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Doominabox

no avatar Level 20 Comments: Peasant
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Date Signed Up:10/08/2010
Last Login:2/01/2013
Funnyjunk Career Stats
Content Thumbs: 44 total,  116 ,  72
Comment Thumbs: 71 total,  645 ,  574
Content Level Progress: 81.35% (48/59)
Level 0 Content: Untouched account → Level 1 Content: New Here
Comment Level Progress: 0% (0/1)
Level 20 Comments: Peasant → Level 21 Comments: Peasant
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Content Views:2923
Total Comments Made:321
FJ Points:127

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#789 - just to watch him die? 01/02/2013 on What's one secret that... 0
#27 - /watch?v=1IKbtBbl3EA 11/19/2012 on too soon? 0
#595 - Obviously a lie, god 11/19/2012 on 10% 0
#166 - obviously im not going to accept your theory, and you mine, so… 11/18/2012 on TODAY 0
#164 - the eqation wrote should have said either 9=10x-.9.. …  [+] (2 new replies) 11/18/2012 on TODAY 0
User avatar #165 - GeorgeBush (11/18/2012) [-]
That still doesn't mean that in the end .999... doesn't = 1, because it does. I can't explain it as well as my teacher could, but it's basic mathematic theory that they, and all other infinitely repeating integers with .999... equals the same number rounded up. 46.999... = 47, etc
#166 - Doominabox (11/18/2012) [-]
obviously im not going to accept your theory, and you mine, so lets both just stop talking
#161 - what you're not getting is that x≠.99999.... because she cha…  [+] (4 new replies) 11/18/2012 on TODAY 0
User avatar #163 - GeorgeBush (11/18/2012) [-]
She didn't change what x equaled, she changed the equation x was used in.

In math you always do to one side of the equation what you do to the other, as I'm sure you know. She simply multiplied the starting equation by 10 on each side.

(1) x = .999... multiplied by 10 is
10x = 9.999... , the value of x did not change.
If you then divide the equation 10x = 9.999... (in which you claim x no longer equals .999...) by 10 you get
(1) x = .999... again, meaning that in the two equations x remains constant.
#164 - Doominabox (11/18/2012) [-]
the eqation wrote should have said either
9=10x-.9..
or 9-x=9x

just because it was writen like that, doesn't mean its true

X=5
50X=50 (times it by 10)
49X=49 (i subtracted one from each side)
X=1 (divided by 49)
and presto 5=1
User avatar #165 - GeorgeBush (11/18/2012) [-]
That still doesn't mean that in the end .999... doesn't = 1, because it does. I can't explain it as well as my teacher could, but it's basic mathematic theory that they, and all other infinitely repeating integers with .999... equals the same number rounded up. 46.999... = 47, etc
#166 - Doominabox (11/18/2012) [-]
obviously im not going to accept your theory, and you mine, so lets both just stop talking
#159 - https://www. [url deleted] ?feature=player_embedded &v… 11/18/2012 on TODAY 0
#158 - Comment deleted 11/18/2012 on TODAY 0
#157 - exactly thats what she did wrong  [+] (6 new replies) 11/18/2012 on TODAY 0
User avatar #160 - GeorgeBush (11/18/2012) [-]
No, she did nothing wrong. Every step of the equation had the same process done unto both sides. Subtracting x and subtracting .999... is the same processes, using the same amount, meaning that it is a perfectly valid. They said x = .999... meaning subtracting .999.... and subtracting x makes no difference as they are the same number. The reason they subtracted x on one side and .999... on one side is because one side (left) already had x as a variable, and the right side did not.
#161 - Doominabox (11/18/2012) [-]
what you're not getting is that x≠.99999.... because she changed it after the first step

you would not subtract x from the other side the actual equation would be
9=10x-.99999....
wich is not the same as
9=9x
User avatar #163 - GeorgeBush (11/18/2012) [-]
She didn't change what x equaled, she changed the equation x was used in.

In math you always do to one side of the equation what you do to the other, as I'm sure you know. She simply multiplied the starting equation by 10 on each side.

(1) x = .999... multiplied by 10 is
10x = 9.999... , the value of x did not change.
If you then divide the equation 10x = 9.999... (in which you claim x no longer equals .999...) by 10 you get
(1) x = .999... again, meaning that in the two equations x remains constant.
#164 - Doominabox (11/18/2012) [-]
the eqation wrote should have said either
9=10x-.9..
or 9-x=9x

just because it was writen like that, doesn't mean its true

X=5
50X=50 (times it by 10)
49X=49 (i subtracted one from each side)
X=1 (divided by 49)
and presto 5=1
User avatar #165 - GeorgeBush (11/18/2012) [-]
That still doesn't mean that in the end .999... doesn't = 1, because it does. I can't explain it as well as my teacher could, but it's basic mathematic theory that they, and all other infinitely repeating integers with .999... equals the same number rounded up. 46.999... = 47, etc
#166 - Doominabox (11/18/2012) [-]
obviously im not going to accept your theory, and you mine, so lets both just stop talking
#156 - also, you have to do exactly the same on both sides, so when t… 11/18/2012 on TODAY 0
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