Will one number be equal to another num?

[Samphire]

Recently my friend proved theorytically 1=2 using a,b.

 

Like this will one num be equal to another?

[insane_alien]

ahh i have seen that one before but it has a division by zero in it which cannot be done so it is invalid.

 

a=b

a^2=ab

a^2 + a^2 = a^2 + ab

2a^2 = a^2 + ab

2a^2 - 2ab = a^2 + ab - 2ab

2a^2 - 2ab = a^2 - ab

2(a^2 - ab) = 1(a^2 - ab) [this is what fails it]

2 = 1

 

at the noted line to cancel out (a^2 - ab) from both sides reqauire division by the equation but since a and b are equal the equation equals 0

 

divisions by zero cannot be done

 

note: i really need to learn how to use latex

 

http://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html

[YT2095]

LOL, there are many little tricks like this (yes they are Tricks).

 

I can Prove you have 11 fingers!

 

count your finger 1 to 10, then count hand number one, 1 2 3 4 5.

now on hand #2 count down from 10, 10 9 8 7 6 averyone know that 5+6=11

and you`ve counted ALL your fingers only Once!

 

 

Smoke and Mirrors!

[Mobius]

Yes I have seen that one before where the mistake made is the divide by 0. It is a good one though.

 

I can't remember which mathematician did this but I'm sure someone will enlighten us.

He proved that if 1=2 you could prove anyone was the pope.

i.e. Fact 1: John and the pope are 2. Fact 2: 1=2 therefore we can deduce that John and the pope are 1.

[YT2095]

effectively, it`s all Sophist Maths.

 

still fun though

[BigMoosie]

Where's the error in this

 

[math]\sqrt{x} = x^{1/2} = x^{2/4} = \sqrt[4]{x^2}[/math]

 

Substitute in -1:

 

[math]\sqrt{-1} = \sqrt[4]{(-1)^2} = \sqrt[4]{1} = 1[/math]

 

There is no imaginary constant, they were tricking us all along!!!

[DQW]

Yes I have seen that one before where the mistake made is the divide by 0. It is a good one though.

 

I can't remember which mathematician did this but I'm sure someone will enlighten us.

He proved that if 1=2 you could prove anyone was the pope.

i.e. Fact 1: John and the pope are 2. Fact 2: 1=2 therefore we can deduce that John and the pope are 1.

I believe it was either Hardy or Bertrand Russell, but I could be remembering wrong.

[DQW]

Where's the error in this

 

[math]\sqrt{x} = x^{1/2} = x^{2/4} = \sqrt[4]{x^2}[/math]

 

Substitute in -1:

 

[math]\sqrt{-1} = \sqrt[4]{(-1)^2} = \sqrt[4]{1} = 1[/math]

 

There is no imaginary constant' date=' they were tricking us all along!!![/quote']That's exactly the same as saying [imath] 2 = \sqrt{4} = -2 [/imath], so 2 =-2. Only, it throws in more steps in between to "cover up" the obvious logical misstep.

[BigMoosie]

But the square root of four does not equal negative 2. The negative suare root of four does. Which step exactly is the one the contains the error?

[BobbyJoeCool]

Where's the error in this

[math]\sqrt{x} = x^{1/2} = x^{2/4} = \sqrt[4]{x^2}[/math]

[math]\sqrt{-1} = \sqrt[4]{(-1)^2} = \sqrt[4]{1} = 1[/math]

 

Here's the error...

 

[math]\sqrt[4]{(-1)^2}=\sqrt[4]{1}=1' date='-1,i,-i[/math']

 

Now, since the original equation calls for plus or minus i, that's our answer (as you have to plug each possible answer into the original equation in order to get an actual solution...)

 

you'll notice of course, that by raising both sides of the equation to the 4th power...

 

[math]1^4=(-1)^4=i^4=(-i)^4=\sqrt[4]{1}^4=1[/math]

 

when you have an nth root, there are n number of answer to the equation... NOT ALL OF WHICH will work in the original equation... which is why you have to plug them in and figure it out.

 

PS: How do you get the braces in Latex to show a set of numbers?

[DQW]

But the square root of four does not equal negative 2.

Yes it does. The square of negative 2 is 4, is it not ?

[5614]

The square root of any positive integer has two values.

 

A positive one and a negative one.

 

Because say you are taking root4 then you've got 2 (because 2x2=4) and -2 because -2x-2=4 remember that a negative multipled by a negative is equal to a positive!

[BigMoosie]

Guys, it is insulting to see you guys think I am ignorant of the fact that (-2)^2 = 4. I was simply poining out this:

 

[math]\sqrt{4} = 2[/math]

 

[math]\pm\sqrt{4} = \pm{2}[/math]

 

A square root sign does not imply that there is more than one solution unless it has a "plus or minus" sign.

There is only one solution to [math]\sqrt[3]{8}[/math] but three for [math]x^3 = 8[/math].

 

I believe the only error to be between these to statements:

 

[math]\sqrt{x} = \sqrt[4]{(x)^2}[/math]

 

It should read:

 

[math]\sqrt{x} = \sqrt[4]{x}^2[/math]

 

Unless the index's numerator is odd it doesnt really make sense to bring the power inside the root.

[BobbyJoeCool]

A square root sign does not imply that there is more than one solution unless it has a "plus or minus" sign.

There is only one solution to [math]\sqrt[3]{8}[/math] but three for [math]x^3 = 8[/math].

 

Yes it does. The square root has TWO solutions. it will always have 2 solutions. whether or not you put the plus or minus sign in front of the square root of 4, the solutions will always be 2 AND -2. Because when you square both sides with either of those solutions, you get 4=4, which is a true statement.

 

And the cubed root of 3 does have three solutions... 2 are in the complex plane. I really can't think of what they are off the top of my head, but they do exist.

 

We do not mean to insult you. Well, at least I don't. But a couple of statements are false (common misconceptions, so don't feel bad).

 

In high-school/college level mathematics, you're taught not to look at the negative solutions of a square root (because if you graph all solutions to nth root, you'll notice that it no longer passes the line test of a function. as at x=4, y=2 AND y=-2). Some school systems even say that you need the plus or minus in front in order to look at both solutions, but to say the don't exist is wrong.

[BobbyJoeCool]

The square root of any positive integer has two values.

The square root of any non-zero number has two values...

 

[math]\sqrt{-1}=\pm i[/math]

[BobbyJoeCool]

I figured it out now...

 

[math]\sqrt[3]{8}=2[/math]

[math]\sqrt[3]{8}=1+\sqrt{3}i[/math]

[math]\sqrt[3]{8}=1-\sqrt{3}i[/math]

 

note that if you cube both sides of any of the above equations you get:

 

8=8

[DQW]

BigMoosie, I meant no insult to your intelligence.

 

Try this (it's the same thing) :

 

[math]-1 = i^2 = i*i = \sqrt{-1} * \sqrt{-1} = \sqrt{-1*-1} = \sqrt{1} = 1 [/math]

 

The logical error is in saying [imath] x = x^{2/2} = \sqrt{x^2} [/imath], since the last term in that expression represents 2 numbers : [imath]x[/imath] and [imath]-x[/imath]

 

While the trigonometric, hyperbolic, exponential, and power functions are all single-valued functions (some definitions require that all functions be single-valued, others don't) their inverses are, in general, multivalued. So applying the function and its inverse does not, in general, get you back where you started.

[BigMoosie]

I see what you guys are saying, but I have been led to believe that if I wanted to make x the subject in this:

 

[math]x^3 = 8[/math]

 

I would only be partially correct saying:

 

[math]x = \sqrt[3]{8}[/math]

 

And the correct answer which gives all three solutions would be:

 

[math]x = \sqrt[3]{8} (sin(120n)i + cos(120n))[/math] (where n is an integer)

 

Edit:

I figured it out now...

 

[math]\sqrt[3]{8}=2[/math]

[math]\sqrt[3]{8}=1+\sqrt{3}i[/math]

[math]\sqrt[3]{8}=1-\sqrt{3}i[/math]

 

The complex solutions are actually:

 

[math]\sqrt[3]{8}=-1+\sqrt{3}i[/math]

[math]\sqrt[3]{8}=-1-\sqrt{3}i[/math]

[BobbyJoeCool]

I see what you guys are saying' date=' but I have been led to believe that if I wanted to make [i']x[/i] the subject in this:

 

[math]x^3 = 8[/math]

 

I would only be partially correct saying:

 

[math]x = \sqrt[3]{8}[/math]

 

And the correct answer which gives all three solutions would be:

 

[math]x = \sqrt[3]{8} (sin(120n)i + cos(120n))[/math] (where n is an integer)

 

Edit:

 

The complex solutions are actually:

 

[math]\sqrt[3]{8}=-1+\sqrt{3}i[/math]

[math]\sqrt[3]{8}=-1-\sqrt{3}i[/math]

 

opps... yea thats right.

 

 

And no, the answer is [math]\sqrt[3]{8}[/math], because in the equation x^3=8, there are three values for x that will give you the answer.

 

x={[math]2,-1+\sqrt{3}i,-1-\sqrt{3}i[/math]}

 

if you plug any of these three into the original equation (x^3), you'll get 8. So, there are three solutions for x^3=8.

 

But saying the [math]\sqrt[3]{x}=2[/math] (or [math]-1-\sqrt{3}i[/math] or [math]-1+\sqrt{3}i[/math]), there is only one solution, x=8, because you can only plug in one x, value to get the solution.

 

Even though [math]\sqrt[3]{8}=2,-1+\sqrt{3}i,-1-\sqrt{3}i[/math], only one of those answers works in the original equation...

 

This type of situation shows how you have to plug all answers into the original equation and if it doesn't work, you know that the answer you got is either wrong (you did something wrong), or it doesn't work (as in you got multiple answers from such a situation, and one of them isn't the right answer)

 

It's a mistake a LOT of people make, sometimes, just because they don't know better. When helping people with their math homework, it's one of the things I usually have to correct people on.

[BobbyJoeCool]

ok. I see what you were saying. Whomever led you to believe that saying [math]x=\sqrt[3]{8}[/math] either didn't know better (unlikely) or didn't want to confuse you by saying that it [math]\sqrt[3]{8}[/math] has three solutions... In elementary math, you're taught the you only look at the positive answer for the square root (and thus only the positive, real value for the nth root) because it's the only one you'll use until you get to advanced mathematics. Even in calculus, I haven't seen much of taking any answer but the positive real value... but's it's a bad habit to get into thinking that sqrt 4=2 but does not equil -2. because it does...

[BigMoosie]

From wolfram

Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of 9 are -3 and +3, since [math](-3)^2=(+3)^2=9[/math]. Any nonnegative real number x has a unique nonnegative square root r; this is called the principal square root and is written [math]r=x^{1/2}[/math] or [math]r=\sqrt{x}[/math]. For example, the principal square root of 9 is [math]\sqrt{9}=+3[/math], while the other square root of 9 is [math]-\sqrt{9}=-3[/math].

 

Notice that the sign before the root symbol does effect the value of the surd.

 

The [math]\sqrt{}[/math] sign means the "principle square root of". I may have contradicted myself before but this is where I am coming from.

[BobbyJoeCool]

The [math]\sqrt{}[/math'] sign means the "principle square root of".

 

Not exactly... but, taking the principle root is what usually happens... the radical symbol doesn't mean principle root, but it's usually what the answer is (because a half of time, the non-principle roots are complex...).

 

I can almost see that they say that in your quote, but they don't... they are just saying [math]x^{1/2}=\sqrt{x}=r[/math] since there are two ways to write square root (to the 1/2 power and using the radical sign).

 

EDIT: also, the sign in front of the radical can tell you which one to use (but in the example we used, YOU put the radical there, it wasn't in the equation, so you need to specify which solution(s) that are useable.)

 

And, in order for that to be true, you'd also have to put [math]+\sqrt{4}=+2[/math]

[Mot]

[math]{e^{2Ii\pi}}=1[/math] and [math]{e^{\frac{(4I+1)i\pi}{2}} = \sqrt{-1}[/math] etc... where [math]I[/math] is an integer. To find the roots of a [math]\R[/math] number, simply put it in this form (polar coordinates). Then divide until you get all the roots.

 

Example: [math]\sqrt[4]{16} = 2e^{\frac{8i\pi}{4}}, 2e^{\frac{6i\pi}{4}}, 2e^{\frac{4i\pi}{4}}, 2e^{\frac{2i\pi}{4}} = 2e^{2i\pi}}, 2e^{\frac{3i\pi}{2}}, 2e^{i\pi}, 2e^{\frac{i\pi}{2}} = 2,-2i,-2,2i[/math]

 

You will generally get more than half your roots in the complex plain. I chose [math]\sqrt[4]{ }[/math] because I could do it in my head.

[ydoaPs]

proofs that require one to divide by zero are invalid.

 

[math]\lim_{x{\to}0^+}\frac{c}{x}=+\infty[/math]

[math]\lim_{x{\to}0^-}\frac{c}{x}=-\infty[/math]

[math]\lim_{x{\to}0^+}\frac{c}{x}{\not{=}}\lim_{x{\to}0^-}\frac{c}{x}[/math]

therefore,

 

[math]\lim_{x{\to}0}\frac{c}{x}[/math] does not exist

[The Thing]

In quadratic equations we take 2 values of the answer all the time. I don't see why when it comes to square roots it changes. After all, it's all the same. You're just solving x^2=4.

I saw this over in the Brainteasers forum, and it has a curious ambivalence to it as well like the 1=2 theory above.

1 RS = 1 RS

1 RS = 100 PAISE

1 RS = (10x10) PAISE

1 RS = (0.1x0.1) RS

1 RS = 0.01 RS

1 RS = 1 PAISE??