Can a = b when a and b are different integers?

[Masterbandit666]

If you think you can make 2 = 3 or 1 = 3 or any other weird arrangement, post it up here. I start this thread because this arguement is going all over the place, it needs to be centralized.

[bugzpodder]

i personnally don't agree with these two arguments:

vsusi:
(-1)^2=1^2
SRBS
((-1)^2)^(1/2)=(1^2)^(1/2)
(-1)^(2*1/2)=1^(2*1/2)
-1^(1)=1^1
-1=1

sql_lall:
only i is defined by its square, namely i^2=-1
but (-i)^2=-1
since the definition of a number squared=-1 is not unique
therefore invalid definition

[Masterbandit666]

Don't you just wish all the rejects on this forum who don't know what they're talking about and who are really just here to get attention by arguing would leave?

From what I've read you're one of the few people who understands basic maths in this place...

[Destined Soul]

Well, I do agree with the first argument, largely because it is a circular argument. I can argue that it's wrong (and the same reasoning):

Say (-1) = (1)
Square both sides (valid) to get (-1)^2 = (1)^2. However, it would be wrong to not evaluate these values and still continue. Our next statment would be 1 = 1. It all has to do with one-to-many "functions" and many-to-one (or 1-1) functions. I can't really explain this much better, but we all know that we're applying rules to things where it doesn't apply.

As for the second argument, I kinda like it, if not from the statement that "y = x^2" is not unique, as for the same value of y, x = a or x = -a. To define i as i^2 doesn't really define i uniquely, so is it two things? Does this mean that i = -i, even though the two numbers: a + bi <> a - bi. Or does it?

Taking this a bit further, suppose i^2 = -1.
Does this imply i = -i?
Can you factor this to become 1 = -1?

Somehow, this is wrong - we know 1 <> -1, so what rule(s) does this invalidate, and where?

Destined

[Destined Soul]

Originally posted by Masterbandit666
Don't you just wish all the rejects on this forum who don't know what they're talking about and who are really just here to get attention by arguing would leave?

From what I've read you're one of the few people who understands basic maths in this place...

Personally, I come here (math forum) because I like to learn. This cannot be done without debate (ever taken any Philosophy?)

However, I learn from people's examples, other people's mistakes, and most importantly, from my own.

Destined

[snakeeyes1000]

i look at things this way:

If there is a word for it, it has it's place in society. Some of these things are good and bad. Personally, I think debate is a good thing. And the beauty of debate is that two people can be saying two different things, and neither one has to be right or wrong(in most cases). Just because people don't have the same views as you doesn't mean you should disregard it. The point I am trying to make is that I would be much happier to help people who don't know what they're talking about than listening to a fool, such as masterbandit, who complains when these people voice their opinions.

PS. I am suggesting that masterbandit, as well as others, change their attitude. If they want to remain as they are, that is certainly their choice and a dissapointing one.

[jim mcnamara]

They can in computer memory - you can have two variables assigned the same value, then printed in C:

#include <stdio.h>

int main(){
	unsigned int i;
	int j;
	j=0xffff;
	i=0xffff;
	printf("unsigned: %u  signed %d \n",i,j);

	return 0;
}

output:

unsigned: 65335 signed: -1

And, no, I'm not asserting that 2=1. ---- People who don't know much about programming get all wrapped around the axle on code like this.

[NotLKH]

It was my viewpoint that the 2=1 thread, or whatever its called, was just a request for feedback on a certain fallacious proof that 2 = 1. NOT that it was meant to be serious, but more into the investigation of how people use misdirection in math to prove an impossibility.

It was not meant to be taken that the thread starter beleived 2 = 1.
At least, I hope not.

-Lou

[Alphanos]

An interesting attempt by my grade 11 math teacher to make 0.99999999999999999999999999999 equal to one:


X = 0.99999999999999999999999999999
10X - X = 9
Therefore 9X = 9; X = 1.

Of course, the flaw with this idea (which I pointed out and he ignored, being proud of showing the class how smart he was) is that you get

09.999999999999999999999999999990
-0.999999999999999999999999999999
----------------------------------------------------
08.999999999999999999999999999991

The number of nines is just an arbitrary number.... it is my opinion that even if the number of nines were infinite, the proof would still be incorrect, because we cannot perform multiplications with numbers containing an infinite number of decimal places; it cannot be done.

[bugzpodder]

lol. you'll know the idea of limits when you are taking calculus. it says that when the limit of x approaches c, of f(x) doesn't necessary equal to f(c).

but you can beleive what you want here. you can say its equal, you can say that it isn't. here is an interesting look at the question though:

there should be no real values expressable between 0.(9) (0.9 repeating) and 1. however let x be the greatest real <1. now the average of the two numbers (x+1)/2 is also real, and is greater than x. so there is no lagest real <1. assume 0.(9) is less than one contradicts with the fact that there is no largest real <1 and so 0.(9)=1.

[NotLKH]

bugz, don't bother.
Alph is in the camp where .9999... <> 1 no matter what you trow at him.

There are a few good threads here about the subject. I might dredge some up.


-Lou

[Alphanos]

there should be no real values expressable between 0.(9) (0.9 repeating) and 1. however let x be the greatest real <1. now the average of the two numbers (x+1)/2 is also real, and is greater than x. so there is no lagest real <1. assume 0.(9) is less than one contradicts with the fact that there is no largest real <1 and so 0.(9)=1.

In order to perform said operation, you would require infinite computational power, since x is effectively infinitely difficult to calculate at all, let alone work with. This is like saying that you want to do a math question with all of the decimal places of pi... well, you can't have all of the decimal places of pi, it simply isn't possible. Since it is not possible to do your proposed question, I am forced to conclude that the resulting number (which would be 'larger') is effectively as much larger than x as infinity +1 is larger than infinity.

[Alphanos]

On a side note, I recognize the difference between two numbers being equal, and two numbers being so nearly the same that it may be best to treat them as though they were; the latter is certainly important, and should be done, but is not the same as the former.

[Destined Soul]

Originally posted by Alphanos
... two numbers being so nearly the same that it may be best to treat them as though they were ...

I was suddenly reminded of all of my physics tests. In the first year or two, you're usually looking for an exact solution, but once you hit upper level classes, you often have to make approximations or, as you said, you could be there forever.

My favourite was when the prof used (invented?) an "approximately proportional to" symbol. (proportional symbol with a tilde (~) over top.) The equation had gone on for about 2.5 classes, so we weren't bothered for being perfect. I think we were quite skeptical of how accurate the solution was after that. hehehe..

Destined

[Masterbandit666]

I think some people misunderstood my previous post:

what I was trying to say was that this forum is kind of ruined by people who claim to know this that and the other, and don't. You're right, people come here to learn too, and they're not going to learn anything if someone else comes on here and says something like "2 = 3 and that's right because I said so". I've only been coming here for a few days now, and it seems to me that there are a few individuals who regularly post incorrect maths and claim it to be correct, and it's this that needs to go. People can't learn anything if they have other people aggressively arguing against each other about a topic. Maybe snakeeyes should lighten up, since he immediately assumed I was being derrogatory (spelt right?) whereas the contrary was actually the case.

[Masterbandit666]

Here's something along the way this thread is going, to keep people thinking. I brought this up in a lecture recently to provoke thought on how valid our numberline really is. Every positive number on the number line has two square roots right? The square roots of 16 are ±4, the square roots of 0.25 are ±0.5.

So what are the two square roots of '0.9' recurring? (In case someone doesn't know, that means 0.99999999999 and an infinite amount more nines.)

Bear in mind that 0.3 recurring NOT 1/3 (if it was equal then 3 * 0.3 recurring = 1, which it doesn't, 3*0.3 recurring = 0.9 recurring, not 1). That's a flaw in the number line as it is.

So without spending hours typing a massive post on why this that and the other, try defining the square roots of 0.9 recurring.

[bugzpodder]

so what if 0.999 recuring is 1? isn't the problem solved? lol.

[Alphanos]

I would think that the square roots would be +/- itself, like 1. Saying that a number should be larger than 0.9 recurring and less than 1 is meaningless, so we must effectively treat 0.9 recurring as its own square root; its not completely correct, but the difference between the 'correct' answer and this is meaningless for our purposes.

[bugzpodder]

lol i missed Alphanos' first post.

In order to perform said operation, you would require infinite computational power, since x is effectively infinitely difficult to calculate at all, let alone work with.

I think you misunderstood me. i am not trying to calculate x. what i am trying to say is that there isn't a real number that can be considered the largest number less than 1 (proven by contradiction).

here is the proof again:

assume x is the largest real number less than 1
so the average of x and 1 (1+x)/2 is greater than x but less than 1
this contradictions with the assumption
so there isn't a largest real number less than 1


but since 0.9 repeating is considered the largest real number less than 1 (since its physically not possible to write another another number that's greater than 0.9 repeating), then this belief also contradicts with the fact that there are no largest real thats less than 1. so i am concluding that 0.9 repeating is not less than 1 (proven again by contradiction). -- although this doesn't mean that 0.9 repeating is 1, but at least it's not less than 1

[Masterbandit666]

Well I can give you a part answer to my question:

ANY positive real number > 1 will give x² > x.
ANY positive real number < 1 will give x{sup]2[/sup] < x.
ANY positive real number = 1 will give x² = x

And since 0.9 recurring is the closest real number to 1, there is no value of x smaller than 1 which will give x² = 0.9 recurring; also since 1² = 1, there is no value of x larger or the same as 1 which will give x² = 0.9 recurring.

As such the square roots of 0.9 recurring are undefined on the number line.

So solve x² = 0.9 recurring, proving your answer is correct (so not just saying "x = blah blah because it does and I'm right). You have to prove it.

[Alphanos]

assume x is the largest real number less than 1
so the average of x and 1 (1+x)/2 is greater than x but less than 1
this contradictions with the assumption
so there isn't a largest real number less than 1

You can't do that. I think you misunderstood me; while I understand what you are saying, it cannot be done. How do you make x = 0.9 repeated in any actual real life situation, in any form? This is like saying that you cannot have infinity because infinity +1 is bigger... it doesn't really work.

[sql_lall]

Just 1 quick thing first:

sql_lall:
only i is defined by its square, namely i^2=-1
but (-i)^2=-1
since the definition of a number squared=-1 is not unique
therefore invalid definition

As Destined Soul pointed out, this was just saying that if i was defined using "i² = -1", then this has two values, i and -i.

Anyway, both proofs to why 0.(9) = 1 are very good ones (both the square and average things.)

You can't do that. I think you misunderstood me; while I understand what you are saying, it cannot be done. How do you make x = 0.9 repeated in any actual real life situation, in any form? This is like saying that you cannot have infinity because infinity +1 is bigger... it doesn't really work.

We are not really talking about "in any real life situation". Just like in real life Pi cannot be exactly found, but circles exist. And i cannot be found, yet i² = -1 ()
You don't have to "make x = 0.9 repeated", it just is.

[bugzpodder]

lol first i was never trying to make x=0.9 repeating. but then again, I don't see that it can't be done. you are saying that 0.9 repeating is not 1. so why you can't make x=0.9 repeating? I can make x=pi can't i? I can work with x=i+7, can't I? so why can't I make x=0.9 repeating? However, just like you know, you can't work with infinity because it is not a number (while i is a complex number, pi is a real number, so is 0.9 repeating)

Masterbandit: And since 0.9 recurring is the closest real number to 1

lol I proved that should be no real number thats closest to 1 that's less than 1. so 0.9 recurring must be equal to 1

[NotLKH]

Originally posted by Alphanos


You can't do that. I think you misunderstood me; while I understand what you are saying, it cannot be done. How do you make x = 0.9 repeated in any actual real life situation, in any form? This is like saying that you cannot have infinity because infinity +1 is bigger... it doesn't really work.

Does a number have to be written down in order to exist?

Even though we cannot ever thoroughly calculate PI, Does not PI
Exist independantly?

Why is an endless series of Nines following a decimal point
impossible?

Certainly, we can conceive of an endless number of nines trailing
a decimal place. We can represent it, with either a properly
defined summation formula, or by .999...

Since there is no such physical thing as a number, can any
number be said to exist in "actual real life"? Of course!

All numbers exist CONCEPTUALLY. So, Numerically, if you
can conceive it with justifiable properties, and it operates on
verifiable rules, then it exists. And I believe that .999... exists,
and is equal to 1.

At least, Thats IMHO,


-Lou

[Gandalf_Grey_]

well with your logic .99999 = 1 so that would mean .9999999 .... 8 = .9999999. So 1 = .99999999 .... 8
if you extrapolate further 1 = .999999 ... 7 = 1. Even though the error is infitesimly small over infinity it adds up.
therefore if you do it enough (initinely) you could prove that 1 = 0

EDIT: I know that is not possible however with the logic used here it is.

[MartinLiss]

Let a = 1
Let b = 1

Therefor
a = b
a² = ab
a² - b² = ab - b²
(a + b)(a - b) = b(a - b)
a + b = b
1 + 1 = 1
2 = 1

[bugzpodder]

lol a-b=0. in math you can't divide by 0 (not that you ever could).

Gandalf:

well with your logic .99999 = 1 so that would mean .9999999 .... 8 = .9999999. So 1 = .99999999 .... 8
if you extrapolate further 1 = .999999 ... 7 = 1. Even though the error is infitesimly small over infinity it adds up.
therefore if you do it enough (initinely) you could prove that 1 = 0

EDIT: I know that is not possible however with the logic used here it is.

no you are mistaken. we are talking about .9999 repeating. not .99999

[Alphanos]

I'm trying to figure out a way in which to make my point clearer, because I can see I was misunderstood.

I am not saying that you have to be able to write down a number to know that it exists, but that you have to be able to write it down in order to use it in an evaluated equation. pi and infinity exist, but they cannot be used in equations that evaluate to a single defined answer. No, we do not use the number pi in equations; we may use 3.1415926 (or more decimal places), but this is only an approximation, because we cannot use the true value of pi because it is too complex. In order to use the true value of pi in an equation, you must write down an infinite number of digits, and it would take an infinite amount of time. Thus we can never perform equations with pi that require a single defined solution; we can either do only algebra with pi, never evaluating what it is, or we can approximate pi. This is the same situation as with 0.999 repeated. We cannot use the true value of 0.999 repeated, because it truly is infinitely close to 1. There can be no larger value than it that is less than 1. This is because its semi-infinite nature excludes it from being used in equations where we need a single exact number as the solution. We can approximate the number, introducing error, or perform algebra with it, like pi. But we cannot solve problems with it. In this case, approximating it introduces so much error relative to the desired precision of the answer (infinite precision) that the equation becomes meaningless. If I asked you to calculate the area of a circle with a radius of 1 centimeter accurate to the 100 trillionth decimal place, you could not do it. The level of precision we would require to evaluate (1+0.9 repeated)/2 is infinitely greater than this.

While you have a theory stating that 0.9 repeated cannot be the largest number less than 1, it cannot be proven, because the necessary procedure to get the result is impossible to do. You may be right, you may be wrong, but you cannot use such an equation as proof, since the amount of error introduced in any attempt is infinite relative to the required accuracy.

If we ignore equations and talk purely in theory, then your proof really makes no sense because there is an infinite number of 9s after the decimal point. It has to be the largest numer less than one, because it is infinitely close to 1.

Conceptual values can only be used when you need algebraic results; otherwise an approximation is necessary.

[Alphanos]

MartinLiss isn't dividing by zero; its done algebraically until the values are substituted in at the end. This has left me boggled, because I fail to see something wrong with this math.

[bugzpodder]

lol. first, dividing an variable assumes that it is not 0.

here is something:

x(x+1)=5x

if you divide by x, you'll get x+1=5, which can be simiplified to x=4.

however, we divided by x, which assumes x!=0. As you can see, x=0 is another solution to this equation. when you divide by a-b, we also assmue a-b!=0. but as Martin clearly pointed out in the beginning, a=b=1, so we can't deivide by a-b because it has a value of 0.

While you have a theory stating that 0.9 repeated cannot be the largest number less than 1, it cannot be proven, because the necessary procedure to get the result is impossible to do.

I proved that ANY REAL NUMBER cannot be the largest number less than 1. It is proven. I've given the proof. What you are talking about a) makes no sense to me and b) does not relate to my proof in any way

You may be right, you may be wrong, but you cannot use such an equation as proof, since the amount of error introduced in any attempt is infinite relative to the required accuracy.

my proof is accurate because algebra is used. why you cannot calculate 100 trillionth decimal place of pi, but pi itself (the algebraic symbol) is understood to be EXACT. that's the power of algebra, you don't need to round off in order to find an answer.

any math student would tell you that 5pi is not the same as 5*3.1415926, because the former is EXACT, latter is NOT.



[/quote]
If we ignore equations and talk purely in theory, then your proof really makes no sense because there is an infinite number of 9s after the decimal point. It has to be the largest numer less than one, because it is infinitely close to 1.[/quote]

my proof proved that there is no largest number less than one. it is a legal proof. does anyone else think that my proof is wrong, and if so, give me the reason? Alphanos, I don't believe what you are arguing here is of any relavence.



btw Although we can't approximate accurately up to a trillion decimal places of pi, wouldn't you agree that we don't need to? I'd be surprised if the most sophisticated project would need pi up to 500th decimal place (of course other than solely determine a closer approximation of pi). What i am trying to say, the approximation is close enough to be used in almost any situation, unlike what you described of the huge error margins.

[NotLKH]

Originally posted by Alphanos
MartinLiss isn't dividing by zero; its done algebraically until the values are substituted in at the end. This has left me boggled, because I fail to see something wrong with this math.

Starting with the identity of a = b, then (a-b) is zero. When you divide it from both sides, then you are dividing by zero.

[Alphanos]

bugz, please try to read the post more carefully, I don't really think I can explain it in another way more clearly.

You have to use ONLY algebra, with NO solving for real numbers if you use conceptual values. If you use an approximation, as is done in almost all uses of conceptual values, the type of proof you're using becomes invalid, because the difference between the approximation and the conceptual value is infinitely more than your desired answer.

Both pi and 0.99 repeated are theoretically real numbers. But you cannot use them as such because their real value is infinitely difficult to work with.

[NotLKH]

Originally posted by bugzpodder
here is the proof again:

assume x is the largest real number less than 1
so the average of x and 1 (1+x)/2 is greater than x but less than 1
this contradictions with the assumption
so there isn't a largest real number less than 1

but since 0.9 repeating is considered the largest real number less than 1 (since its physically not possible to write another another number that's greater than 0.9 repeating), then this belief also contradicts with the fact that there are no largest real thats less than 1. so i am concluding that 0.9 repeating is not less than 1 (proven again by contradiction). -- although this doesn't mean that 0.9 repeating is 1, but at least it's not less than 1

I find this to be completely logical and without fault, although overly cautious.

C'mon, jump off the cliff! You'll bounce!


if .999... is not less than 1, and certainly its not greater than 1, then I'm 99.999...% sure its equal to 1, unless it doesn't exist.

[sql_lall]

1:

well with your logic .99999 = 1 so that would mean .9999999 .... 8 = .9999999. So 1 = .99999999 .... 8
if you extrapolate further 1 = .999999 ... 7 = 1.

Now, we were talking about 0.(9), which =0.99999999......etc (there is no end to the number)
However, you mention 0.99999...8 This number does have last digit, it is rational, so you can't use the same proof that you did with 0.(9) and 1

2:

Both pi and 0.99 repeated are theoretically real numbers. But you cannot use them as such because their real value is infinitely difficult to work with.

Ok, answer this. You have circle 1 of diameter D₁, and circle 2 of diameter D₂, What is the ratio of the area of circle 1 to circle 2??
The area of the big circle = Pi*D₁²
The area of the little circle = Pi*D₂²
The ratio of the areas = (Pi*D₁²)/(Pi*D₂²)
= D₁²/D₂²
Now, surely this is using pi in a real situation?? AND using its exact value (in this case, "Pi" stands for the greek letter Pi, which itself stands for the EXACT value of pi)

[Gandalf_Grey_]

Originally posted by bugzpodder

No you are mistaken. we are talking about .9999
repeating. not .99999

i am also talking about .999 repeated. If .99 repeated is equal to one then .9999 repeated with an 8 on the infinited decimal point is equal to .99 repeated which is equal to one. If done an infinite number of times you could prove that any number is equal to one with NotLKH's logic

[Gandalf_Grey_]

Originally posted by sql_lall
1:


Now, we were talking about 0.(9), which =0.99999999......etc (there is no end to the number)
However, you mention 0.99999...8 This number does have last digit, it is rational, so you can't use the same proof that you did with 0.(9) and 1

but that last digit would be at infinity making it infinitely less than .99 repeated, which is infinitely less than 1. so if .99 repeated is infinitely less than one and .999 .... 8 is infinitely less than .999 repeated then .999 repeated = .999 ... 8 so they both equal 1. If done an infinited amount of times 1 could equal 0. This is with your logic though. I agree that it isn't possible. Even though the difference is infitesimally small it adds up

[Alphanos]

Ok, answer this. You have circle 1 of diameter D1, and circle 2 of diameter D2, What is the ratio of the area of circle 1 to circle 2??
The area of the big circle = Pi*D12
The area of the little circle = Pi*D22
The ratio of the areas = (Pi*D12)/(Pi*D22)
= D12/D22
Now, surely this is using pi in a real situation?? AND using its exact value (in this case, "Pi" stands for the greek letter Pi, which itself stands for the EXACT value of pi)

bugz, you used pi algebraicly only, with no evaluating in that case. Just the same, if x= the conceptual 0.9 repeated, you COULD say that 2x-2x = 0x = 0 because the number that you use for x is irrelevant. To restate, conceptual values can be used a) algebraically or b) with approximations instead of the real infinitely complex value.

You got a ratio of 4/2, which has no relation to the real value of pi, because it was canceled out.

Now, we were talking about 0.(9), which =0.99999999......etc (there is no end to the number)
However, you mention 0.99999...8 This number does have last digit, it is rational, so you can't use the same proof that you did with 0.(9) and 1

I fail to see how you can look at these two situations differently! If 0.9 repeated is irrational, and there is NO last digit, then (1+0.9 repeated)/2 could not possibly be greater than 0.9 repeated!

[bugzpodder]

lol i didn't post that message. you got your names wrong Alphanos!

but exactly, as you yourself pointed out

then (1+0.9 repeated)/2 could not possibly be greater than 0.9 repeated!

so what is it equal to? it can't be greater than 1. it can't be less than 0.9 repeating. so its either 1, 0.9 repeating or undefined.

undefined won't make any sense.

so if its 1 or 0.9 repeating, it would indicate that the three numbers are the same, like:

if the average of 4 and x is 4, what is x? you would say 4 wouln't you?

the truth is, if there is indeed a difference between two numbers x and y, their average should be in between them. however as indicated, this would happen iff there isn't a difference between x and y, so x=y, or in other words, 0.9 repeating =1

i understand that conceptual numbers can be represented algebraically or can be used as an approximation, but i don't see how it is relevant to anything we've talked about. btw i am representing those conceptual numbers algebraically, not using an approximation, in case thats what u r saying.

Gandalf:

but that last digit would be at infinity making it infinitely less than .99 repeated, which is infinitely less than 1. so if .99 repeated is infinitely less than one and .999 .... 8 is infinitely less than .999 repeated then .999 repeated = .999 ... 8 so they both equal 1. If done an infinited amount of times 1 could equal 0. This is with your logic though. I agree that it isn't possible. Even though the difference is infitesimally small it adds up

you cannot have .999...8 where there is an infinate number of 9's. you can't just say at the end of infinity I will add a 8, because there is no end to infinity!!! That was what I am trying to say. sorry about the bold, i was just illustrating my point, or in NotLKH's words:IMHO, lol!

[NotLKH]

Bugz, As you can see, the level of argument when it comes to .999...
lacks here at vbf.
I mean, they misquote you, Gandolf says I have a proof here, when I have None?!?

***?

lol...

-Lou

[NotLKH]

Gandy,

Originally posted by Gandalf_Grey_


i am also talking about .999 repeated. If .99 repeated is equal to one then .9999 repeated with an 8 on the infinited decimal point is equal to .99 repeated which is equal to one. If done an infinite number of times you could prove that any number is equal to one with NotLKH's logic

And, WHAT are you talking about??? My Logic Can you point it out???