[Dantheman]

This topic is in continuation of the "1=2" post. I decided to put up some, very famous, math questions for some people to ponder. Yes, I know that 99% of all population already seen them and knows the answer. I'm posting this for the other 1%.

Please, don't post the answers, I'm sure some people will enjoy figuring it out by themselves :-)

1. 1 = 2?

a = b #multiply by b
ab = b² #subtract a² 
ab - a² = b² - a² #factor
a(b - a) = (b - a)(b + a) #divide by (b - a)
a = a + b #since a = b, substitute
a = a +  a
a = 2a #divide by a
1 = 2

2. 0.99999... = 1?

NOTE: 0.99999 is a repeating decimal that has infinite number of 9's.

1/3 = 0.33333.... #multiply both sides by 3
3/3 = 0.99999....
1 = 0.99999....

3. 1 + 1 = 0?

This one is a bit more complicated, it uses an imaginary unit i which defined as √ -1.

Note that i x i = -1

  1 + 1
= 1 + √ 1
= 1 + √ (-1 x -1)
= 1 + √ (-1) x √( -1)
= 1 + i x i
= 1 + -1
= 0

If you know other math tricks, please post them!

[NickDMax]

#1 is division by zero (a=b so b - a = 0, so when you "cancel" out the common multiple you are dividing by zero which is undefined).

#2 Is actually true... Its one of these little oddities of mathematics. Its causes quite a few feathers to ruffle but indefinite repeating 9's round up.
so 1.0001999999999... is 1.0002 -- funky but true.

#3. Actually at the moment I am ashamed to say that I am stumped... I know there is a flaw but I can't see it. AH! sqrt(a * b ) == sqrt(a)*sqrt( b ) IIF a or b is positive else it is -1*sqrt(a)*sqrt( b ) which end up with 1 + -i x i

[Locke]

NickDMax, on 29 May, 2009 - 11:16 PM, said:

#2 Is actually true... Its one of these little oddities of mathematics. Its causes quite a few feathers to ruffle but indefinite repeating 9's round up.
so 1.0001999999999... is 1.0002 -- funky but true.

Not true. We only round up because we know we will only have a VERY slightly wrong answer. It makes everything easier, still getting a VERY accurate result. 1.0001999999999 != 1.0002 ... 1.0001999999999 = 1.0001999999999

We just accept it as right, because we're only off by some really, really small value.

[Dantheman]

Locke, on 29 May, 2009 - 10:29 PM, said:

NickDMax, on 29 May, 2009 - 11:16 PM, said:

#2 Is actually true... Its one of these little oddities of mathematics. Its causes quite a few feathers to ruffle but indefinite repeating 9's round up.
so 1.0001999999999... is 1.0002 -- funky but true.

Not true. We only round up because we know we will only have a VERY slightly wrong answer. It makes everything easier, still getting a VERY accurate result. 1.0001999999999 != 1.0002 ... 1.0001999999999 = 1.0001999999999

We just accept it as right, because we're only off by some really, really small value.

No, you're wrong. 0.9999.... is PRECISELY 1. You're making a common mistake of treating infinity as a number.

[born2c0de]

Quote

0.9999.... is PRECISELY 1

No it's not. It is tending towards 1 but it's still not discrete as 1.

We do assume that it's 1 and that's the basic principle behind limits but 0.9999 != 1.

[paperclipmuffin]

Are we sure that there even is a difference between 0.9r and 1? Will the ever decreasing gap ever become 1? As before mentioned, infinite is not a number. No-one ever has and ever will test this. Our laws of number may not apply at that scope.

[NickDMax]

You see this always happens. Don't feel bad that you are too dumb to realize that 0.99999... is equal to 1 -- many very prominent mathematicians have tried to come up with a way of fixing this bug in our notation but alas every "fix" leads to more trouble -- it is a FACT plain and simple and not even the great minds in the history of mathematics have been able to disprove it (though there are a number of proofs that it is true).

Oddly though, since it IS a fact, it has been used in a number of proofs.

The thing you have to keep in mind is that 0.99999... is not dynamic, it is not "approaching the limit" -- it IS the limit of the series of numbers { .9, .99, .999, .9999... } and the limit of that series is 1.

or in more lay terms: 1/3 = .3333... multiply both sides by 3 and you get: 3/3 = .9999... now this does not really work as a "proof" since here the .9999... IS dynamic because it is our endless multiplication operation -- so in terms of a symbolic operation we can never really complete the multiplication we just get closer and closer to 1.

[NeoTifa]

This isn't fun....

[NickDMax]

lol fun is a relative term... its fun for me. I once sat up all night trying to prove that 0.9999.... was NOT 1.

[Dantheman]

born2c0de, on 30 May, 2009 - 12:29 AM, said:

Quote

0.9999.... is PRECISELY 1

No it's not. It is tending towards 1 but it's still not discrete as 1.

We do assume that it's 1 and that's the basic principle behind limits but 0.9999 != 1.

Sorry, but you're wrong. 0.999r and 1 represent PRECISELY the same number. To the casual person this may be shocking, but I'm a math major, so I'm used to this. Do some research if you don't believe me.

Quote from Wikipedia:

Quote

In other words: the notations 0.999… and 1 actually represent the same real number. This equality has long been accepted by professional mathematicians and taught in textbooks. Proofs have been formulated with varying degrees of mathematical rigour, taking into account preferred development of the real numbers, background assumptions, historical context, and target audience.

Just like NickDMax and Wikipedia have already stated, this fact has already been proven. You may think what you want, but 0.999r is precisely 1.

This post has been edited by Dantheman: 30 May 2009 - 09:03 AM

[NickDMax]

paperclipmuffin, on 30 May, 2009 - 03:39 AM, said:

Are we sure that there even is a difference between 0.9r and 1? Will the ever decreasing gap ever become 1? As before mentioned, infinite is not a number. No-one ever has and ever will test this. Our laws of number may not apply at that scope.

Actually mathematics routinely makes infinite proofs using a variety of techniques. It is true that you can't go out and out and test an infinite number of cases, but you can use logic to prove an infinite result.

For example take the proof that there are an infinite number of prime numbers. We can't start collecting primes together until we have an infinite set -- but we can show that the set of primes can not be finite. This is done by assuming that there are only a finite number of primes and then proving that our assumption is false.

BTW @Locke 1.0001999999999 != 1.0002 is true statement -- I was talking about an infinite repetition of the 9 so not 1.0001999999999 or 1.000199999999999999999999 or 1.0001999999999999999999999999999999 but 1.0001999<bar_over>9</bar_over> which is 1.0002

[CTphpnwb]

Dantheman, on 30 May, 2009 - 11:58 AM, said:

born2c0de, on 30 May, 2009 - 12:29 AM, said:

Quote

0.9999.... is PRECISELY 1

No it's not. It is tending towards 1 but it's still not discrete as 1.

We do assume that it's 1 and that's the basic principle behind limits but 0.9999 != 1.

Sorry, but you're wrong. 0.999r and 1 represent PRECISELY the same number. To the casual person this may be shocking, but I'm a math major, so I'm used to this. Do some research if you don't believe me.

I haven't seen a proof either way, but this strikes me as being the same as Euclid's incorrect assumption that it is axiomatic that given a line and a point not on that line, there exists one and only one line through the point that is parallel to the given line. We simply don't know what happens at infinity, and it's likely that the assumption that 0.999r = 1 is based on the same idea:

y = mx + b would seem to imply that parallel lines never meet or diverge, but that's because our mathematics is based on Euclidean geometry. It's possible, even likely, that any proof that 0.999r = 1 is flawed for the same reasons.

Take a course in non-euclidean geometry and you'll be a lot less confident about the value of 0.999r.

This post has been edited by CTphpnwb: 30 May 2009 - 11:21 PM

[firebolt]

0.9r = 1
Proof:

0.9r = x
9.9r = 10x

-------------------

10x - x = 9x
9.9r - 0.9r = 9

-------------------

Therefore, 
9x = 9
x = 1

-------------------

Thus, 0.9r = 1

Discuss.

This post has been edited by firebolt: 31 May 2009 - 03:04 AM

[gothik12]

@Dantheman let me disagree with you at this:

Quote

This one is a bit more complicated, it uses an imaginary unit i which defined as √ -1.

This is a frequent mistake in maths - chapter complex numbers. We shall never say that i is defined as √ -1 (sqrt(-1) = i)...never ever . We only know that i x i = -1, and that's all.

That trick is very good for students which take their first math test at complex numbers. Our teacher gave us something similar when we first made complex numbers.

We can say that this is paradoxical mathematics. And yes, is funny to see those guys who don't find the evident mistakes.

This post has been edited by gothik12: 31 May 2009 - 04:10 AM

[firebolt]

Quote

No, you're wrong. 0.9999.... is PRECISELY 1. You're making a common mistake of treating infinity as a number.

0.9r does equal 1, if you go by my proof shown above. ^

#13