redngold
Posts: 6673
Joined: Wed Mar 22, 2000 12:26 pm

Another Math Mind-bender

Fri Sep 29, 2000 8:01 am

0.99999... = 1

Let x = 0.99999...
Therefore,
10x = 9.99999....

10x - x = 9x
Therefore,
9.99999... - 0.99999.... = 9

9x = 9
Therefore,
x = 1

That's the Algebraic conundrum.
If you want to know how to solve this seeming paradox, take Calculus!

redngold
Up, up and away!
 
Pacific
Posts: 1043
Joined: Fri Mar 10, 2000 2:46 pm

RE: Another Math Mind-bender

Fri Sep 29, 2000 9:23 am

This is in the British curricilum for secondary 5 or 6. People taking GCSE maths have to learn it so I know how to do it!  
 
redngold
Posts: 6673
Joined: Wed Mar 22, 2000 12:26 pm

RE: Another Math Mind-bender

Fri Sep 29, 2000 11:22 am

That's cool. But can you prove it correct or incorrect? We can see easily that algebra proves a seeming paradox. What does calculus have to say about this? One word: LIMITS!

redngold
Up, up and away!
 
Pacific
Posts: 1043
Joined: Fri Mar 10, 2000 2:46 pm

RE: Another Math Mind-bender

Fri Sep 29, 2000 11:38 am

I've just started my British A-Level maths course in September so I've not learnt calculus yet! We don't do calculus in GCSE!
 
oxygen
Posts: 633
Joined: Tue Sep 14, 1999 12:27 am

RE: Another Math Mind-bender

Sun Oct 01, 2000 12:24 am

x=0.999999999999999........= lim[x>1] x
10x-x
= 10 lim[x>1] x - lim[x>1]
= 9 lim[x>1]
= 9 (1)
= 9

So 9x Tends to 9.
 
flyf15
Posts: 6633
Joined: Tue May 18, 1999 11:10 am

RE: Another Math Mind-bender

Sun Oct 01, 2000 1:40 am

I loved learning that back in Calc 1. Aren't limits fun?  

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