Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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That's a fair comment. If the ship was a mile high, you'd be able to see the top long after the bottom had disappeared below the horizon. The method also disregards the curvature of the Earth and where the person's eyes are in that height figure. But it gives a good approximation for most practical purposes.

Bob

Yes. It follows from the following

If one of n events must occur then

P(event 1) + P(event 2) + ... + P(event n) = 1

Bob

Not quite sure about the m conversion. You just need everything in the same units.

That formula works. Pythag is a part of trigonometry. I think it's the easiest and quickest method. d comes out as the straight line distance, not the curvature distance.

Bob

Fair point. At the top of each post there's a list of symbols you can copy and paste. ≠ is available there. Does your phone allow you to copy any of those.

Whatever, I was happy with your answer as it had the essential bits.

Bob

Dictionary.com defines rogue as "no longer obedient, belonging, or accepted ". 6 doesn't obey the rules as it's not in the universe.

Let's get more practice by making C = {1, 3, 4, 5}

That does change your previous answer.

Bob

As C complement is the elements not in C 6 doesn't feature whether it's there or not. But If you change 6 to 0 for example it would change the answer.

Bob

OK

B

Yes, OK.

I would prefer

Bob

Again we have a rogue 6 in C.

Bob

Yes, that's ok. But you have typed

C = {1, 3, 4, 6}

That 6 cannot be an element as it's not in the universe. Fortunately it doesn't effect your answer.

Bob

A set and its complement divide the 'universe' into two non overlapping sets with no elements left out of being in one or the other. So when you unite them you get everything, ie the universe.

And when you look to find what's in both a set and its complement you find nothing hence the empty set.

Normally when I'm showing union and intersection in a picture I draw two overlapping circles (inside a rectangle for the universe). But for a set and its complement a better picture is a rectangle for the universe divided by a single border making the two sets that don't overlap and together make the whole of the universe rectangle.

Bob

Me too.

Bob

Yes, that's got it!

Bob

The first is correct for set builder notation. Your second answer is the roster answer.

Bob

Yes that's right. I've never heard it called the roster method though. Always good to learn something new.

Bob

I should find A U C first.

A U C = {0, 1, 2, 3, 4, 5} U {2, 4, 6, 8, 10}

A U C = {2, 4}

Yes you should but the union is {0,1,2,3,4,5,6,8,10}

Bob

It is correct.

Bob

Yes, that's correct.

Bob

I only have the same search as you. The keyword can throw up loads of hits. The trick is to remember a word you used that is unlikely to be in anyone else's posts.

I tried 3 people A,B and C and got pages and pages. I think it was finding every post that had any of those in it so then I tried "3 people A,B and C" hoping to find only that specific sentence but still didn't get to it.

Bob

I used the search of users looking for cool* where the * acts as a wildcard. Quite a few hits, but none look like your old account.

Anyone can do this type of search so maybe you can find the exact username yourself. I can then give you access to all your old posts if you wish. There doesn't appear to be a way to link the two accounts however.

Or we can just leave everything as it is.

Bob

hi Turbo

Welcome to the forum and thanks for the puzzles.

Two things I need to check:

y/y isn't that just 1 ?

You've copied these from somewhere. Is there a copyright issue?

Bob

That's excellent , well done!!!

Bob

hi Zach,

Welcome back!

I do remember that question but I cannot find your old username in our lists, nor the thread. Ho hum, not to worry.

Bob

When photons pass through a cloud they are scattered and so arrive in my back garden from a variety of directions. If I knew these directions I could backtrack and, by statistical analysis, determine the most likely direction from which they came.

So it should still be possible to deduce where the shadow would have fallen if only the Sunlight had had an unrestricted path to the dial. Building the supercomputer that can do this analysis in my garden is a project in the early stages of development. As the detectors would have to be sensitive to individual photons, it would also be possible to determine the positions of some of the brighter stars. As the sidereal day is a more reliable time keeping device, not only would it become possible to tell the time at night, but without the annoying correction factor known (incorrectly) as the equation of time.

Bob

Even and Odd are just words for something. So, having thought more about this, they are defined as having the properties ( 2n, and 2n+1) rather than axioms.

A major part of mathematics is concerned with building models to 'describe' something useful. Euclidean geometry, for example, enables us to work with triangles and circles and angles (and lots more) and leads to useful stuff like trigonometry. It's an idealised view because you cannot have a line with zero thickness, nor can you measure an angle with absolute accuracy. Nevertheless, it's very useful.

Number theory is built upon axioms such as x + y = y + x where x and y are real numbers. We can use the axioms to build theorems and then the theorems plus axioms to build new theorems. Eventually you end up with calculus and complex numbers.

The M-W definition looks ok to me.

If you can prove an axiom (presumably from some of the other axioms) then it isn't needed as an axiom, so you could develop your model with one less axiom.

For example. One axiom is "There exists a number (let's call it p for the moment) such that n + p = n. This number is called the additive identity. There is no axiom stating that p is unique because we can prove it.

Proof: suppose there are two identities p1 and p2. Consider the sum p1 + p2. It equals p1 because p2 is an identity, but it also equals p2 because p1 is an identity, thus p1 + p2 = p1 = p2 so the two identities are the same number.

Bob