Can anyone spot the error in this fallacy

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In summary, the conversation discusses how taking the square root of numbers can be erroneous and that all solutions must be +/- unless given as an absolute value. However, when doing this with the 5th and 6th lines, the results end up being the same - 0.5625. This is due to the inconsistency in the 6th line.
  • #36
mugaliens said:
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:
[tex]\sqrt(1-2)^2[/tex] or
[tex]\sqrt(-3)^2[/tex]
(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:
[tex]\sqrt(-3)^2=
\sqrt(9)=+/-3[/tex]

Where do we take the square root of any negative number?
 
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  • #37
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.
 
  • #38
jimmysnyder said:
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.

:-p

jkz :)
 
  • #39
Sorry! said:
jkz
ndrstd
 
  • #40
By the way, if you quote this post, it will show you how to extend the line of the square root symbol

[tex]\sqrt{(-3)^2}=
\sqrt{(9)}=\pm3[/tex]
 
  • #41
jimmysnyder said:
By the way, if you quote this post, it will show you how to extend the line of the square root symbol

[tex]\sqrt{(-3)^2}=
\sqrt{(9)}=\pm3[/tex]

OHHHH ok thanks. I tried using those [] brackets but it was just leaving a huge space. Thanks jimmy :smile:
 
  • #42
I am locking this thread. The OP has not been back since his initial post.
 

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