How to equate 2=3?

actually, vsusi, I'm not completely clear on this, but you have to simplify

((-2)^2)^(1/2)

into

(4)^1/2

You must start from the inside out. Then you get

4^1/2 = 4^1/2.

Which, of course, is exactly correct. If you don't follow the rules of math, like order of operations, you can get whatever you want.

ok, first of all, do not ever check a mathematical term in a english dictionary, lol. use a math dictionary (the Square root thing in ditionary.com is absolutely hilarious and of course wrong). sqrt(x) is defined to be the *NON-NEGATIVE square root of x*. it may seem logical to include the negative sqrt as the answer to sqrt(x), but it is most definately wrong!!!!

if you want to solve x^2=c, or x^2=y^2, do difference of squares
if c<0, there is no real solutions for x.
otherwise (x+sqrt(c))(x-sqrt(c))=0 => x=sqrt(x),-sqrt(x)
x^2=y^2
x^2-y^2=0
(x+y)(x-y)=0
x=y, -y

what you fail to realize that x is a VARIABLE and 2 and 3 are constant values. you can not treat a 2 as a variable.


some common math newbie mistakes (these are the mathematical definations, not some made up stuff, so don't bother argue them)

1. you can't divide by 0, period.
2. you can only add 0 or mulitiply by 1 to an equation.
3. constants are NOT variables. don't treat them like one!
4. function by definition returns exactly one value. otherwise it is not a function. domain restrictions are placed to ensure that this happen!!
5. sqrt root is a function. sqrt is defined to be "The NON-NEGATIVE square root of a number". its domain restriction of f(x)=sqrt(x) is that x>=0
6. infinity is not a number. don't treat them like one!
7. rules that applies to real numbers does not necessary mean that they also applies in imaginary and complex numbers

btw for your cutting up the sphere thing, there is this thing thats called spherical triangles, namely triangles that are drawn on spheres (so the side would be curves). rules of triangles does not apply to spherical triangles. i can draw a spherical triangle with 3 right triangles. i think that if you place two of the vertices would be the north and south pole and the third would be any point along the equator to form such a spherical triangle (i am not 100% sure). there are no 3 dimensional triagles. there are only tetrahedrons in 3D.

SnakeEyes

Mathematics says.

If there is a Power to a Power, you can multiply both powers.

(ie) (a^X)^Y = a^{X x Y}

This is also correct according to mathematics.

BTW you said you were not clear but where did you find the error?
:D

Yeah, 'crap' is an English synonym for '$h1t' or 'poo' which is ruder than poo so you feel better after saying it, but not as rude as $h1t so you can use it in public restaurants.

you treated the constant as a variable. you should do the squaring first. so snakeeyes is right

when it comes to x^(2/2), the answer will still be x.

fractional exponents law says you take the sqrt first, then square it.

so it would be (sqrt(x))^2
since x could be less than 0, sqrt(x) could be imaginary. but squaring the value would get your original x back.

Values (Numbers) should be substituted for variables.

Variable is defined in Mathematics as Having no fixed Quantitative value.

IMHO Equations with Variables are templates for substituting with any value (numbers).

(eg) Take any calculator and you will see X^Y. That means X has to be substituted by first number and Y by the next number.

Same way creating functions (with parameters) in a computer program. Parameters are substituted by values.

If you are right then all the Theorems and Equations in the world will be non-sense.

Don't get confused with Constants and Variables.

lets look at derivitives:

f(x)=a^x
g(x)=x^a

so according to you, there is no difference between variables and constants so f'(x)=g'(x)
but in reality is not.

looking at another function:

f(x)=0/x

Quote:

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Equations with Variables are templates for substituting with any value (numbers).

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according to you i can substitute x=0 into the function. but in reality i can't because you can't divide by 0. Why can't I? because the function itself have domain restrictions.

same with the square root function.

my point is that if you got a bunch of constants, evaluate it to avoid trouble cause if you treat it as a variable, you might cross a boundary and does something you can't do. usually you won't have trouble replacing a variable with a constant. but at some special cases you will.

for this specific problem,

if you just look at the algebra:
(x^2)^(1/2)
since x^2 is non-negative, and taking it the principal square root of that number would also yield in a non-negative number - so the correct answer is abs(x)

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f(x)=0/x

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is valid even if you substitute 0. Only thing is you will get an error (In Calculator). Or Indeterminate or Infinity or -Infinity if you do manually. Thats it. It doesn't mean the function itself is wrong.

Acceleration = Velocity / Time

If you want to substitute 0 to Time, it is absurd. Atmost you will get an error (in calculator) even if you do that for fun. It doesn't mean that the formula itself is wrong.

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according to you i can substitute x=0 into the function. but in reality i can't because you can't divide by 0. Why can't I?

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You can substitute any number you want to a variable. That is why it is called as variable. If you cannot substitute then variable will also become constant. Then there is no meaning in calling it as variable. In my case I am not using 0. So we should stop this kind of argument.

Please don't try to insist your own theory here.

In Science(reality) they use infinity and 0 for doing lot of calculation. Infinity in science means a very big value and 0 means very small value. (eg) Even 2 miles can be infinity for ordinary Camera lens if it is not able to focus.

lol 0/0 (indeterminate) is not infinity nor -infinity. get your facts straight. you are talking about undefined.

it is a mathematical fact that you can't divide by 0. infinity is not a value. you restrict the domain so actions like that doesn't happen. why do you think calculator gives you an error? why not give you infinity? because infinity isn't a value to play around with. it is not possible to divide by 0. you can divide by
1X10^(-1X10^(1X10^999999999))) that would result in a really big number. but infinity is not a value!!

u can substitute these numbers. I can't stop you. math can't stop you. your boss can't stop you. but the answer you get has no meaning and you cannot work with them.

you can work with 1X10^(1X10^(1X10^999999999))) on the other hand, but what can you do with infinity? oh right, as C++ says it, NaN, or Not a Number.

right i should agree that we all live in a infinity time where an infinity things are possible. lol this is america you can have your own theories and beliefs. we are on the internet so there are even less rules to be bounded by so don't lecture me about it.

Cool Bugz

Infinity is based on context and not based on just value.

Infinity cannot have a constant value.

In the following link you said

http://vbforums.com/showthread.php?s=&threadid=188573

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having a size or absolute value that is greater than any natural number (or growing out of all bounds).

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It need not be greater than any natural number.

Take 1 drop of poison and mix in a pool of 10,000 Cubic meters. Even though 10,000 is finite nobody is going to die if he drinks it because for 1 drop of of poison 10,000 is infinite(very huge value).

For a camera lens or eye lens even 2 or 5 miles can be infinite even though they are finite.

Dish or DirectTV and all satellite stations works with the same principle too. Distance between your dish and satellite is finite and not infinite. But for your dish that distance is infinite.

BTW I already know that 0/0 is indeterminate (I have already posted in one of the math threads long back). I was just giving you an example. :D

If you do not consider Infinity even after this please consider atleast now IMHO.

I think we are going out of topic. This is just a discussion about a real mathematics and unreal fact (like 2 = 3).

Thanks for your participation.

Infinity is not a number, it's a concept.

1/0 does not = infinity or anything else, it is undefined.

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for 1 drop of of poison 10,000 is infinite

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That's incorrect: 10,000m³ is not infinite, it's just a lot bigger than 1 drop. 10000¹⁰⁰⁰⁰m³ isn't infinite either, relative to the drop or otherwise. If infinity could be measured (which it can't because it's not a value), it wouldn't be measured by how much a drop of poison needs to dilute in water to not kill someone, where do you come up with that crap?

Vsusi most of what you say is bollocks: stop arguing, for your sake and ours.

I admit. That example may be wrong or I don't know how to explain. But not all.

You guys may be Math & Science genius. Definitely not me.

But 1/0 is not undefined. It is +Infinity for my knowledge(Even though limit for infinity is undefined). But the answer is +ve and huge.
http://mathworld.wolfram.com/d2img1616.gif

Even based on limit concept

http://mathworld.wolfram.com/iimg972.gif

Similarly

http://mathworld.wolfram.com/iimg971.gif

There are some more concepts related to infinity like
Countably Infinite, Continuum, Finite, Infinite, Transfinite Number, Uncountably Infinitem, etc.,

The following is just for fun
Pour 1 x 10^-Infinity ml of poison in that pool instead of 1 drop.:p :p :p

1/0 is undefined, and it definitely is NOT infinity. Infinity cannot be evaluated. If 1/0 = infinity then what's 2/0? Bigger 2x infinity?

OK consider this:

if 1/0 = infinity, then that's equivalent to saying 1 = infinity x 0.

Think about that one then!

Mathematica refers to divisions by zero as 'Complex Infinity'.... I dunno. But, Masterbandit666, even if 1/0 = infinity, 2/x will share a similar limit as 1/x, it will still approach infinity as x->0. Same goes for all positive real numbers as a numerator (I would think).

I just try to avoid devisions by zero for all the reasons stated here.

Good Catch by Masterbandit666.

For me Infinity x 0 will lead to indeterminate. (Since Infinity = 1/0)

1/0 x 0 = 0/0 x 1 = Indeterminate (which should mean any +ve number. Can be 1, 2 or n) because any +ve number divided by 0 is infinity.

So 1/0 = infinity, 2/0 = infinity and n/0 = infinity.

I guess Parents yeild Children and not vice versa.

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1/0 x 0 = 0/0 x 1 = Indeterminate

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you cannot treat 1/0 as a number because it is, and take it at face value: UNDEFINED!

I've been at work a lot lately... not much time spent here. excuse me. Let's consider what division is. Division is actually just multiplying by the inverse, correct. Look at the following statement.

x/y

can ALWAYS be rewritten(for all real numbers) as

x*(1/y)

The only exception to this rule is that y must have a real number multiplicative inverse. Therefore, as it is plain to see, mathematics disallows the value of y from ever being 0. You can see why as the value of y decreases, the quotient increases. Therefore, as y approaches 0, the product grows without bound. But it can never actually be 0. That would be undefined. If in a problem, you correctly reach a solution divided by 0, you automatically know that the question has no answer. If anyone wishes to argue with this, argue with my logic, not the math. The math is correct; I might not be.

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Originally posted by snakeeyes1000
Therefore, as it is plain to see, mathematics disallows the value of y from ever being 0.

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Is it maths or is it humans? lol, just kidding. :D

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2. there are no bugs in math. the only bugs that exist are the ones people come up with

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Am I the only one who reacted to this?
Mathematics seems to be flawless, and for more than two millenniums (as long as everyone keep out of the "When does the millennium start-question) people thought it was beyond doubt that pure logic reigns over the math.
This thought means that once a mathematical theory is proved, using only the ten postulates from Euclides and logical rules derived from them, nothing could threaten this theory. Similarly, two theories that both have been proved according to the rules cannot contradict each other - if they do, [at least] one of them must have an error we have overlooked.
This thought, or rather way of thinking, also opens up for an "end" of mathematics - with enough thinking power, possibly with the help of logical and advanced computers, all mathematical theorems can be proved.

However, the Czech-born mathematician Kurt Gödel in 1931 proved, that we cannot prove the mathematics to be flawless!
He didn't say there are flaws in mathematics, but he did say we cannot rule them out - only pray we never find any.
This is because every set of rules and derived proofs have to rely on each other, and can therefore not be proved independently.

This means that we cannot be sure we will never find a set of rules for right angled triangles which contradicts Pythagoras but nevertheless is true. And if that happens, some math teachers are going to have some serious rethinking of marks to do...

So what's the point? We have still not found any bugs in mathematics, there seems to us there are none, but we cannot prove it. There might be bugs in mathematics!

To add to the misery of all this, Gödel later developed paranoia and starved to death when his wife underwent treatment for some disease in hospital. He was afraid of getting poisoned, and refused to eat anything she hadn't cooked.

That's what happens if you meddle with Mathematics.
Pentax

I must say I can't argue with that!

I have an idea, lets let the people who think 2 = 3 live their lives spending $2 and buying something worth $3, and develop an overdraft. According to their logic, the bank will begin to owe them money and they will get more money. Of course, the more sensible ones already know that 2 != 3, so whenever we go into overdraft, we know that we owe the bank money, rather than vice versa.

Which group of people will be better off? The people who can count? Or the people who can't?

If you have a sguare root function it can only return the positive root, or the negative root. Never both, since it is a function.
Therefore, if you apply a square root function to 4 = 4 you will get -2 = -2 or +2 = +2, depending on which root the function returns.

Alternatively, you can stray away from functions, and talk about the one-to-many relation of square root. Then the only way it makes sense is to say 4 = 4 => {2, -2} = {2, -2} and get the sets. The sets can be said to "equal" each other (where equal is being used in the form of set equality, rather than numeral equality).

I can't believe we are having this discussion again! Ok, a function on LHS has nothing to do with the same function on RHS. who says LHS has to give u the positive one just because RHS gave it? functions exists by itself, it does not have to rely on the other side of the equation!

just like sqrt(4)=sqrt(4) according to you, sqrt(4) can only return the positive root OR the negative root, never both.

so the LHS returns positive, RHS returns negative. poof!

you may say that sqrt(4) gives you ONLY positive/negative root at all times.

so when does sqrt(x) gives you negative root?

it doesn't work that way, thats my first point.

Second point, sqrt(x) deotes the principal sqrt of x, meaning that it *ALWAYS* return the non-negative square root of x! so sqrt(4)=2 no matter what. In another thread you indicated that you were always taught that sqrt(x) extends to every real numbers (it doesn't mean that its right. your teacher may be wrong. just like i was always taught this way.). No, sqrt(x), when x<0 is not properly defined. you may argue that i is the "positive" square root of -1, but the fact is, there is no positive and negative about square roots. the modulus of i and -i both equal to 1, meaning thay they are the same "size", although not equal to each other. while you can get away with sqrt(-1) by saying the above. can you get away with sqrt(-5)? maybe 10th root of -1? 11th root? there are 11 roots to choose from, which one is it? Just like what I talked about, functions aren't condictional. given a specific x, it will always return one y, and same y every time. you can't say that in the question they are asking for the negative root, i'll give you the negative root. in the other question they want positive root, so i'll give you the positive one (what happens if the questions want both roots?) and thus thats why i don't like use sqrt(x) on x<0 because it is not properly defined.

To the other post that talks about Incompleteness Theory,

Mathematics is also full of theories. Thus if one of the theories is indeed wrong, then everything else based on it is wrong. HOWEVER, once it is found wrong, mathematicians will fix it as well as everything thats based on it. It CAN be fixed.

You seemed to have missed the point I was trying to make on functions. I was talking about a mathematical function, which is a one-to-one or many-to-one relation. My point was that you could define a "square root function" to return the positive root or the negative root, but it would always return that root. So if you defined it as the positive root (which is standard) then it would always return the positive root on both sides of the equation. If you defined it to return the negative root it would always return the negative root on both sides. I didn't mean that you could take your pick of which root you returned. Sorry.
Functions with computers are different since they have state, and can return different answers depending on when called.
My point was that if you say it can be either answer, it only makes sense to say it must be both, either with the +/- notation, or as a set. With the +/- notation, it must always be in that order, so when you compare +/- 2 = +/- 2, you see if it holds for +2 = +2 and if it holds for -2 = -2. +2 = -2 isn't covered by the abovr equation. That would have to be +/- 2 = -/+ 2 which just isn't true.
sqrt(x) where x < 0 is defined when you have the definition i = sqrt(-1). Then all x<0 are defined by:
sqrt(-x) = sqrt(-1.x) = sqrt(-1).sqrt(x) = i.sqrt(x)
i is the principal square root of -1 which you continually tell us is denoted as sqrt(-1). I'm not saying anything about the 10th root. It is a definition, any conclusions drawn about the tenth root must be supplied with a derivation. I supplied the derivation of sqrt(-x), you show me how it leads to ambiguities over the tenth root.

right, it is not only standard, it is always like that -- for I have never heard of a sqrt(X) that returns the negative root. I don't know about functions on computers, but it isn't relevant -- I am just talking about mathematical functions. of course, if you have a function of f(x)=x+y. if y is different, then f(x) would be different. but really it is f(x,y)=x+y.

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My point was that if you say it can be either answer, it only makes sense to say it must be both, either with the +/- notation, or as a set.

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there is only one answer for sqrt(x). if you are talkin' about x^2=c, where c>=0, there is always two answers. so you always have to use +/-.

when c<=0 you can either say no [real] solutions, or you can always find 2 solutions, as a polynomial of degree n have n solutions (real and complex). I know that you know that but I am just stating this.

in the other threa you said sqrt(x), x extends to any real numbers.

but the domain x also extends to complex numbers. but as i mentioned, sqrt(x) is not defined properly for x<0 and for a complex domain, so i would use demoivre's theorem to find all the roots (easy and simple, nothing about the derivation. if you really want to argue, you can say that the "principal" n-th root would be the root where +360k, k=0 other roots are when k=1,2,...,n-1). lets see your way of doing sqrt(i) and sqrt(-i)! what do they equal to?

btw i know that 2=-2 isn't true.

lastly, your derivation is wrong:

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sqrt(x) where x < 0 is defined when you have the definition i = sqrt(-1). Then all x<0 are defined by:
sqrt(-x) = sqrt(-1.x) = sqrt(-1).sqrt(x) = i.sqrt(x)

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ok let x=-1
sqrt(-(-1))=sqrt(-1*-1)=sqrt(1)=1

if I sub x=-1 into your formula (which is i*sqrt(X)), i would have obtained i*sqrt(-1)=-1

but anyone should know that sqrt(1)=1 (don't tell me sqrt(1)=-1 in this case!)

so my point is, if you want i=sqrt(-1), fine, but you gotta watch out for the trap from the example I've provided in the other thread (you just fall right in that trouble), but here is it again:

i*i=sqrt(-1)*sqrt(-1)=sqrt(-1*-1)=sqrt(1)=1, exactly what you did.

i don't want to argue about "i" again (i am saying that i^2=-1 is a better definition but you guys seem to be arguing if i=sqrt(-1) exists or not -- i don't think you even read the whole thread before you posted the reply since what I said in this post is a mere repeating of myself from the other thread what is i?). you can i=sqrt(-1) if you properly define it and avoid the trap. but i^2=-1 is a better definition -- you can't fall into the trap this way.

Nice try, I really had to think about that one, but there's a flaw in your argument, resulting from me being slightly sloppy in my notation (I don't mind taking the blame for that - just read what I meant in the future, not what I wrote ;))
This is being used to extend the domain of x from Z+ (positive reals) to Z (all reals). Therefore, you can only start off with +ve x, since that is all that sqrt() is defined for. Then define i = sqrt(-1) [the principal square root of -1 as I think we're mostly agreed to now].
Now for +ve x, we have sqrt(x) defined. We have also defined sqrt(-1). We can then say that for x>0 [i.e -x < 0 - this is where we started talking about different things last post. My fault]:
sqrt(-x) [this is not defined yet!] = sqrt(-1*x) [okay, this is probably a better notation on a computer than "."] = sqrt(-1)*sqrt(x) [oh look, these have a defined answer] = i.sqrt(x)
I would have expected it to follow through for sqrt(x) = sqrt(-1)*sqrt(-x), but it obviously doesn't, but if I hadn't messed up my definition it wouldn't have been an issue.
Now I think about it though, it makes sense that you can't do that for sqrt(+ve x) because otherwise you could end up with an ever increasing series of i's just by flipping the sign of x. It's directly analogous with the phase of sin's and cos's. cos(x) = sin(90+x), but sin(x) = -cos(90+x). It's because we're in two dimensions now, and we're doing rotations of 90 degrees. So we go x, i, -x, -i, then back to x, i, -x...

lol you've came up with a pretty reasonable explaination but I wonder if its true or not. I have never heard of that before (does not mean that its wrong) did you make it up or can you find a professional proof that states what you stated elsewhere?

Although your claim is reasonable, but I wonder, why didn't the algebra fall through for (-x)? after all, there is nothing wrong with the algebra of your proof!

since x=(-x)*(-1) for all x, x (E) R (by the way, R denotes the set of real numbers, Z denotes the set of integers)

right now, lets say that all sqrt(-x), x(E)R+ are defined.
then if it is indeed a correct definition, we can use it to define the sqrt(X), x(E)R+ back!

then technically it should work, but it didn't. Let me tell you why, in the end, when you arrive at sqrt(1), sqrt(1) here is no longer the principal square root of 1, but actually the negative square root of 1. and thus my point as to why sqrt(x) is not well defined for complex numbers, therefore not recommended.

btw as i said before sqrt(x), x can extend to complex numbers. then you couldn't really pick one and say its the principal root. like:

sqrt(i)=cos(45)+isin(45), cos(225)+isin(225)
sqrt(-i)=cos(135)+isin(135), cos(315)+isin(315)

right now you still could see the angle, you might just pick the smaller one. but if you convert them to real numbers, you can't tell the difference (except for the signs).

my point? sqrt(x), x (E) R- is not well defined and thus should not use it (fall into the +ve, -ve square root trap)! and i^2=-1
I think I've illustrated my point well enough. if you still insist on using square roots, just don't do what Evil Giraffe did!

I'm just an idiot so I can't really contribute constructively to this discussion; perhaps it's better that way. Just to say: A theory can never be proven but it can be disproved - so the odds are stacked against it from the very beginning. If therefore a theory passes the test of time, we can be very confident in that theory. Each time the theory is correct our confidence in this theory can grow. Never though can we say that a theory is utterly correct...then it wouldn't be a theory but fact. My theory is that you're going to laugh at this silly post but I can never prove this...and again…perhaps it's better that way :)

There are actually four cases to the quadratic formula, but we can eliminate two because they are identical to the other two.

lol never heard of this before, so enlighten me, TT! Show me a way to derive the four cases!

(hey remember a question on pascal/cayley that involves a Mr. Cow Cu Later taking cube roots and add to a number -- when he does it many times and the display on calculator doesn't change anymore he got an answer of 5!)

btw member status are determined by the number of posts lol -- can't believe ur still browsing maybe u urself is addicted even ur status says otherwise lol

I'm also interested... if it's called a quadratic, there must be four of something ;).

Here's your answer Dreamlax:
http://mathforum.org/library/drmath/view/57456.html

Oh! Silly me! Well I suppose 'quadratic' sounds better than 'squaratic'.

When you take an equation and square root both sides, it should look like this:
x^2=16
SRBS
|x|=4
x=4, or x=-4
However, it is possible to do it this way:
x^2=16
SRBS
+/-x=+/-4
x=4, or x=-4, or -x=4, or -x=-4
The third case is equivalent to the second case, and the fourth case is equivalent to the first case, so there are four cases, which can become two.

In the proof for the quadratic formula, there is a step where you must SRBS, and that is where the four cases arise.

i would use difference of squares:

x^2-16=0
(x+4)(x-4)=0
x=4, -4

and it doesn't matter how many cases there are in the middle case, as long as we so agree that there are 2 roots in the end.

-x=4 is the same as x=-4. -x=4 just hasn't been simplified all the way.

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Originally posted by Kalkewl8ter
When you take an equation and square root both sides, it should look like this:
x^2=16
SRBS
|x|=4
x=4, or x=-4
However, it is possible to do it this way:
x^2=16
SRBS
+/-x=+/-4
x=4, or x=-4, or -x=4, or -x=-4
The third case is equivalent to the second case, and the fourth case is equivalent to the first case, so there are four cases, which can become two.

In the proof for the quadratic formula, there is a step where you must SRBS, and that is where the four cases arise.

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Wrong.
Since X is a variable, saying +x or -x is redundant, since x has no value. Now, saying sqrt(16) = +/- 4, well, thats acceptable. and so, if sqr(x^2) = sqr(16), then there are only two possible statements allowable, and that is x=4, or x=-4. saying any statement of what -x is equal to, in combnation with x=, is mathematically stuttering.

:(

What's SRBS? Surely you don't mean: http://www.acronymfinder.com/af-quer...t&Acronym=SRBS !!!

square root both sides

Oh, my bad. I had a feeling the BS part was both sides. I dunno why I didn't pick it up before... oh well, thanks for that bugzpodder.

Correct me if I am wrong

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i*i=sqrt(-1)*sqrt(-1)=sqrt(-1*-1)=sqrt(1)=1, exactly what you did.

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sqrt(-1) * sqrt(-1) should be = (-1)^1/2+1/2 = -1¹ = -1

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Originally posted by vsusi
Correct me if I am wrong

sqrt(-1) * sqrt(-1) should be = (-1)^1/2+1/2 = -1¹ = -1

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I'd argue the same thing, except it gets messy with something like
sqrt(-10.315)*sqrt(-.70711).

And Then!
What about this:
sqrt(-x³-2)*sqrt(-y^4/5-y^1/3+1)

:)

Sorry I couldn't understand what u r trying to say. (like sqrt(-10.315)*sqrt(-.70711) ???).

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Originally posted by vsusi
Sorry I couldn't understand what u r trying to say. (like sqrt(-10.315)*sqrt(-.70711) ???).

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I was leading towards a generalization of this:

Quote:

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Originally posted by vsusi
Correct me if I am wrong

sqrt(-1) * sqrt(-1) should be = (-1)^1/2+1/2 = -1¹ = -1

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Basically, you are implying to do the following:

Given two negative numbers, a and b, if b = a^2*m+1
then:
sqrt(a)*sqrt(b) =>
==> a^(1/2)*(a^(2m+1))^1/2
==> a^(1/2)*a^(1/2)*(2m+1)
==> a^(1/2)*a^m+(1/2)
==> a^m+2*(1/2)
==> a^m+1
and when we have the form a^m+1, then we have gotten away from sqrt'ing a negative number.

My point was, thats extremely difficult and lengthy when a <> b.

:)

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Originally posted by NotLKH
then we have gotten away from sqrt'ing a negative number.

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Strange. When m is odd, your answer is positive, and when it's even, the answer is negative. I can't really find what's wrong with it, but there's something we're missing (I hope.. :D)

I guess another way to write vsusi's math is as follows, of which this is almost always the method I prefer to use.

Let a, b be negative numbers.
Let x, y be positive numbers, such that x = -a, y = -b

Thus sqrt(a)*sqrt(b) = sqrt(-x)*sqrt(-y)
==> i*sqrt(x) * i*sqrt(y)
==> i*i * sqrt(x)*sqrt(y)
==> -1 * sqrt(x*y)

Which is what you want. This does work for a = b = -1, but it basically goes with the definition or equality of sqrt(-1) = i.

Destined

:confused:
Yes...
When M is Odd, M+1 is even, so
a^m+1 IS positive,
and when M is Even,
a^m+1 IS negative.

Using the Defined property of "a" being negative, I see no strangeness.
Am I missing something?

there is nothing wrong with:

Quote:

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sqrt(-1) * sqrt(-1) should be = (-1)^(1/2+1/2 )= -1^1 = -1

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yes but it doesn't solve the problem of the other case. there is no single "correct" way of doing things -- as long as i follow the rules mathematics then its correct. however as i have stated, right there we are looking for the negative square root of 1 (instead of principal square root of 1), which would be -1

now there is nothing wrong with what vsusi's quote either, its another power rule:

it (a^b)*(a^d)=a^(b+d)

like (2^3)*(2^4)=2^(3+4)=2^7

I am not justifying anything. But the below articles are interesting.

http://www.mathmojo.com/interestingl.../1equals2.html

http://mathworld.wolfram.com/Fallacy.html

1 + 1 = 10 now hush n go back to programmin...

Now that's a way to get the last word - wait for 4 years until all other participants in the original thread are likely to have died.

It's interesting to see the lengths people will go to to prove themselves wrong.

Sorry. hadn't realised there were several pages - someone else said this

You make even less sense than the guy who started this thread.

Charming. That's because there was a deleted post inbetween.

I was hoping for some joke proofs

you know like 1+1=3 for extreamly large values of 1.