one equal to two ?
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a=0 never means a/a=0 (a can never be 0). a=a <=> a/a=a/a <=> 1=1 I could fix the post as : Given a C ]--;0[ U ]0;++[ (meaning 0 excluded)
Which means that you have an equation system and not a single equation...It's a different thing to solve...
Skipper: We'll fix it. Alex: Fix it? How you gonna fix this? Skipper: Grit, spit and a whole lotta duct tape.
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Yes, but the damage is done before that.
You mean, when OP went to learn math?
Skipper: We'll fix it. Alex: Fix it? How you gonna fix this? Skipper: Grit, spit and a whole lotta duct tape.
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Let read this :
1=1 a=a a²=a² a²-a²=a²-a² a(a-a)=(a+a)(a-a) a=a+a a(1)=a(1+1) 1=1+1 1=2Where is the error ? -
Abbott and Costello said it better.
PIEBALDconsult wrote:
Abbott and Costello_, who were on first,_ said it better.
It was missing something.
I wanna be a eunuchs developer! Pass me a bread knife!
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Let read this :
1=1 a=a a²=a² a²-a²=a²-a² a(a-a)=(a+a)(a-a) a=a+a a(1)=a(1+1) 1=1+1 1=2Where is the error ?welcome to the lounge. I know this isn't quite your first post - but nearly. And I would like to apologise for the negativity your post received. For someone who hadn't seen that 'proof' before it may have been interesting - as you can see, not only have the majority here seen it (more than once!) but they like to stuff it down your throat - whether to big-note themselves or simply in an attempt to belittle you we cannot tell. They should be ashamed. Merry Xmas
PooperPig - Coming Soon
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Let read this :
1=1 a=a a²=a² a²-a²=a²-a² a(a-a)=(a+a)(a-a) a=a+a a(1)=a(1+1) 1=1+1 1=2Where is the error ?How about this... ;P Start with this: 1/9 = 1/9 Then convert one side to decimal equivalent (which is infinitely recurring) 1/9 = 0.11111111111111111111111111111111111111111111111111111111...(etc etc) Then multiply both sides by nine 1 = 0.9999999999999999999999999999999999999999999999999999999999...(etc etc) Therefore, 1 is equal to 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...(where the 9's are in infinite recursion). And yes, this actually is mathematically correct.
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welcome to the lounge. I know this isn't quite your first post - but nearly. And I would like to apologise for the negativity your post received. For someone who hadn't seen that 'proof' before it may have been interesting - as you can see, not only have the majority here seen it (more than once!) but they like to stuff it down your throat - whether to big-note themselves or simply in an attempt to belittle you we cannot tell. They should be ashamed. Merry Xmas
PooperPig - Coming Soon
Thank you Max and happy new Xear
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How about this... ;P Start with this: 1/9 = 1/9 Then convert one side to decimal equivalent (which is infinitely recurring) 1/9 = 0.11111111111111111111111111111111111111111111111111111111...(etc etc) Then multiply both sides by nine 1 = 0.9999999999999999999999999999999999999999999999999999999999...(etc etc) Therefore, 1 is equal to 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...(where the 9's are in infinite recursion). And yes, this actually is mathematically correct.
I agree but here 1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9 I could write 1/9~= 0.111111111111111111111111...... then 1 ~= 0.9999999999999999999999999...... ??
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I agree but here 1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9 I could write 1/9~= 0.111111111111111111111111...... then 1 ~= 0.9999999999999999999999999...... ??
tayoufabrice wrote:
1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9
Wish I could agree, but I can't... read all about it[^] :-D Even google 0.999999999999999 = 1[^] if you're still unconvinced.
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tayoufabrice wrote:
1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9
Wish I could agree, but I can't... read all about it[^] :-D Even google 0.999999999999999 = 1[^] if you're still unconvinced.
Ah là là :laugh: Mathematics !! (French laughing)
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Let read this :
1=1 a=a a²=a² a²-a²=a²-a² a(a-a)=(a+a)(a-a) a=a+a a(1)=a(1+1) 1=1+1 1=2Where is the error ?a(a-a) = (a+a)(a-a) // divide by (a-a), i.e. divide by 0 a = a+a Division by zero is a no-no because it can lead to "impossible" results like the above.
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a(a-a) = (a+a)(a-a) // divide by (a-a), i.e. divide by 0 a = a+a Division by zero is a no-no because it can lead to "impossible" results like the above.
I could fix the post as : Given a C ]--;0[ U ]0;++[ (meaning 0 excluded) Now ??
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I could fix the post as : Given a C ]--;0[ U ]0;++[ (meaning 0 excluded) Now ??
The value of a is irrelevant; a - a == 0, and factoring out a - a is division by 0, which is forbidden. I am not a mathematician, so I don't know if it is possible to create a self-consistent arithmetic in which division by 0 does not result in nonsensical results. All I know is that in the arithmetic I learnt in school it is forbidden.
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The value of a is irrelevant; a - a == 0, and factoring out a - a is division by 0, which is forbidden. I am not a mathematician, so I don't know if it is possible to create a self-consistent arithmetic in which division by 0 does not result in nonsensical results. All I know is that in the arithmetic I learnt in school it is forbidden.
Sure ! :laugh: number can never be divided by zero 0 ; even 0/0 :confused: It is the real error of my process
