You are not logged in.
What do you need to multiple the denominator by so that it is an integer? Multiple the numerator by the same.
For infinity discussion, I refer you to my post: https://www.mathisfunforum.com/viewtopic.php?id=31062
I agree - long division. Synthetic, while a cool idea that it works, is unnecessary, and for students with so much else to learn in the course, just a lot of clutter.
I'll add a bit more to Bob's excellent exposition.
The expression under the radical (the radicand) is called the discriminant.
Note that even before completely solving for the roots, the value (<0, =0, >0) of the discriminant tells us something about the roots and the graph.
=0: both roots are the same, because the formula has ±0. The parabola's vertex touches the x-axis there.
<0: both roots are complex; they are called complex conjugates: a+bi and a-bi. (Any two expressions x+y and x-y are called conjugates.) The parabola does not intersect the x-axis.
>0: both roots are real. The parabola has two distinct x-intersects.
You say the circle is shaded, and that you want the area of the shaded part.
So - you want the area of the circle, not the difference between the areas of the square and the circle.
Also, technically, you can calculate this only if the circle is specifically inscribed in the square, not just inside it. An infinite number of circles of various radii are inside the square.
I'm not being picky here, just using mathematical terminology correctly.
Well, I do apologize. I was expecting a thanks for pointing forum members to what I consider a most interesting field.
In short: two sets are considered to have the same size (or cardinality) if they can be put into a 1-1 relationship.
E.g., the naturals N = {1,2,3,...} would seem naively to be twice as big as the evens E = {2,4,6,...}. However, as they can be matched up 1-1: (1,2), (2,4), (3,6), and so on, they have the same size. Georg Cantor called this smallest transfinite size by the first Hebrew letter aleph, with a subscript zero, read as aleph-null.
It turns out that many infinite sets are in fact aleph-null sets.
Take the rationals Q = {p/q where p and q are integers}. Although there are an infinite number of rationals between any two integers, Q is in fact aleph-null. Here's a listing that contains all rationals (some may be duplicated): {1/1, -1/1, 1/2, -1/2, 2/1, -2/1, 2/2, -2/2, 1/3, -1/3, 2/3, -2/3, etc.}. You should get the point here. Since they can all be listed, they match 1-1 with N, hence aleph-null (also called countably infinite).
BUT - the reals are > aleph-null.
Proof by reductio ad absurdum. Assume you have a list of all reals. Then here's a real not in your list: Construct it by creating a number that differs from the nth number in your list in its nth position. Since this new number is not in your list, you did not in fact provide such a list, hence it cannot be done. Even if you add that number to the list, we'll just do this all over again.
Here's the really cool part: In 1900 or so David Hilbert created his famous list of 23 problems for the next century. First on the list: the Continuum Problem: is the cardinality of the reals aleph-one, i.e., the very next transfinite size?
In 1963 Paul Cohen proved (get ready for this) - You can have it either way. That is, the statement "size of the reals is aleph-one" and its negation are independent of the generally accepted axioms of set theory, ZFC (Zermelo-Fraenkel with Axiom of Choice).
I see numerous (and in my opinion repetitive and meaningless) posts about doing various arithmetic with infinity: ∞.
The theory of transfinite arithmetic was initiated mainly by Georg Cantor around the 1890's.
The theory explores in a definitive and axiomatic way the various (differently sized) infinite sets of different types of numbers (e.g., Natural Numbers, Rationals, Irrationals, Transcendentals).
This is too involved to say more here, but a good starting point are these articles:
https://en.wikipedia.org/wiki/Aleph_number
https://en.wikipedia.org/wiki/Transfinite_number
https://en.wikipedia.org/wiki/Continuum_hypothesis
But 2.151515 is not in your set N.
Follow each line of the "proof." When you get to a division line, evaluate the divisor. If it = zero, then the following lines of the "proof" are invalid if they assume that the quotient is a real number, as opposed to the division being undefined.
Have you found which line has the division by zero?
The point is that there is most likely a division which is not obviously by zero, but close analysis shows that it is, and yet the subsequent lines rely on the quotient being a real number.
But your post has a lower case i, so as it stands it may be considered wrong, assuming the rational/irrational distinction applies to reals only.
I guess it's too late to correct, because then these comments will make no sense.
I didn't know that i was considered an irrational.
Even if so, wouldn't the same "reductio ad absurdum" proof be better using a real irrational?
A definition does not have a proof.
What it does need is to be consistent with other statements in a given theory, and to be useful in the various formulas.
Bob's and KerimF's explanations show how it fits with the rules of arithmetic, as odd as it seems for a^0 = 1.
In my opinion this is not per se a "proof," but rather several explanations for why a^0=1 (for a not= 0) is a reasonable definition.
And I accept your kind apology.
Good luck on your self-study. Bob has clearly assisted you greatly.
Thanks Bob for jumping in.
Is this your homework?
Sure, it is correct: (x, y) = (x, f(x)).
We say y = f(x) because, worldwide, we agree on the convention that x is the independent variable (i.e., pick a value for x), and y is the dependent variable, because when we write y = f(x), or, e.g., y = 2x+5, the value of y depends on the value of x we picked.
Some algebra books have a few sections where we consider x = f(y). All graphs are turned sideways, vertical line tests become horizontal line tests, etc.
But the convention as noted above allows us to share our math. Nothing magical about y = f(x) - we just all agree to do it that way.
That help?
Set theory and the continuum hypothesis.
Fermat's Last Theorem.
Too bad this was finally resolved - it ruined a good joke:
A mathematician was asked to give a talk at the local college. He told them ahead that he would be discussing his proof of Fermat's Last Theorem.
But at the symposium he actually discussed a totally different topic.
When asked later why he changed topics, he said that he told them the original topic in case he got run over by a bus on the way to the talk.
In general, to solve f(x)/g(x) when their ratio becomes 0/0 for x=a, we calculate f'(a)/g'(a).
The rule you mention here is L'Hopital's rule, which does NOT calculate "f(x)/g(x) when their ratio becomes 0/0 for x=a"
but rather calculates the LIMIT of f(x)/g(x) as x approaches a where f(a)/g(a) is indeterminate.
That limit may well exist, even if f(a)/g(a) is undefined.
E.g., The graph may be smooth but with a hole at f(a)/g(a).
The graph of (x^2 -3x + 2)/(x-2)
looks like the straight line y = x-1, but (e.g., in Desmos) hover over the line at x=2 and it will tell you (2,undefined).
This is an example of two functions equal at all but a finite number of points.
I've got degrees in Mathematics and Computer Science, and I've never heard this.
Just curious - what's your source for: I guess lots of computer people are saying that 0/0 should be 1
I'm sure k/0 will always be undefined for any value of k, including zero.
As a math student - I suggest you do it the "messy" way also. Since you know the answer from the "easy" way - you'll know if you got it right. One more technique learned.
(Disclosure - Of course, as a FORMER math student and teacher/tutor, I have no intention of doing that myself. I presume I can, and that's good enough for me! I'm going out to play Frisbee.)
This should get you started.
I assume you mean each car's speed is 5m/h. To simplify your thinking, it is as if one is stationary and the other's speed is 10m/h.
Distance is 15 miles, so total time till collision is 1.5 hours.
Bird flies at 30m/h for 1.5 hours.
You should be able to finish the problem now.
Think of it this way..
The shortest distance between 2 points is a straight line.
If the sum of the lengths of 2 side of a triangle = the length of the third, then the "triangle" would collapse to 2 overlapping straight lines.
If the sum of the lengths of 2 side of a triangle is < the length of the third, then the shortest distance between the third line's endpoints would not be that straight line.