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# This is mathematics

Discussion in 'JF Chit-Chat' started by Decapitator, Dec 7, 2011.

1. ### DecapitatorJF-Expert Member

#1
Dec 7, 2011
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What is an error in these calcns?

a+b=c

4a-3a+4b-3b=4c-3c

4a+4b-4c=3a+3b-3c

4(a+b-c)=3(a+b-c)

deviding by (a+b-c) throughout
4=3

2. ### SMUJF-Expert Member

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But a+b-c=0!

3. ### KongoshoJF-Expert Member

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4. ### Graph TheoryJF-Expert Member

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5. ### mdaumieMember

#5
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from a+b=c,means a+b-c=0.
then
4(a+b-c)=3(a+b-c) will be 4(0)=3(0),you can't divide by 0,but you can multiply by 0.
so the end result will 0=0

6. ### mafiakisiwaniJF-Expert Member

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7. ### sijui niniJF-Expert Member

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a+b=c

4a-3a+4b-3b=4c-3c

4a+4b-4c=3a+3b-3c

4(a+b-c)=3(a+b-c)

deviding by (a+b-c) throughout
4=3

so ..

8. ### ShineJF-Expert Member

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9. ### Tz-guyJF-Expert Member

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Source?

10. ### NEW NOELJF-Expert Member

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11. ### MapondelaJF-Expert Member

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12. ### Wa kusomaJF-Expert Member

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13. ### HusninyoJF-Expert Member

#13
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