beenjammin2lp
Registered Member
i heard somewhere that a professor found a way through a long drawn out process how 1+1 does not equal 2....
has anybody seen this/ heard about it?
has anybody seen this/ heard about it?
1+1=?
- Thread starter beenjammin2lp
- Start date
no. but i have seen .9999 repeating proven to equal 1. :m:
I vaguely remember something . . .
Something about factorials. Like putting an exclaimation point next to a number to mean multiply that number with all the numbers before it. For example, 5!=5x4x3x2x1.
I don't remember exactly how this related to 1+1=?. I don't even remember if it relates. I'm very tired, so I'm sorry I couldn't help out.
Something about factorials. Like putting an exclaimation point next to a number to mean multiply that number with all the numbers before it. For example, 5!=5x4x3x2x1.
I don't remember exactly how this related to 1+1=?. I don't even remember if it relates. I'm very tired, so I'm sorry I couldn't help out.
One and 1 make 2 in the base-10 number system, not in all number systems
1 cup plus 1 cup does not always equal 2 cups.
Mix a cup of vinegar with a cup of a baking soda solution. The result will be less than 2 cups of liquid, as some molecules are transformed into carbon dioxide and released into the air as gas.
1 cup plus 1 cup does not always equal 2 cups.
Mix a cup of vinegar with a cup of a baking soda solution. The result will be less than 2 cups of liquid, as some molecules are transformed into carbon dioxide and released into the air as gas.
Blue_UK said:Try this one
a = b
aa = ab { Multiply by 'a' }
aa + aa - 2ab = ab + aa - 2ab { Add 'aa - 2ab' }
2(aa - ab) = aa - ab { rearrange }
2 = 1 { existance crashes }
Can't parse the last line, I keep getting a divide by 0 error.
F
fetus_fajitas
Guest
2+2=5
If anyone's read 1984
If anyone's read 1984
AndersHermansson
Registered Senior Member
Blue_UK said:Try this one
a = b
aa = ab { Multiply by 'a' }
aa + aa - 2ab = ab + aa - 2ab { Add 'aa - 2ab' }
2(aa - ab) = aa - ab { rearrange }
2 = 1 { existance crashes }
at the second last row you have 2(aa-ab)=aa-ab, which is true only if aa-ab=0, the last line then becomes 2*0=1*0, which is obviously true. If aa-ab is not equal to 0, then the second last row is not true, which then means that your last line 2=1 is not true, because they are both equivalent.
And this is the answer to this and many other impossible math problems. I had a teacher that showed me an equation that 'proved' that 1=-1 and I was awed and promptly forgot the exact formula but remembered the concept and had this idea in my head of antimatter and such
p). Well, one day when I was older, I went back to the teacher (actually sent my younger brother who was in his class at the time) to go fetch the formula for me and I looked it up on the net. Guess how he did it? Divide by zero.
When looking for it, I found a site that was filled with 'impossible' equations. That was a few years back though and I don't remember it. Maybe someone else knows a link to a similar site?
beenjammin2lp said:
If '1' & '1' represent units that start off inqeual then adding them
will not equal two identical units.
Been doing a bit of digging. Found a few erroneous proofs.
Theorem : 3=4
Proof:
Suppose:
a + b = c
This can also be written as:
4a - 3a + 4b - 3b = 4c - 3c
After reorganising:
4a + 4b - 4c = 3a + 3b - 3c
Take the constants out of the brackets:
4 * (a+b-c) = 3 * (a+b-c)
Remove the same term left and right:
4 = 3
Theorem : All numbers are equal to zero.
Proof: Suppose that a=b. Then
a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
a + b = b
a = 0
Theorem: 1$(dollar) = 1c(cent).
Proof:
And another that gives you a sense of
money disappearing...
1$ = 100c
= (10c)^2
= (0.1$)^2
= 0.01$
= 1c
Theorem: 1 = -1 .
Proof:
1/-1 = -1/1
sqrt[ 1/-1 ] = sqrt[ -1/1 ]
sqrt[1]*sqrt[1] = sqrt[-1]*sqrt[-1]
ie 1 = -1
Theorem: 4 = 5
Proof:
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5
Source.
Theorem : 3=4
Proof:
Suppose:
a + b = c
This can also be written as:
4a - 3a + 4b - 3b = 4c - 3c
After reorganising:
4a + 4b - 4c = 3a + 3b - 3c
Take the constants out of the brackets:
4 * (a+b-c) = 3 * (a+b-c)
Remove the same term left and right:
4 = 3
Theorem : All numbers are equal to zero.
Proof: Suppose that a=b. Then
a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
a + b = b
a = 0
Theorem: 1$(dollar) = 1c(cent).
Proof:
And another that gives you a sense of
money disappearing...
1$ = 100c
= (10c)^2
= (0.1$)^2
= 0.01$
= 1c
Theorem: 1 = -1 .
Proof:
1/-1 = -1/1
sqrt[ 1/-1 ] = sqrt[ -1/1 ]
sqrt[1]*sqrt[1] = sqrt[-1]*sqrt[-1]
ie 1 = -1
Theorem: 4 = 5
Proof:
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5
Source.
Is this what your looking for?
Examine the sequence:
1 - .9
1 - .99
1 - .999
1 - .9999
.
.
.
This sequence can be rewritten as:
.1
.01
.001
.0001
.
.
.
So, we can use induction to deduce that the limit of both sequences is 0.
The limit is achieved when n (the number of digits right of the decimal) reaches infinity.
So, as n reaches infinity, the sequences are as follows:
1 - .9999999999999999999999999999999999.......................
And:
0
We've stated the sequences' equilinity, such that we can deduce (by induction, once again):
1 - .999999999999999999999999999999999999.................. = 0
The Arithmetic property of addition gives us the equality:
1 = .999999999999999999999999999999999999999999999999........... ....
~ Stax
Examine the sequence:
1 - .9
1 - .99
1 - .999
1 - .9999
.
.
.
This sequence can be rewritten as:
.1
.01
.001
.0001
.
.
.
So, we can use induction to deduce that the limit of both sequences is 0.
The limit is achieved when n (the number of digits right of the decimal) reaches infinity.
So, as n reaches infinity, the sequences are as follows:
1 - .9999999999999999999999999999999999.......................
And:
0
We've stated the sequences' equilinity, such that we can deduce (by induction, once again):
1 - .999999999999999999999999999999999999.................. = 0
The Arithmetic property of addition gives us the equality:
1 = .999999999999999999999999999999999999999999999999........... ....
~ Stax
beenjammin2lp
Registered Member
well..ok...but i got another question... why dont we base our number system on negative numbers, numbers between .000~~~001 and .999~~~999, and all positive intergers...it eems easier to organize them like that seeins how all decimals/ fractions draw numbers toward them...and vice versa for whole numbers...only whole numbers dont really have an end...come to think...neither do decimals...hmmm
I have heard of (1+1 !== 2).
But it was a trick question. It was in base 2, not base 10. In the binary (base 2) system, 1 + 1 = 10. I think. I am a bit rusty on my binary calculations. Anyway, it doesn't equal 2.
But it was a trick question. It was in base 2, not base 10. In the binary (base 2) system, 1 + 1 = 10. I think. I am a bit rusty on my binary calculations. Anyway, it doesn't equal 2.
Anyway, I have also heard of .999999999999 (repeating) equaling one. In my 7th grade math class, we learned the procedure to change repeating decimals into exact quantities. This was years ago, so I can't remeber it. When .33333 was put into the procedure, it came out in fraction form as exactly 1/3. This worked for all others, and strangely enough, .9999 = 1. I find this very odd.
It would seem that if you could get a function to eventually output .99999999 it would have a vertical asymptote at 1. Its limit would be 1, but the function would never be defined there. I find this extremely odd.