wilbald
JF-Expert Member
- Dec 17, 2007
- 1,810
- 1,378
In football we have two types of CARDS:
1st-YELLOW CARD
2nd-RED CARD
NOW THEN here i go.....(1 yellow card)+(1 yellow card)=(1red card)
Therefore, (1) + (1) = (1)
Hence proved.
Nadhani amekusoma(garbage in,garbage out)
If a = 1 and b = 1 therefore i could say that .....
a = b and that
a^2 = ab where
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b) dividing both sides by (a-b) we get:
a+b=b substituting that in equation 1 makes
1+1=1 as in 2=1 ?!!
A concrete proof about this is here under:
In football we've two types of CARDS.
A YELLOW CARD AND A RED CARD.
1 yellow card + 1 yellow card = 1 red card.
this is to say that 1+1=1.
Hence proved.
If a = 1 and b = 1 therefore i could say that .....
a = b and that
a^2 = ab where
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b) dividing both sides by (a-b) we get:
a+b=b substituting that in equation 1 makes
1+1=1 as in 2=1 ?!!
In football we have two types of CARDS:
1st-YELLOW CARD
2nd-RED CARD
NOW THEN here i go.....(1 yellow card)+(1 yellow card)=(1red card)
Therefore, (1) + (1) = (1)
Hence proved.
[9x2=18,1+8=9],[9x5=45,4+5=9],[9x7=63,6+3=9],[9x11=99,9+9=18,1+8=9],[9x29=261,2+6+1=9],[9x219=1971,1+9+7+1=18,1+8=9] and so on and on and on...keep trying with other intergers.
This is only true in base ten system, and not universal.
Because it is not universal, it is not as impressive as it is made out to be.
There is more to Mathematics than base ten mathematics.
If a = 1 and b = 1 therefore i could say that .....
a = b and that
a^2 = ab where
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b) dividing both sides by (a-b) we get:
a+b=b substituting that in equation 1 makes
1+1=1 as in 2=1 ?!!
A concrete proof about this is here under:
In football we've two types of CARDS.
A YELLOW CARD AND A RED CARD.
1 yellow card + 1 yellow card = 1 red card.
this is to say that 1+1=1.
Hence proved.
ok lets come to the first case: if a=1 and b=1, i agree with all the assuptions and equations up to
(a+b)(a-b)=b(a-b), but when U say u, divide by (a-b) both side is meaningless coz a-b=0, so 2 can not be equal to 1!!
On 2nd case bout football cards.. Yes we have two cards, yellow card and red card!
When a player shown a yellow cards twice results in a red card! Still 1 + 1 can not be equal to 1! Since 2 yellow cards can be equivalent to 1 red card! And i can conclude that 2 can be equivalent to 1 and 2 can not be equal to 1. Sijui nyinyi mlikuwa mna-prove nini ati?
ok lets come to the first case: if a=1 and b=1, i agree with all the assuptions and equations up to
(a+b)(a-b)=b(a-b), but when U say u, divide by (a-b) both side is meaningless coz a-b=0, so 2 can not be equal to 1!!
Relax Dude.. we aint try to prove anything! we are just having fun with math! thts all! every one knows 1 +1 = 2 ... ni kiridi hapo kwenye Red my purpose were to make B a subject so that i can force my answer and prove 1 + 1 = 1 .. i created an equation based on a^n â b^n = (a â b)(an â 1 + an â 2b + an â 3b2 + ··· + abn â 2 + bn â 1) btw it is not possible to finish such an equation (a+b)(a-b)=b(a-b) easy like the way u did ..
anyways i can be wrong!
Allah Knows Best!
Hii kitu mathematically haikubaliki manake unagawanya na sifuri (0) kwa sababu a-b=1-1=0If a = 1 and b = 1 therefore i could say that .....
a = b and that
a^2 = ab where
a^2 - b^2 = ab - b^2
Code:(a+b)(a-b) = b(a-b) dividing both sides by (a-b) we get:
a+b=b substituting that in equation 1 makes
1+1=1 as in 2=1 ?!!
Hii kitu mathematically haikubaliki manake unagawanya na sifuri (0) kwa sababu a-b=1-1=0
Na hii je?,1^1=1,1^0=1,therefore:1^1=1^0.equating the bases,1=1.equating the exponents,1=0,hence proven that 1=0.
Na hii je?,1^1=1,1^0=1,therefore:1^1=1^0.equating the bases,1=1.equating the exponents,1=0,hence proven that 1=0.
From FIFA laws,two consecutive yellow cards are equivalent to a red card,that is right.From which mathematique rule/principal/law is 1 equivalent to 0?.1 is not equal to 0 and neither equivalent to 0,we are just making fun of maths.We proved wrong that proof telling us 2=1 by saying that,if a=b,then a-b=0.so that proof was not valid.Now we need to prove wrong this fallacy proof that 1=0.coz we know that,1^sin 30=1^cos 60,then sin 30=cos 60.Also 1^1=1^0,then 1=0.I know practically 1 is not equal to 0,now prove this fallacy proof wrong mathematically!mmh, that equation based in logic..
b'coz, logically one can say, if 1^1=1 and 1^0=1 then equating the bases comes 1=0. It makes sense huh!
Chukua huu mfano: Kuku alitamia mayai, logically tulitegemea atotoe vifaranga vya kuku, lakini yeye alitotoa vifaranga vya bata coz mayai yalikuwa ya bata, je vifaranga vya kuku ni sawasawa na vifaranga vya bata?
It's the same story about EQUAL and EQUIVALENT. So if 1^1=1, 1^0=1, then theoletically one may think 1=0, but pratically 1 may be equivalent to zero as it was to vifaranga vya bata to be equivalent to vifaranga vya kuku! So 1 can't be equal to zero except if you are making funny of maths!