use the negative or the positive values!! Because velocity is a vector so it has a magnitude and direction (not like a scalar quantity ,as the distance, which has a magnitude only) so in general the answer depends on the directions which you take as positive. But in this case they ask only for the "rate of change" hence you can use any value (negative or positive) for calculation and then drop the sign (you will get the same answers).
To solve problem number "3" let

to be the upstream travel while

denotes the downstream travel, so from the provided data we have:

But we know also that:

Now since the rquired value is the speed (the magnitude) of the upstream travel and the problem discribes travel in both directions as if they were parallel, we will take the positive value and discard the negative (the problem has been transulated into the above equations with this idea in mind). Therefore:

or

Hence the required answer is

mi/h

Q.E.D

Good!

Alternative solution:

If we let θ to be the angle which the bug made with the vetical when it lost contact with the surface. Then the Work done by the weight of the bug will be:

This Work will be converted into kinetic energy hence:

But at the moment when the bug lost contact with the surface the vetical component of the weight was equal to the centripetal force preduced by the circular motiom on the surface. Therefore:

From the previous two arguments we have:

Wence:

Q.E.D