The Maths Topic

Lucid Lounge
21

And Sandra is right too: b² = 0.01$ [/b] and there are even three different ways of proving that! :shock:

/me geeft Sandraatje een biscuit :cookiemon: en een knuffel :tongue: :hugs:

Now, Kenneth just made another puzzle and this one is slightly more difficult to prove:

b² = 0.01 $²[/b]

But, as Sandra proved,

b² = 0.01$ [/b]

Therefore,

0.01$ = 0.01 $²

1$ = 1$²

But oh look, Sandra’s law develops as follows:

b² = 0.01$[/b]

b² = 1 c[/b]

100 c² = 1 c

1 c² = 0.01 c

The Number Theory freak who manages to tell me why 1 $² = 1 $ whilst 1 c² = 0.01 c gets two biscuits. :cookiemon: :cookiemon:

Oh, by the way, I found another way to answer the π problem and I think it’s actually the easiest available so far. I’ll try to post it later because right now I have to go. :wave:

22

You seem almost a bit more nerdy than me! :eek:

That problem is simple.

1$ ^2 = 1$ * 1$ = 1$
1c ^2 = 0.01$ ^2 = 0.0001$ = 0.01c

23

Eh, these kind of tricks are as old as the algebra itself.
I still remember how you can prove 2+2=5, had a little fun in middle school.

Engeneering is good enough, but real use of advanced math is programming. And that IS something you can really have fun with. Coding games, for example. Writing your own shaders and then admiring how shiny the armor on that trooper is…

24

Good, but that’s demonstrating, and I asked you to explain. :wink:

25

Gnarhl looks at clock let me prove instead for you that 2 = 1.

Let a = 1, and b = 1. Then it is true that we have:

a^2 - b^2 = a^2 - ab
(a+b)(a-b) = a(a-b)
a+b = a
2 = 1

End. I can also prove that sin x / n = 6.

sin x / n = sin x / n = si x = six = 6 :content:

26

Alright, I’ll explain it myself. :tongue: I was expecting someone to say it’s because $ = 1 and c = 0.01, which is wrong.

Lets schedule a couple of those linear equations regarding the relation between $ and c:

1$ - 100c = 0

0.01$ - 1c = 0

↓ THEREFORE

[code] – –
| 1 -100 | 0 |
| 0.01 -1 | 0 |

L2 => (-L1/100 + L2)

| 1 -100 | 0 |
| 0 0 | 0 |

0$ + 0c = 0

0 = 0[/code]

“Sweet, Bruno, what does this mean?”

That you can’t define $ if you don’t have c, and you can’t define c if you don’t know $. That means $ isn’t necessarily 1: all we know is, it’s worth 100 c. So if 1 c is worth 5 in pure numbers, 1 $ equals 500.

“Alright, and what’s the difference?”

The difference is, since $ and c can’t be defined in terms of pure numbers, they’re said to be a number system. As a number system, it doesn’t matter how much $ is worth, what matters is $ is the basic unit of its system, so it’s worth 1 of whatever we’re measuring with it. Although $ doesn’t have to equal 1, $ α 100% (α indicates proportionality) of it’s system, whilst c α 1%.

Thus, the issue can be demonstrated as follows:

1 $ α 100%
1 c α 1%

↓ TO THE POWER OF 2

1² $² α (100%)²
1² c² α (1%)²

↓ THEREFORE

1 $² α 100%
1 c² α 0.01%

↓ THEREFORE

1 $² = 1 $
1 c² = 0.01 c

There’s a lot of nice psyched metaphysical stuff one can find out from that, a real lot, but I really don’t have the time for the crazy what the metaphysics.

Now, as promised:

x = ³√8

x³ -8 = 0 → 2³ -8 = 0

2 = 2∙(cos 0 + i sen 0)

By Euler’s fomula,

2 = 2∙e^(i 0)

Since the other two solutions must be spread around the complex field with argument 2 and equal angular distances, we divide 2π/3 and get the three solutions:

[b]α = 2∙e^(i ⅓∙2π)

β = 2∙e^(i ⅔∙2π)

γ = 2∙e^(i 2π)[/b]

↓ THEREFORE

S = { 2∙e^(i ⅓∙2π), 2∙e^(i ⅔∙2π), 2∙e^(i 2π) }

Wow that was fast! :cool:

Someone else propose a problem!

27

a) (easy) 1 + 1 + 1 + 1… + 1 (x times) = x

take derivatives:
0 + 0 + 0 + 0 + … + 0 = 1
0 = 1

b) (moderate) sqrt(1) = 1; but 1 = -1*-1;
sqrt([-1]*[-1]) = 1;
sqrt[-1]sqrt[-1] = 1;
ii = 1;
-1 = 1.

OR: -1 = sqrt([-1]^2) = sqrt(1) = 1.

c) (hard but involves a tiddy bit of physics) a ladder is leaning with one end of it on the wall, and the other end on the ground. It touches the ground at a distance x from the wall, and touches the wall at a distance y from the ground. If L is how long the ladder is, it’s obvious then that: y = sqrt(L^2 - x^2).

so now we grab the end of the ladder that is touching the ground and we pull it away from the wall at a constant speed vx. vx is obviously dx/dt. the other tip of the ladder slides down the wall towards the ground. let’s see how fast the other end of the ladder moves towards the ground:

vy = dy/dt = - x* vx / sqrt(L^2 - x^2). (no need to check this calculation, it’s correct).

ah but notice, the ladder hits the ground when x = L. but in the above equation, setting x to L gives us: vy = infinity. So the other tip of the ladder hits the ground at an infinite speed?

28

:shock: I think I might have figured this one out after extensive making of desperate notes in me notebook. I’ll post the solution tomorrow if no–one else posts anything about it by then. I’ll write a couple of tips, some of them might be useful and some might not. :grin:


[color=#333333][size=100]The pink β is actually β[/size][size=67]2[/size]. :peek:[/color]

For any α and β, α + β = π/2 so that [color=#ff9900]α[size=67]1[/size][/color] + [color=#ff9900]β[size=67]1[/size][/color] = [color=#ff0099]α[size=67]2[/size][/color] + [color=#ff0099]β[size=67]2[/size][/color].

Given ∂x/∂t = constant, can we obtain ∆α/∆t?

Finally, what’s particular about the system when ∆y = −y[size=67]1[/size]?

Just some food for thought. :wink:

29

Plain old regular numbers? How boring. Let’s use these numbers instead:

s is the number defined by z^y^x^s = (z^y)^z^x for all x, y, z (^ is right-associative)
k is the number defined by y^x^k = x for all x

Of course, now we have to throw away things like 1^x = 1 for all x. It’s not like it was useful anyway, right?

So yeah, we have those two “special numbers”. Now find (1^1^s)^1^1^s for me.

30

Eh… Either b^1^1^s = 1[/b] or s = ±∞ right? I’m not sure, I’m not good with analysis.

31

Wow if I am going to learn this I am destined to work at Mcdonalls for life. :cry:

32

Wow, it’s fun to see that this topic got stickied :smile: in that case, I better show you some problem solving techniques and the like. Now, here is a problem for you.

There is a school with exactly 733 students in it. Can you prove that there are at least 2 boys or 2 girls whose birthday occurs at the same date?.

Happy solving :peek:

33

Sure. The number of gender/birthday combinations is 732, or the number of genders (2) times the number of birthdays (366, counting February 29). Since there are more students than combinations, one combination has to be repeated: two people with the same gender and the same birthday.

According to the Law of Enaggeration (which seems to be the opposite of “exaggeration”), this shared birthday is most likely to be February 29.

Edit: oh, right.

s and k aren’t real numbers.

Let’s use the definitions:

(1^1^s)^1^1^s
((1^1^s)^1)^(1^1^s)^1
(1^1^s)^(1^1^s)^1
(1^1^s)^1^1^s
((1^1^s)^1)^(1^1^s)^1
(1^1^s)^(1^1^s)^1
(1^1^s)^1^1^s
((1^1^s)^1)^(1^1^s)^1
(1^1^s)^(1^1^s)^1
(1^1^s)^1^1^s

Continue as long as you want. You might get somewhere eventually.

34

I’m sorry, I refuse to believe pie is squared :no:

35

Yep, that’s basicly right. It goes off something called “the box principle”.

Imagine that we have 366 boxes, one for each day of a year with 366 days. Now, we tell all the children in the school to jump down into the box with the date when they were born. Now, the “worst” possibility would be that the first 366 “jumpers” all have a different birthday from eachother. But the next one will share a box with someone else.

Let’s say that in the worst, worst chance each box has exactly 2 children when 732 children have hopped in (this possibility is near nothing :tongue: ), logically, the next child jumping in (child no. 733) has to get in a box shared with two other children. We now have a box with three children in it.

Now, we make two new boxes, one named “Boys”, and one named “Girls”. We tell the three children in the box to jump down in the box with their gender. Logically, one box will get at least two children in them :wiske:

36

Wow :cool_laugh:… and I thought I was nerdy at math, haha.

Some interesting stuff. If Im ever stuck with math homework, Imma coming here! :tongue:

37

Exact the same thought as mine xD

38

Of course…pies are round, aren’t they? :eh:

Ah! I’m going to get sucked into this topic now…I hated math in high school so I avoided it wherever I could…but this year I had to take a course in statistics, and I loved it :grin:
Everything made so much sense…I suppose it’s the way the professors teach in post-secondary as opposed to high school, or maybe I just wasn’t into math back then.
Anyhow, I have the option of taking a second stats course which will give me more advanced standing when I go to University in the near future, so I’ll definitely be taking that, because it’s sure to be lots of fun :yes:

39

Theorem: 1 is the greatest natural number.

Proof:
We assume the absurd that 1 is not the greatest natural number. Let N be the greatest natural number. So N > 1. Than we multiply by N. So N^2 > N. But N^2 is still a natural number and N^2 != N (not equal). So we have a contradiction (N is not the greatest natural number because N^2 is greater). So the assumption is false. So 1 is the greatest natural number.

qed

Do you agree with this proof?

40

That’s a very badly formatted proof :content:

This because even if you say that N is the greatest natural number, multiplying as you said so N^2 > N just shows there is a natural number that is higher than N that is larger than N. Therefore, you cannot assume that it’s a contradiction.