1 = 0
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Take your pick: How Can 0.999... = 1? | Purplemath[^] For example:
x = 0.999...10x = 9.999...
10x - x = 9.000...
9x = 9
x = 1EDIT: As Nagy said[^], 2+ hours ago.
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
1/6 = 0,166666 ->*6 0,9996
x = 0,9996
10x = 9,996
10x -x = 8,9964
9x = 8,9964
x = 0,9996?
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HobbyProggy wrote:
1/6 = 0,166666 ->*6 0,9996
Nope -
6 × 0.1666... = 0.9999...It's simple multiplication:1 x 6 = 6 (0.6)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.96)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.996)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.9996)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.99996)
etc.Because you're repeating the operation an infinite number of times, there's no point where you stop and leave the last digit as
6. There's always another digit to the right which needs to be multiplied by6, carrying the3into the current column.
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
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HobbyProggy wrote:
1/6 = 0,166666 ->*6 0,9996
Nope -
6 × 0.1666... = 0.9999...It's simple multiplication:1 x 6 = 6 (0.6)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.96)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.996)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.9996)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.99996)
etc.Because you're repeating the operation an infinite number of times, there's no point where you stop and leave the last digit as
6. There's always another digit to the right which needs to be multiplied by6, carrying the3into the current column.
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
So this works only for 0.999999 ? Because the 6 at the end is in this case important to show that it wont be a 1 in the end :) But nevermind, this is all just playing with numbers :)
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So this works only for 0.999999 ? Because the 6 at the end is in this case important to show that it wont be a 1 in the end :) But nevermind, this is all just playing with numbers :)
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}HobbyProggy wrote:
Because the 6 at the end is ...
But there isn't a
6at the "end", because there isn't an "end"! :laugh: Think of it like this:using System;
static class Program
{
static void Main()
{
while (true)
{
Console.WriteLine("9");
}Console.WriteLine("Squirrel!"); }}
If you run that program, how long will you have to wait before it prints "Squirrel"?
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
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HobbyProggy wrote:
Because the 6 at the end is ...
But there isn't a
6at the "end", because there isn't an "end"! :laugh: Think of it like this:using System;
static class Program
{
static void Main()
{
while (true)
{
Console.WriteLine("9");
}Console.WriteLine("Squirrel!"); }}
If you run that program, how long will you have to wait before it prints "Squirrel"?
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
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Do you mean they are imaginary? :doh: :doh: :doh: This is getting more and more difficult by moments. I think I am grabbing a bier, maybe seeing it double makes it easier :rolleyes: ;P :laugh: :laugh: I'll get my coat
M.D.V. ;) If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about? Help me to understand what I'm saying, and I'll explain it better to you Rating helpful answers is nice, but saying thanks can be even nicer.
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So this works only for 0.999999 ? Because the 6 at the end is in this case important to show that it wont be a 1 in the end :) But nevermind, this is all just playing with numbers :)
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}mmm working by 9, not working by 6 what about 69? does it works? Oh, wait... I think I stop now since I am about to break the KSS rule :laugh: :laugh: I am still on the way getting my coat
M.D.V. ;) If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about? Help me to understand what I'm saying, and I'll explain it better to you Rating helpful answers is nice, but saying thanks can be even nicer.
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Ah, the ol' divide by zero trick. I think my math teacher did that when I was in 8th grade. Almost 40 years ago. And I'm sure it's older than that. ;) Marc
Imperative to Functional Programming Succinctly Contributors Wanted for Higher Order Programming Project!
Since I remember it from math about that time it has to be old. :-D How can I remember this but cannot find my car keys?
Mongo: Mongo only pawn... in game of life.
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Dominic Burford wrote:
x2 - y2 = xy - y2. Dividing by (x-y), obtain x + y = y.
Wrong... aint it? x2 -y2 / (x-y) => x+y xy -y2 / (x-y) => y+y xy / x-y => 1*1/1-1 => (actually how could you divide this when there is a minus...) y2 / x-y => y/x-1 => (actually how could you devide this when there is a minus...) But i guess thats bullshit 2 because it "should"? be x2/x-y - y2/x-y which ruins everything :) Okay screw everything... you are wrong, thats all i can say :)
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Take your pick: How Can 0.999... = 1? | Purplemath[^] For example:
x = 0.999...10x = 9.999...
10x - x = 9.000...
9x = 9
x = 1EDIT: As Nagy said[^], 2+ hours ago.
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
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The only thing i can agree with is that 1 != 3/3 (at least not exactly) because 1/3 is 0,333... and multiplied with 3 it is just 0,99999.... which is technically 1 but not 100%
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}Actually, yes 0.9999..... is in fact 1: x = 0.99999.... 10 x = 9.999999.... multiply by 10 10 x = 9 + 0.99999.... split right side into arithmetic expression 10 x = 9 + x replace 0.9999.... with "x" 9 x = 9 subtract "x" from both sides x = 1 divide by 9. without dividing by "0"
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x = y. Then x2 = xy. Subtract the same thing from both sides: x2 - y2 = xy - y2. Dividing by (x-y), obtain x + y = y. Since x = y, we see that 2 y = y. Thus 2 = 1, since we started with y nonzero. Subtracting 1 from both sides, 1 = 0. :wtf:
"There are two ways of constructing a software design: One way is to make it so simple that there are obviously no deficiencies, and the other way is to make it so complicated that there are no obvious deficiencies. The first method is far more difficult." - C.A.R. Hoare Home | LinkedIn | Google+ | Twitter
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x = y. Then x2 = xy. Subtract the same thing from both sides: x2 - y2 = xy - y2. Dividing by (x-y), obtain x + y = y. Since x = y, we see that 2 y = y. Thus 2 = 1, since we started with y nonzero. Subtracting 1 from both sides, 1 = 0. :wtf:
"There are two ways of constructing a software design: One way is to make it so simple that there are obviously no deficiencies, and the other way is to make it so complicated that there are no obvious deficiencies. The first method is far more difficult." - C.A.R. Hoare Home | LinkedIn | Google+ | Twitter
The second statement is false. if x = y, then x2 does not equal xy; x to the second power equals xy. For example: if 1=1 then 1(2) does not equal 1(1) Bullshit from the start, not to mention the division by zero
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The only thing i can agree with is that 1 != 3/3 (at least not exactly) because 1/3 is 0,333... and multiplied with 3 it is just 0,99999.... which is technically 1 but not 100%
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}Am I the only one that doesn't see the issue that we are not dividing 1 by 3.... 1 = 3/3... If you're going to divide by 3 on the 1... wouldn't you need to also need to divide 3/3 by 3 also... 1/3 = (3/3)/3.... which would both be .03333... ??
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x = y. Then x2 = xy. Subtract the same thing from both sides: x2 - y2 = xy - y2. Dividing by (x-y), obtain x + y = y. Since x = y, we see that 2 y = y. Thus 2 = 1, since we started with y nonzero. Subtracting 1 from both sides, 1 = 0. :wtf:
"There are two ways of constructing a software design: One way is to make it so simple that there are obviously no deficiencies, and the other way is to make it so complicated that there are no obvious deficiencies. The first method is far more difficult." - C.A.R. Hoare Home | LinkedIn | Google+ | Twitter
Sorry guys but second line is false its only true when x=1 or 2 x = y. Then x2 = xy is wrong if x=3 , xy=9 x2=6 6 !=9 you can't divide this way if its proper algebra left side is x2 - y2 2(x-y) ------- = ------ = 2 <- left side of the equation (x-y) (x-y) and obviously you cant separate (x-y) from the right side (xy - y2)/(x-y) != y sorry guys example is flawed and 1 != 0
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x = y. Then x2 = xy. Subtract the same thing from both sides: x2 - y2 = xy - y2. Dividing by (x-y), obtain x + y = y. Since x = y, we see that 2 y = y. Thus 2 = 1, since we started with y nonzero. Subtracting 1 from both sides, 1 = 0. :wtf:
"There are two ways of constructing a software design: One way is to make it so simple that there are obviously no deficiencies, and the other way is to make it so complicated that there are no obvious deficiencies. The first method is far more difficult." - C.A.R. Hoare Home | LinkedIn | Google+ | Twitter
Unfortunately, this "proof" falls apart at the 2nd line. x2 = xy only holds true for 0, 2, and -2. The correct equation is x^2 = xy. Still, a pretty interesting use of mathematics in an attempt to destroy all of our beliefs in what is thought to be true. :thumbsup:
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x = y. Then x2 = xy. Subtract the same thing from both sides: x2 - y2 = xy - y2. Dividing by (x-y), obtain x + y = y. Since x = y, we see that 2 y = y. Thus 2 = 1, since we started with y nonzero. Subtracting 1 from both sides, 1 = 0. :wtf:
"There are two ways of constructing a software design: One way is to make it so simple that there are obviously no deficiencies, and the other way is to make it so complicated that there are no obvious deficiencies. The first method is far more difficult." - C.A.R. Hoare Home | LinkedIn | Google+ | Twitter
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x = y. Then x2 = xy. Subtract the same thing from both sides: x2 - y2 = xy - y2. Dividing by (x-y), obtain x + y = y. Since x = y, we see that 2 y = y. Thus 2 = 1, since we started with y nonzero. Subtracting 1 from both sides, 1 = 0. :wtf:
"There are two ways of constructing a software design: One way is to make it so simple that there are obviously no deficiencies, and the other way is to make it so complicated that there are no obvious deficiencies. The first method is far more difficult." - C.A.R. Hoare Home | LinkedIn | Google+ | Twitter
huh? I don't think x = y implies x2 = xy. How did you get that?
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The only thing i can agree with is that 1 != 3/3 (at least not exactly) because 1/3 is 0,333... and multiplied with 3 it is just 0,99999.... which is technically 1 but not 100%
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}HobbyProggy wrote:
because 1/3 is 0,333
No. 1/3 does not equal 0.333. 1/3 is APPROXIMATELY 0.333. and 0.99999 is not technically 1. It is APPROXIMATELY 1 There is even a different symbol for it. 1/3≈0.333 (1/3 is approximately equal to 0.333) 0.99999≈1 (0.99999 is approximately equal to 1)
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