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 Author Topic: 1 = 1  (Read 4985 times)
whitetiger0990
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Posts: 2964

 « on: September 01, 2003, 10:08:41 PM »

Code:
1 + 1 = 2
2 - 1 = 1
1 * 1 = 1
1 / 1 = 1
1 + 1 - 1 * 1 / 1 = 1

it all makes sense!

Isn't it wierd?
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Ninkazu
Been there, done that

Posts: 1169

 « Reply #1 on: September 01, 2003, 10:13:35 PM »

This doesn't even deserve a response, but I just have to say this:
yu0 isz r3+4|2d
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am an asshole. Get used to it.
whitetiger0990
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Posts: 2964

 « Reply #2 on: September 01, 2003, 10:22:33 PM »

ALEIN!!!!
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SCM
Wandering Guru

Posts: 311

 « Reply #3 on: September 01, 2003, 11:49:30 PM »

Code:
3 + 3 = 6
6 - 3 = 3
3 * 3 = 9
9 / 3 = 3
3 + 3 - 3 * 3 / 3 = 3

Most of what you have there would be true for any number.  The second line is just the inverse of the first, and the fourth is the inverse of the third. In the fifth the added and subtracted terms are equal, so you are adding 0, and you don't change the original value.

You have more ones than I have threes because one is the multiplicative identity (this just means that any number is left unchanged when multiplied it by one)
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Ninkazu
Been there, done that

Posts: 1169

 « Reply #4 on: September 01, 2003, 11:52:06 PM »

3 + 3 - 3 * 3 / 3 = -1
(6 - 9) / 3 = -1
-3 / 3 = -1

jeez.
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SCM
Wandering Guru

Posts: 311

 « Reply #5 on: September 02, 2003, 12:02:57 AM »

Ninkazu,
use proper order of operations.  Multiplication and division come before addition and subtraction (you may have been taught PEMDAS).
Code:
3 + 3 - 3 * 3 / 3
= 3 + 3 - 9 / 3
= 3 + 3 - 3
= 3 + 0
= 3

Code:
IF 3 + 3 - 3 * 3 / 3 = 3 THEN PRINT "It's true"
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hrist Jesus came into the world to save sinners, of whom I am first.(I Timothy 1:15)

For God so loved the world, that He gave His only begotten Son,
that whoever believes in Him should not perish, but have eternal life.(John 3:16)
whitetiger0990
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Posts: 2964

 « Reply #6 on: September 02, 2003, 12:05:26 AM »

so you see my observation is correct
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Moneo
Na_th_an

Posts: 1971

 « Reply #7 on: September 02, 2003, 12:44:04 AM »

WhiteTiger,

You forgot:
N^0 = 1
where N can be any number.
*****
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Diroga
Been there, done that

Posts: 1087

 « Reply #8 on: September 02, 2003, 02:53:07 AM »

uh..what are you trying to get at that i dont see
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oracle
*/-\*

Posts: 3652

 « Reply #9 on: September 02, 2003, 03:25:22 AM »

Well done, you've just proved that 1=1, I'm sure, however, that that fact has already been proven by far smarter brains than ours.

Here, lets prove something a little more complicated: .9(recuring) = 1

n = .9(recuring)
10n = 9.9(recuring)
9n = 8.9(recuring) (subtract this line from line 2)
n = 1

Now, who can prove that the square root of 2 cannot be written in the ratio of two integers? That's a little more tricky, but let's do the easy ones first.
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ak00ma
Ancient Guru

Posts: 669

 « Reply #10 on: September 02, 2003, 03:55:20 AM »

if n is equal to .9 and you multiply n with 10 then it equals 9 and not 9.9, or did I missunderstand something???
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B 4 EVER
SCM
Wandering Guru

Posts: 311

 « Reply #11 on: September 02, 2003, 04:28:57 AM »

n = 0.99999999999999999999999999999999999999999999...
It keeps repeating forever.  The formal mathematical method for proving it involves showing that you can't find any number between 0.9... and one.  If no number exists between two numbers then they are really the same number.
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hrist Jesus came into the world to save sinners, of whom I am first.(I Timothy 1:15)

For God so loved the world, that He gave His only begotten Son,
that whoever believes in Him should not perish, but have eternal life.(John 3:16)
oracle
*/-\*

Posts: 3652

 « Reply #12 on: September 02, 2003, 05:02:55 AM »

Quote from: "SCM"
n = 0.99999999999999999999999999999999999999999999...
It keeps repeating forever.  The formal mathematical method for proving it involves showing that you can't find any number between 0.9... and one.  If no number exists between two numbers then they are really the same number.

Correct. I think it is called "dense" when there is no number between 2 numbers.

So, anyone know how to prove that root 2 aint a fraction?
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Flexibal
Forum Regular

Posts: 128

 « Reply #13 on: September 02, 2003, 05:50:19 AM »

yes, tis simple:

let the square root of 2 be a rational number, which we'll represent as a/b.

so

sqr(2) = a/b
2 = a^2 /  b^2
2 * b^2 = a^2

now, according to the rules of even numbers, any number multiplied by two is even. so (2*b^2) is even, and since (a) equals it, (a^2) is even as well.

another rule of even numbers is that an even number squared is an even number. since we already found out that (a^2) is an even number, it means (a) is an even number.

because a is even, we can represent it an a multiple of 2 and a natural number:

a = 2*c

we get

2*b^2 = (2*c)^2
2*b^2 = 4*c^2
b^2 = 2*c^2

hence (b^2) is also even, and therefore (b) is even as well.

now, if (a) and (b) are even numbers, they are both multiples of 2, which means we can reduce them into a simpler form.

a = 2*c
b = 2*d
a/b = (2*c)/(2*d) = c/d

so

a/b = c/d = sqr(2)

and this whole process will start over - we'll see that both (c) and (d) are even numbers.... etc. this means that the fraction that represents sqr(2) can be always reduced by 2, which contradicts one of the rules of rational numbers (can be reduced only a finite number of times).

therefore, sqr(2) cannot be represented as a rational number.

---

also, i came up with some hypothesis of my own.

now rational number squared can result in an integer, except for integers. e.g.

integer^2 = integer

rational-that-is-not-an-integer ^ 2 != will-never-be-an-integer

irrational ^ 2 = may-be-an-integer

like the square root of 2 is irrational, but squaring it results in an rational number (2)... while (a/b)^2, where a, b are integers, a!=b, and b!=1, will never result in an integer.

this hypothesis is quite interesting, because it will allow you to prove that any square root that is not an integer, is necessarily irrational.

so we can prove that sqr(2) is irrational, but how could you prove that sqr(3) is irrational just as well? that even-numbers trick will not work here....

so if my hypothesis is correct, you can say that any non-integer square root is irrational.

----

[Flexibal>
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Agamemnus
x/ \z

Posts: 3491

 « Reply #14 on: September 02, 2003, 12:48:15 PM »

Quote

3 + 3 = 6
6 - 3 = 3
3 * 3 = 9
9 / 3 = 3
3 + 3 - 3 * 3 / 3 = 3

is:

(a + a - a) * a / a

which is:

a + a - a

which is:

a.

Erroneously assuming it is:

"a + a - a * a / a = a + a - a = a" gives the same answer. But it's WRONG!
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