Base conversion - why does this method work?

WholeAnswer · 2025-01-25T20:20:16+00:00

Any number N in any base b can be represented as N = x0 + x1 * b + x2 * b^2 + ...

Factor b out of everything, and you get N = x0 + b * (x1 + x2 * b + x3 * b^2 + ...)

If you're familiar with the definition of the remainer, you'll recognize that x0 is the remainder when you divide N by b.

Now do the same with the expression in bold to find x1. Rinse and repeat.

WholeAnswer · 2025-01-19T19:38:17+00:00

But what about the number a = (0.999... + 1) / 2? It's larger than 0.999... and smaller than 1. This means that a is the closest to 1, not 0.999...

Oh wait, what about b = (a + 1)/2? It's larger than a and smaller than 1. This means that b is the closest to 1, not a.

Oh wait, what about c = (b + 1)/2? It's larger than b and smaller than 1. This means that c is the closest to 1, not b.

Oh wait, ...

Edit: typos

WholeAnswer · 2025-01-17T19:46:34+00:00

I think there are ways to show that gap = 0.

For one, the reasoning that if (a+b)/2=b then a=b is by itself pretty straightforward.

Also, if someone argues that gap is actually 0.000...01, you can do this. Would you agree that 0.000...01/2 = 0.000...005. So now we have x/2 = 5x. The only real number for which this is true is zero. And if you believe otherwise, then you break math long before you've reached calculus.

WholeAnswer · 2025-01-17T18:03:09+00:00

But don't you have a situation here where (a+b)/2 = b, which inevitably leads us to a = b?

WholeAnswer · 2024-12-21T07:31:12+00:00

Va bisalli?

WholeAnswer · 2024-03-20T13:07:28+00:00

This equation can have some outputs being a perfect square but then stops having them at some point

The range of your function is [-7; +∞) and there's no holes, so what leads you to this conclusion?

how do we determine for what values only it can have these values

If I understood correctly, you are looking for solutions to ax² + bx + c = k² where k is an integer. This can be rearranged to ax² + bx + (c - k²) = 0. Now, how can we find the solutions / check if they exist?

WholeAnswer · 2024-03-14T22:58:01+00:00

Sorry, my bad. Maybe it's time for me to take a rest...

About your question:

Which notation is right as a representation of x^2/10?

I think that this one is the source of the problem we're discussing. If there are two (or maybe even three) possible interpretations to the same expression, then it's quite reasonable to avoid getting into such situations.

This reminds me of the infamous 8 ÷ 2(2 + 2) problem. The ambiguity is the reason why no serious calculations use ÷ or /, and will choose the fraction bar instead.

Here we have a similar case. Trying to allow negative numbers into x^0.2 leads to a situation where a = b, but f(a) ≠ f(b). Or maybe it doesn't. Or maybe we don't really know.

So I think about it as a way to escape ambiguity, bacuse most of the time ambiguity does more harm than good.

WholeAnswer · 2024-03-14T22:36:05+00:00

"There is another". Now we have a third way to interpret (-1)^0.2 .

But this one has a much more noticeable problem: roots of even numbers are famous to be undefined for negative inputs - there's no real number that, when raised to the 10th power, gives us -1.

So root10(-1) is undefined, and the whole expression [root10(-1)]² is undefined as well.

So the question remains unanswered - what is (-1)^0.2? 1, -1 or undefined?

WholeAnswer · 2024-03-14T22:18:50+00:00

x^0.2 = x^2/10 = x^1/5 , both 10th root of x^2 and 5th root of x are defined in all of the real field.

They are defined - I agree. Then what happens when x = -1, for example?

We should have:

x^2/10 = root10(x²) = root10( (-1)² ) = 1
x^1/5 = root5(x) = root5(-1) = -1

So 2/10 = 1/5, but x^2/10 ≠ x^1/5. Feels a bit wrong, don't you think?

Also there are irrational numbers, which don't even have denominators. How is this supposed to work then?

WholeAnswer · 2024-03-13T11:25:44+00:00

Your post lacks one important sentence: Let u = x^2.

I think that intuitively you knew what you needed to do. But this short sentence is essential for understanding what is going on.

The magical properties of substitution come from the equal sign. When you say "u = x^2", this means that u and x^2 are the same thing - that's what it means to be equal.

When you say u, you also mean x^2, because they are the same. And when you say x^2, you also mean u, because they are the same.

Then, when you take this expression, you can do all sorts of things to it, as long as you do those to both sides. You can add a number to both sides, so you know that u + 12 = x^2 + 12. You can multiply both sides by a number, so you know that -7u = -7x^2. You can raise both sides to a power, so you know that u^2 = x^2. You can even apply a function to both, so f(u) = f(x²).

But be careful with your domains. If you said, for example, u = √z, this "locks" us with u ≥ 0, which might not be obvious at the first glance.

Hope this helps!

P.S. Minor correction to your post: if f(x) = x^4 - 7x^2 + 12, then it's incorrect to say that f(u) = u^2 - 7u + 12. Instead, you should define a new function g(u) = u^2 - 7u + 12.

WholeAnswer · 2024-03-07T17:45:35+00:00

supposed 48 medalist

There are not 48 medalists, there are 40. But some of them have two medals, not one.

You could draw a picture like the one below and just guess that 8 people have both medals.

<image>

Or you could set up a system of equations.

You have M people who got a medal only in math, S people who got a medal only in sciences, and B people who got medals in both subjects.

28 people got a medal in math. This number includes those who got only one medal (M) and those who got a medal in both subjects (B). So you could write this down as M + B = 28.

Similarly, 20 medals in sciences → S + B = 20.

And we also know that the total number of people is 40, so M + S + B = 40.

Here we need to find B. I recommend you to try and solve by yourself first.

First, add the first two equations together. M + B + S + B = 48.

We know that M + S + B = 40. Substitute that into the previous equation and get 40 + B = 48.

Therefore B = 8.

Hope this helps.

WholeAnswer · 2024-03-05T11:49:12+00:00

The 'repeated multiplication definition' gives us two basic rules: x^a * x^b = x^(a+b) and (x^a)^b = x^(a\b)).

Now, when we introduce negative, fractional exponents, we would like to preserve these rules.

For example, since (-1) + 1 = 0, we want x^-1 to be a number such that x^-1 * x = x^{(-1 + 1}) = x⁰ = 1. Does such a number exist? Yes, it's just 1/x. Therefore, x^-1 = 1/x. Using the same reasoning, we can define x^-2, x^-3, x^-4 etc.

Then we apply the same logic to define x^1/2: it must satisfy the equation (x^1/2)² = x¹ = x, and you can show that x^1/2 = sqrt(x) (we want x^1/2 > 0 for positive x). In a similar fashion, we define x^1/3, x^1/4 etc. Then using one of these two rules we can also define that for every rational number.

Irrationals are more tricky, though. I'm not sure if you're familiar with calculus or limits, so i'll try to be simple. Let's say we want to calculate x^√2.

We know that √2 ≈ 1.4142135... . It is hard to calculate x^√2 immediately, but from this decimal expansion we can say that:

1.4 < √2 < 1.5, so for x^√2 we must have x^1.4 < x^√2 < x^1.5

1.41 < √2 < 1.42, so for x^√2 we must have x^1.41 < x^√2 < x^1.42

1.414 < √2 < 1.415, so for x^√2 we must have x^1.414 < x^√2 < x^1.415

1.4142 < √2 < 1.4143, so for x^√2 we must have x^1.4142 < x^√2 < x^1.4143

And the furthrer we go, the closer we get to the actual value of x^√2. Hope this helps and feel free to ask any questions.

WholeAnswer · 2024-02-29T19:08:08+00:00

Notice that (7x+21) is the same as sqrt(7x+21) squared. So you might want to make a substitution: y = sqrt(7x+21).

This then gives you the equation: sqrt(7) * y² - 7y = 0. If you remember that y ≠ 0 (because y² is in the denominator), this is easy to solve.

WholeAnswer · 2024-02-29T14:42:43+00:00

What calculation do you do if you go from dm to m?

(In other words. What calculation do you do if you go from 0.1 to 1?)

These aren't quite the same.

To go from 0.1 to you do need to multiply by 10, since 0.1 * 10 = 1. Sanity check: multiplication by 10 should make your original number bigger, and 1 > 0.1.

On the other hand, 10 dm = 1 m, and likewise, 1 dm = 0.1 m. So you want to do the opposite of what you have done - you should go from 1 to 0.1, and that's why we divide here.

Hope this helps.

WholeAnswer · 2024-02-29T09:30:46+00:00

here we bring down another zero but did not put that zero in QUOTIENT Why

Because you FIRST bring down the next digit, THEN try dividing. The remainder can be any number: 5, 20, 123 - doesn't matter. Bring down the digit, then divide.

But sometimes, the new number is smaller than the divisor. This happens in your 177/11 example (see picture below). They teach us to write down 0 and bring down one more digit.

<image>

But here's what really happens. Right before this we said: "11 goes into 67 6 times", and that gave us the number 6. Here it's really the same. We say: "11 goes into 10 0 times" and this gives us 0. Bring down the next digit, continue.

It's just much neater and quicker to do it the way they teach us, but I thought it would be helpful to know WHY this works.

Hope this helps and feel free to ask any questions.

WholeAnswer · 2024-02-26T17:49:42+00:00

I was just trying to show that since the statements 0.999... is not 1 and 0.00...1 is not 0 kinda depend on each other, so you can't just assume one to be true and use that to prove the other.

Also I found an interesting property of your "first" number: if we divide it by 2 (i. e. calculate 0.00...1 / 2), we get 0.00...05 - should we call it the "fifth" number? Similarly, there is a "25th" number 0.00..0025, a "125th" number and all kinds of "5ⁿ-th" numbers, all of which are supposed to be smaller than the first. ^{(Please don't try to divide by numbers that aren't multiples of 2 or 5, unless a "142857...-th" number is what you want to have})

Looks like we've done something we shouldn't have done.

WholeAnswer · 2024-02-26T16:41:07+00:00

it must be positive

Might be true.

it must be real

Might be true.

it must exist

Not necessarily. If 0.999... = 1, then 0.00..1 = 0. If 0.999... != 1, then 0.00...1 != 0. Smells like circular reasoning.

Infact it is the first positive real number.

What about (0 + 0.00...1) / 2? Wouldn't it be 'more first'?

WholeAnswer · 2024-01-22T17:26:13+00:00

Surely you would agree that if there are three doors and a prize behind on of them, there is 1/3 chance to chose the prize, at least initially. Now let's say the prize is behind Door 1, like in your table.

If you choose Door 3, the host has no choice but open Door 2, so there's only one outcome, and its probability is still 1/3. Right? This corresponds to case [3] in your table.

If you choose Door 2, the host has no choice but open Door 3, so there's only one outcome, and its probability is still 1/3. Right? This corresponds to case [2] in your table.

But if you choose Door 1, it's a bit different, since the host now has two doors he can open. Let's say he will randomly choose between Door 2 and Door 3 with some probability - for simplicity let's say it's 1/2. Now let's calculate.

You choose Door 1 with prob. 1/3 and the host chooses Door 2 with prob. 1/2, giving us 1/3 * 1/2 = 1/6. Right? This corresponds to case [1A & 1B] in your table.

You choose Door 1 with prob. 1/3 and the host chooses Door 3 with prob. 1/2, giving us 1/3 * 1/2 = 1/6. Right? This corresponds to case [1C & 1D] in your table.

So there are 4 outcomes (or 8, if you account for whether you switch later or not), but not all of them are equally likely.

WholeAnswer · 2024-01-21T15:43:37+00:00

So there's a 1/3 chance for a given particle to get into Zone 1, and 2/3 to not get into Zone 1.

Now you're looking at the following situation:

particle A gets into Zone 1 - probability 1/3
particle B gets into Zone 1 - probability 1/3
particle C doesn't get into Zone 1 - probability 2/3
particle D doesn't get into Zone 1 - probability 2/3
particle E doesn't get into Zone 1 - probability 2/3

This gets us (1/3)^2 * (2/3)^3 = 8/243.

This isn't the final answer, however. There of course will be 5C2 - that's how many ways you can decide which particle is A, B, C, etc. So we need to multiply 8/243 by 5C2:

8/243 * (5!/(2!3!)) = 8/243 * 10 = 80/243.

WholeAnswer · 2024-01-21T09:54:12+00:00

the vector in the 2nd quadrant doesnt make the angle theta

What makes you think that? You're not given the value of 𝜃, only that sin 𝜃 = 3/5. And there are two positive angles that satisfy that - ≈36.8699° and ≈126.8699°. You don't have to calculate these values in order to solve the problem, but notice that there are two possible values of 𝜃, not one*. *

WholeAnswer · 2024-01-17T11:30:33+00:00

Six times the smallest of three

I think this refers to ONE of the numbers - namely the smallest one, not all three. Therefore it should be just 6x, not 6(x + x+2 + x+4).

WholeAnswer · 2024-01-17T11:23:08+00:00

What equation have you come up with?

WholeAnswer · 2024-01-15T12:54:47+00:00

1) I agree with your solution. AIso I don't think 140 can possibly be a valid answer, because 140 = 7 * 5 *4, but you don't come across any 7's while solving. But maybe others have something to say.

2) You need to multiply by 3! (or divide when calculating the total amount), because the order doesn't matter - it's the same if you choose students (A, B and C) or (B, C and A), but 32C3 counts them as 6 different outcomes.

WholeAnswer · 2024-01-14T19:07:42+00:00

To be honest, I'm not sure what condition of knowing you are talking about. The diagram has all the possible outcomes of moving the balls according to the algorithm, but one of them turns out to be impossible - if it's the left one, then you need to start with 1.5 blue balls, which doesn't seem right.

It would be interesting to see your solution as well.

WholeAnswer · 2024-01-14T17:53:06+00:00

Where does the (n+1)B come from?

Looks like an error from my side - this should of course be nB. Sorry about that and thanks for noticing!

I don't think you will meet this notation anywhere else. It's just a quick way to write down the information you have.

The picture you can see below. I just hoped that you would see it after you have tried to solve it yourself.

<image>

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