Lolsmith
kill all boomers
Yeah, limiting sums in my 2U exam next week will make me giggle. Thanks for that guys. 
-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
Very interesting (1 Viewer)
- Thread starter hscishard
- Start date
ohexploitable
hey
I get like mathematically why it equals one but I just can't get my head around it
- Joined
- Feb 16, 2005
- Messages
- 8,391
- Gender
- Male
- HSC
- 2006
That is correct and legitimate. The reason being is this idea of limiting sums. The notion of 0.999... represents an INFINITE number of decimal places. In theory this means it should equal 1 "at the point of infinity".
The reason why most people find this hard to accept is because they always see 0.999... with a finite number of decimal places. If it has a finite number of decimal places it would obviously not be equal to 1.
It's only equal to 1 when there is an INFINITE number of decimal places.
Infinity is NOT a number. It is meant to be treated as a limit when something increases positively without bound. The argument that since the y-axis and x-axis are perpendicular therefore "∞ x 0 = - 1" is enough to incite pure mathematicians to murder you lol
Here is a contradiction to disprove the argument, if
∞ x 0 = - 1
=> ∞ x 0 x - 1 = 1
But 0 x - 1 = 0
=> ∞ x 0 = 1
.: 1 = - 1
which is clearly false, so the initial assumption that ∞ x 0 = - 1 is clearly incorrect
Perhaps a more appropriate definition of two lines with gradients m1 and m2 being perpendicular would be:
m1 = - 1 / m2 if m2 =/= 0
OR
m2 = - 1 / m1 if m1 =/= 0
This should be a better formula which works when one of them equals zero and the other approaches infinity. This means the easier to remember high school result:
m1m2 = - 1
only holds when both m1 and m2 are non-zero and finite
Last edited:
untouchablecuz
Active Member
- Joined
- Mar 25, 2008
- Messages
- 1,693
- Gender
- Male
- HSC
- 2009
shame on you
trebla always has the best mathematical posts
trebla always has the best mathematical posts
Who says 1 = -1 is false, afterall:
<img src="http://www.codecogs.com/eq.latex?\begin{align*}i&=i\\\sqrt{-1}&=\sqrt{-1}\\\sqrt{\frac{-1}{1}}&=\sqrt{\frac{1}{-1}}\\\frac{\sqrt{-1}}{\sqrt{1}}&=\frac{\sqrt{1}}{\sqrt{-1}}\\\sqrt{-1}&=\frac{1}{\sqrt{-1}}\\(\sqrt{-1})^{2}&=1\\\therefore -1&=1\end{align}">
For those who take maths a bit too seriously, yes there is a false statement in the arguement but it is pretty hard to find (it took me a few years before I could find someone to point it out)
The problem is that the square root function is multi-valued. sqrt(-1) = i or -i for instance. If you're always choosing i as "the" square root of negative 1, then 1 = sqrt(1) = sqrt(-1 * -1) != sqrt(-1) * sqrt(-1) = i * i = -1 is an obvious counterexample to the "rule" sqrt(ab) = sqrt(a)sqrt(b), which you've invoked in your third to fourth lines.
Choosing the "correct" value -i on one of the sides, the right side say, in line four fixes your working, though anyone seriously working through a calculation like this would just note the multi-valuedness of sqrt and deduce the "correct" value if enough information is given to do so (as we have done).
Choosing the "correct" value -i on one of the sides, the right side say, in line four fixes your working, though anyone seriously working through a calculation like this would just note the multi-valuedness of sqrt and deduce the "correct" value if enough information is given to do so (as we have done).
hscishard
Active Member
Nice but how the hell do you think like that?
hscishard
Active Member
I guess only I found it interesting.
- Joined
- Feb 16, 2005
- Messages
- 8,391
- Gender
- Male
- HSC
- 2006
This is how pure mathematicians think with lots of rigour to ensure technical precision. That being said, I'm more of an applied person though lol
frenzal_dude
UTS Student
undefined can be thought of as being infinity, since infinity is sort of like an undefined number.
frenzal_dude
UTS Student
We know that 1/9 = 0.111...., 2/9 = 0.22... etc.
But the pattern obviously stops at 9, because we all know that 9 divided by 9 is EXACTLY 1 and not 0.9 repeated even if you repeat it for ever.
- Joined
- Feb 16, 2005
- Messages
- 8,391
- Gender
- Male
- HSC
- 2006
If it is repeated forever to infinity it will equal exactly 1. Perhaps the reason that this seems hard to accept is that you keep thinking of it with finite decimal places.
Perhaps a geometric series argument may be more convincing intuitively:
Suppose 0.9999.... had a FINITE number of decimal places (say n decimal places), then
0.9999... = 9 (1/10 + 1/102 + 1/103 + ..... + 1/10n)
The bracket is a geometric series with a = 1/10 and r = 1/10, hence
0.9999... = 9 (1/10) (1 - (1/10)n) / (1 - 1/10)
= 9 (1/10) (1 - (1/10)n) / (9/10)
= 1 - (1/10)n
As n --> ∞, then we reach an INFINITE number of decimal places. But (1/10)n --> 0 when this happens thus 0.9999... becomes 1 when it has an infinite number of decimal places.
Think about it. For example, there is no way 0.2222.... will ever equal 2/9 unless there are an INFINITE number of decimal places.
Consider the proof for 0.22222... = 2/9
Let x = 0.222222....
10x = 2.22222....
10x - x = 2
Why is it equal to 2 under subtraction? Because there is an INFINITE number of decimal places. If there were a finite number of decimal places then the subtraction would clearly not be equal to 2. It will actually be 1.99999...98 if it had a finite number of decimal places.
Last edited:
frenzal_dude
UTS Student
Yeh you're right.
I guess it's hard to comprehend since we don't experience infinity in our reality. But theoretically what you're saying is true.
ohexploitable
hey
But...Who says 1 = -1 is false, afterall:
<img src="http://www.codecogs.com/eq.latex?\begin{align*}i&=i\\\sqrt{-1}&=\sqrt{-1}\\\sqrt{\frac{-1}{1}}&=\sqrt{\frac{1}{-1}}\\\frac{\sqrt{-1}}{\sqrt{1}}&=\frac{\sqrt{1}}{\sqrt{-1}}\\\sqrt{-1}&=\frac{1}{\sqrt{-1}}\\(\sqrt{-1})^{2}&=1\\\therefore -1&=1\end{align}">
For those who take maths a bit too seriously, yes there is a false statement in the arguement but it is pretty hard to find (it took me a few years before I could find someone to point it out)
(-1)^2 =1 ... so yes, the square root of 1 is 1.But...
I;m pretty sure there is a difference between
Last edited:
ohexploitable
hey
I;m pretty sure there is a difference between
^2)
But does that apply to imaginary numbers? I'm only in Yr. 11...
hscishard
Active Member
But...But...
If that was true then from what you said,
Maths can be soo contradictory.