Let us take a+b=c where a,b,c are constants
=>3a+3b=3c
=>3a+3b-3c=0
now if a+b=c
then 4a+4b=4c
=>4a+4b-4c=0
3a+3b-3c=4a+4b-4c
=> 3(a+b-c)=4(a+b-c)
=> 3=4 dividing both sides with a+b-c.
Thanks
2007-09-28 06:32:38
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answer #1
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answered by Anurag M 1
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If the question mark(?) after 4 would have the value of 0.75
i.e., 3 = 4?
=> 3 = 4 x 0.75
=> 3 = 3
=> hence proved mathematically......
2007-09-28 07:50:37
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answer #2
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answered by Anonymous
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Peter's proof is technically improper because the only time 2x=x is when 0=0, and division by zero is implicitly improper.
However, if you do permit division by zero then it does work, but you should be explicit about it. The only time you can get to 3=4 is when division by zero is permitted.
A more explicit division by zero proof goes as follows:
Let x=0.
Since x=0 then x+x=0+x and since you want 3=4 it is also true that x+x+x+x=0+x+x+x and it can further be shown to be true that nx=(n-1)x+0...still let us stick with 3=4
So, since x+x+x+x=0+x+x+x it is true that 4x=0+3x.
Since 4x=0+3x, 4x=3x.
Since division by 0 is explicitly permissible
4x/x=3x/x
It follows that 4=3.
More generally in the n case, it can be shown that if division by zero is permissible it can be shown by varying the number of explicit zeros, that all numbers equal all other numbers.
2007-09-28 08:23:51
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answer #3
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answered by OPM 7
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yes, very much possible
3=4 (mod 7)
check:25 divided by 7 =3 rem 4
so in modular arithmetic or congruence (in # theory) 3=4 (mod 7). to have a very clear view of my point, study modular arithmetic or congruence in number theory which is credited to one of my favourite mathematicians pierre de fermat,
also 5=0 (mod 3) 15/3 =5 rem. 0
2007-09-28 10:17:04
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answer #4
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answered by if not jehovah 2
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let x=1 and y=1
now x=y
subtract y on both sides
x-y = y-y => x-y = 0
devide by x-y on both sides
(x-y)/(x-y) =0/(x-y)
=> 1=0
add 3 to both sides
3=4
this is the only way to do
but actually we are deviding by (x-y) which is 0
we should not devide by 0 so we are getting abrupt answers
2007-09-28 13:11:33
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answer #5
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answered by shuaib 2
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Well, let's start
x^2 (squared)= x+x+x+x+...(x times)... +x
Let us take a derivative
2x=1+1+1+1+1+...(x times)...+1
2x=x
2=1
2+2=1+2
4=3
That's it!
2007-09-28 08:05:31
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answer #6
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answered by Peter Lustmolch 2
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3 = George Bush
4 = Osama bin Laden
3+4 = WAR
4-WAR = George Bush
Inverse of 4 = Good
So if George Bush = Good...
3=4.
Have a nice day.
2007-09-28 07:50:25
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answer #7
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answered by Anonymous
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Multiply both sides by 0!!!
2007-09-28 08:17:37
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answer #8
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answered by rohit s BOSS... 2
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its possible only if you consider complex roots....
and in that case you cannot prove it using real nos.......
which implies that the proof is only imaginary......
2007-09-28 10:02:41
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answer #9
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answered by KG the Play Maker 1
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You and your maths ---------- Thou art grear.
2007-09-28 07:55:21
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answer #10
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answered by brij_26pal 3
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