Can anyone spot the error in this fallacy

[Sorry!]

In order to go from:

(a - t/2)^2 = (b - t/2)^2

To:

a - t/2 = b - t/2

Where a, b, and t are variables for all real numbers, then yes, you are...

Well, you're excluding have the input set, the half where either a-t/2 is negative, or b-t/2 is negative.

So I guess when I take I take:
$$\sqrt(1-2)^2$$ or
$$\sqrt(-3)^2$$
(the line should be over everything but I'm not great with latex.
I'm taking the square root of a negative number... interesting; never learned that in math.

Let's see what happens when we follow order of operations:
$$\sqrt(-3)^2= \sqrt(9)=+/-3$$

Where do we take the square root of any negative number?

 

[Jimmy Snyder]

Where in the OP did it say real number? If a is i (imaginary) and t is 0, then a - t/2 is i and its square is -1. When you take the square root you will be taking the square root of a negative number. But that is not the flaw in the proof.

 

[Sorry!]

Where in the OP did it say real number? If a is i (imaginary) and t is 0, then a - t/2 is i and its square is -1. When you take the square root you will be taking the square root of a negative number. But that is not the flaw in the proof.

lol, i think i hate you.

:tongue:

jkz :)

 

[Jimmy Snyder]

jkz

ndrstd

 

[Jimmy Snyder]

By the way, if you quote this post, it will show you how to extend the line of the square root symbol

$$\sqrt{(-3)^2}= \sqrt{(9)}=\pm3$$

 

[Sorry!]

By the way, if you quote this post, it will show you how to extend the line of the square root symbol

$$\sqrt{(-3)^2}= \sqrt{(9)}=\pm3$$

OHHHH ok thanks. I tried using those [] brackets but it was just leaving a huge space. Thanks jimmy

 

[Integral]

I am locking this thread. The OP has not been back since his initial post.