I need someone to tell me if this is possible or not and is it true or not

[GodOfPower]

I am not sure if this should be in this section or not but lets see what happens...

Ok, first I will show you my proof and then I will ask the questions:
1^2=1
Sqrt(1)=1
Sqrt((1^2))=1

• (-1^2)=1

If x = (-1^2)
Then: sqrt(x)=1?

Sqrt(-1)=??? we are told that we can’t do that, but we could do this:
Sqrt((-1^2))= like we did above with the “x” so it = sqrt(1) from 1. and then it = 1
So we have that no matter if we have sqrt((1^2)) or sqrt((-1^2)) we still got the same result = 1
If we do it with another number like 3 it will look like this:

3^2=3.3=9
Sqrt(9)=3
Sqrt((3^2))=3

Then we take (-3)
(-3)^2=9
Sqrt((-3)^2)=3

So again we have the same thing… no matter if it is sqrt((3^2)) or sqrt((-3)^2) it has the same result =3
So does that means that -1=1 or -3=3?
No, not in any real and practical way but as we don’t have any use for the equation x^2+1=0 and we have indefinite integrals to make it work… so why don’t we have this…
My question to all mathematic geniuses out there is this proof real? Is it possible -1 to equal 1 or any other number to equal its opposite one?

 

[stapel]

1^2=1
Sqrt(1)=1

By definition of "square root", being the "principal" second root, yes, the above is true.

Sqrt((1^2))=1

• (-1^2)=1

No:

. . . . .\(\displaystyle -1^2\, =\, -(1^2)\, =\, -(1)\, =\, -1\, \neq\, +1\, =\, 1\)

Did you perhaps mean \(\displaystyle (-1)^2\), being "the square of minus one" (rather than the negative of the square of one, as you'd posted)?

If x = (-1^2)
Then: sqrt(x)=1?

If you mean \(\displaystyle x\, =\, -(1^2)\), then, no, the square root of negative one will be \(\displaystyle i\), the "imaginary", which is defined as \(\displaystyle \sqrt{-1}\, =\, i.\)

If you mean \(\displaystyle x\, =\, (-1)^2\), then you need to expand your context into the imaginaries, which will allow you to bring in the various rules for working with complex numbers.

Sqrt(-1)=??? we are told that we can’t do that

Within the context of the real (rather than complex) numbers, no, you cannot.

Is it possible -1 to equal 1 or any other number to equal its opposite one?

Obviously not. The whole of mathematics would clearly fall apart as hopelessly faulty, were values allowed to equal their additive inverses. As suggested above, you appear possibly to be attempting to apply the rules of complex-valued functions to a real-valued context. This won't work.

 

[JeffM]

I am not sure if this should be in this section or not but lets see what happens...

Ok, first I will show you my proof and then I will ask the questions:
1^2=1
Sqrt(1)=1
Sqrt((1^2))=1

1. (-1^2)=1

If x = (-1^2)
Then: sqrt(x)=1?
Sqrt(-1)=??? we are told that we can’t do that, but we could do this:
Sqrt((-1^2))= like we did above with the “x” so it = sqrt(1) from 1. and then it = 1
So we have that no matter if we have sqrt((1^2)) or sqrt((-1^2)) we still got the same result = 1
If we do it with another number like 3 it will look like this:
3^2=3.3=9
Sqrt(9)=3
Sqrt((3^2))=3
Then we take (-3)
(-3)^2=9
Sqrt((-3)^2)=3 So again we have the same thing… no matter if it is sqrt((3^2)) or sqrt((-3)^2) it has the same result =3
So does that means that -1=1 or -3=3?
No, not in any real and practical way but as we don’t have any use for the equation x^2+1=0 and we have indefinite integrals to make it work… so why don’t we have this…
My question to all mathematic geniuses out there is this proof real? Is it possible -1 to equal 1 or any other number to equal its opposite one?Ok, first I will show you my proof and then I will ask the questions:
1^2=1
Sqrt(1)=1
Sqrt((1^2))=1

1. (-1^2)=1

If x = (-1^2)
Then: sqrt(x)=1?
Sqrt(-1)=??? we are told that we can’t do that, but we could do this:
Sqrt((-1^2))= like we did above with the “x” so it = sqrt(1) from and then it = 1
So we have that no matter if we have sqrt((1^2)) or sqrt((-1^2)) we still got the same result = 1
If we do it with another number like 3 it will look like this:
3^2=3.3=9
Sqrt(9)=3
Sqrt((3^2))=3
Then we take (-3)
(-3)^2=9
Sqrt((-3)^2)=3 So again we have the same thing… no matter if it is sqrt((3^2)) or sqrt((-3)^2) it has the same result =3
So does that means that -1=1 or -3=3?
No, not in any real and practical way but as we don’t have any use for the equation x^2+1=0 and we have indefinite integrals to make it work… so why don’t we have this…
My question to all mathematic geniuses out there is this proof real? Is it possible -1 to equal 1 or any other number to equal its opposite one?

No, it is not possible (with one possible exception). What would you wear if it was + 70 degrees F? What would you wear if it was - 70 degrees F? If the two temperatures are the same, why would you not dress in shorts and sandals when the temperature was - 70 degrees F?

Possible exception \(\displaystyle + 0 = - 0\), but I am not sure whether signed zero is a valid concept.

As for your proof, it is wrong:

\(\displaystyle -1^2 = -(1 * 1) = - (1) = - 1 \ne 1 = (-1) * (- 1) = (-1)^2.\)

You are in fact assuming what you claim to be proving.

\(\displaystyle x = \sqrt{- 1} \implies x * x = - 1.\) There is no such real number.

\(\displaystyle x = \sqrt{1} \implies x * x = 1.\) There are two such real numbers, + 1 and - 1.

Just because two numbers have the same square does not mean that they are the same number.

 

[pka]

sqrt(1) = +/- 1, not just 1. It's a multi-valued function. It's traditional to take the positive square root but that doesn't negate the fact that both values are equally valid.

That is simply not the case. The case is \(\displaystyle \sqrt{1~}=1 \) BUT \(\displaystyle \sqrt{1~}\ne -1~. \)

There are two square roots of one: one is \(\displaystyle \sqrt{1~} \) and the other is \(\displaystyle -\sqrt{1~} \).

This is the classic counterexample to those who say:
if \(\displaystyle a^2=b^2 \) then \(\displaystyle a=b\)

Also that is why we know that if \(\displaystyle a^2=b^2\) then \(\displaystyle |a|=|b| \)

Students are confused enough about the use of the radical symbol. We need not confuse them.

Moreover, there are no multi-valued functions \(\displaystyle \Re\overset{f}\to\Re \).

 

[pka]

yeah ok. This is why I was an engineer and not a mathematician.a
my point was that -1 and 1 are both valid roots of x^2 = 1.

I have long known that engineers are not literate in theoretical mathematics .
This is a story forty+ years old but it is true.
My university required that a senior professor from some related subject area be on the committee examining a PhD candidate. My thesis was on a very obscure new definition of the integral.
When the committee assignment came out my thesis adviser ask me to do this. "Give your introduction of the unified integral then simply say "this is a more natural way to teach integration to students". Well I did exactly that, how could I do otherwise? Well all **** broke loose. Little did I know that the engineering professor and my adviser had an on going feud on how calculus should be taught. So I sat down an simply watched. I passed 4 to 1.