o-level maths theory...ruined by chem master

[Chem_master]

x=0.999999......
Multiply 100
100x=99.99999......
Subtract:
99x=99
x=1----> X is 0.99999 not 1....theory FAILED!

In case 2
x=1
in an eq
2x-2=x-1
2(x-1)=(x-1)
2(x-1)/(x-1)=1
2=1 ----------> 2=1!!! Basic maths FAILED!

comin up with other cases so be ready...........

 

[XPFMember]

:?

 

[Chem_master]

confused!!!!???

 

[Silent Hunter]

in case 1 , subtract what???

 

[Chem_master]

subtract x from 100x
subtract 0.999999..... from 99.9999999

 

[TSZ]

dude proving this wrng iz so easy...cum on...grow up

 

[abcde]

but how can maths fail!!!???????

 

[OakMoon!]

First of all in case 2, x-1 is equal to 0 since it is 1. And you can write 0 in whatever way you want to, it remains a 0. You can't divide anything by 0 or if you multiply its 0. So it is 0=0 and not 2=1. Maths won't fail that easily.
Case 1 is crap.

 

[abcde]

wait a sec, i was just about to say the same!

 

[Chem_master]

case 2 has been correctly proved wrong:
case 1 is not crap .....study carefully if u wud......case 1 is a method to convert decimals such as 0.67676767.....or 0.345345..... into fractions.....

 

[abcde]

In case one, the 9's after the decimal point of x and of 100x are technically not equal! Only we can't see it so we imagine that even after multiplying x with 100 the recurring 9's will remain same but they don't. Actually, in 100x two 9's are missing after the decimal point. In other words x has two more 9s after d.p. than 100x. So when you subtract 0.9999.. from 99.9999 you should be doing this: 99.99999.. - 0.9999999.. = 98.9999901. Then 98.9999901/99 = 0.999999...

Maths emerges victorious again! Does it not?

 

[OakMoon!]

Actually, I don't get how you proved that x=1.
99.9999-1 is 98.9999 and not 99...

 

[XPFMember]

In case one, the 9's after the decimal point of x and of 100x are technically not equal! Only we can't see it so we imagine that even after multiplying x with 100 the recurring 9's will remain same but they don't. Actually, in 100x two 9's are missing after the decimal point. In other words x has two more 9s after d.p. than 100x. So when you subtract 0.9999.. from 99.9999 you should be doing this: 99.99999.. - 0.9999999.. = 98.9999901. Then 98.9999901/99 = 0.999999...

Maths emerges victorious again! Does it not?

Assalamoalaikum!!

well done @abcde

 

[Chem_master]

abcde ..im afraid ur wrong, the number 0.999...... means 0.999999999999999999999999999999999999999999 and so on. In other words an endless chain of 9's so it does not depends on the number of 9's there r after the decimal..
hamidali391: 99.99999........ is subtracted from 0.99999 not 1

 

[XPFMember]

see..99.999....... is obtained by multiplying 0.9999....with 100...so that means decimal moved 2 places ahead..hence it'll always have 2 9s less than 0.999...

isnt it?

 

[abcde]

@chem master: yes but that series of 9s in x have two more 9's than 100x! that is obvious.

 

[Chem_master]

yeah that is but the decimal shows about million 9's ....removing to 9's gives us a minute change....which is ignorable..

 

[abcde]

If you can ignore that "minute" change, i dont see why you cant ignore the same small change between 1 and 0.999999999999............
If you dont ignore it, then kindly accept that your theory breaking has failed. Thankyou again for giving us the opportunity to exercise our brains. i would like to see more of this coming from you in future.

 

[Chem_master]

well.....u guyz almost got it......the problem is with the step when we multiply the decimal value by 100.....it will always have two 9's less then the original one ...gud work

 

[Nibz]

U better ruin your CHEMISTRY theories