book on proofs

>>8978688
131 is the limit. Anyone under that will not be able to understand anything in the book and everyone over that are going to understand it with relative ease.

>>8978688
The 4th step isn't accurate.

sqrt(a*b) only equals sqrt(a)*sqrt(b) for positive and real numbers.

>>8978759
[citation needed]

>>8978688
Hmm good idea op.. to answer your question no, no one has had the idea to write a Book of Proof teaching one How to Prove It just yet...

>>8978764
>sqrt(a*b) only equals sqrt(a)*sqrt(b) for positive and real numbers.

Incorrect. Here's a counterexample:

sqrt(i * i) = i
sqrt(i) * sqrt(i) = i

The right way to say it is this:

If you take the square root of both sides of an equation (like occurs in step 4), then it is ALWAYS MANDATORY that you split the analysis into two separate branches, like this:

Step 1: Assume a = b

Step 2: Therefore: sqrt(a) = sqrt(b) OR sqrt(a) = -sqrt(b)

You then continue analyzing both branches, independently, with the understanding that:

- A correct result might appear on both branches of the OR.
- A correct result might appear only on the left branch of the OR.
- A correct result might appear only on the right branch of the OR.

When doing this, it helps to remember the truth table for the OR operator:

true OR true = true
true OR false = true
false OR true = true
false OR false = false

Note that there are three cases where the outcome is true (as reflected by the three possible outcomes of analyzing the two branches above). Understanding and applying this boolean algebra is MANDATORY when taking the square root of both sides of an equation. THERE ARE NO EXCEPTIONS.

>>8978688
Step seven doesn't make sense.

Shouldn't it be
i =1/i
(i)2 = (1/i)2
-1 = 1/-1
-1 = -1

>>8978759
>t. 131 IQ brainlet

>>8978989
x * x = x^2 not 2x

>>8978994
4chan apparently replaces the Unicode squared sign with the number two. It's clear from context anyway.

File: 1497588122205.png (1MB, 1032x772px) Image search: [Google]

1MB, 1032x772px

>>8978688
Complex numbers are a mess, kid. I suggest you avoid the Common Math and try to work on your own stuff.

>>8978798
This is giant and overly pedantic. In fact, in a very real sense the person you're responding to is correct. Let me explain.

Until you specify a branch or declare that you're thinking of sqrt as multivalued, \sqrt{a} doesn't even make sense. The only exception is when a is real and nonnegative. Then the convention is to use the principal branch. In that case sqrt{ab} = \sqrt{a} \sqrt{b}.

tl;dr \sqrt{ab} only equals \sqrt{a} \sqrt{b} for nonnegative real a,b unless you specify a branch. Then you might get other (nonreal or negative) solutions.

>>8978989
What you said is a correct bit of algebra. However, if you assume that i = 1/i, then it is also correct to multiply both sides by i to get i^2 = 1 and hence -1 = 1. So the problem must occur before step 7. In fact, the incorrect step is either going from line 3 to 4 or going from 4 to 5 (depending on how you define the square root function).