Can anyone spot the error in this fallacy

[Jimmy Snyder]

I think what you mean to say is all numbers within the sets are the same. Not all numbers are the same.

The set of numbers is a set.

 

[Sorry!]

The set of numbers is a set.

Yes but saying that set A has the same numbers as set B is far different from saying the numbers contained within the sets are the same. 1 is not the same as 2, however the set A {1,2} is the numbers as set B {2,1}

 

[Jimmy Snyder]

Yes but saying that set A has the same numbers as set B is far different from saying the numbers contained within the sets are the same. 1 is not the same as 2, however the set A {1,2} is the numbers as set B {2,1}

I did not say that A has the same numbers as B. For instance, in the example I gave, A had 4 elements and B had 3. What I did say is that if you assume ... Hey wait a minute. You do know what mathematical induction is right? I am doing induction on the number of elements in the set. Do you understand what that means? Can you do an inductive proof that the sum of the first n positive integers is equal to n(n + 1) /2?

 

[mugaliens]

We are not taking the square root of any negative numbers here.

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.

 

[Sorry!]

I did not say that A has the same numbers as B. For instance, in the example I gave, A had 4 elements and B had 3. What I did say is that if you assume ... Hey wait a minute. You do know what mathematical induction is right? I am doing induction on the number of elements in the set. Do you understand what that means? Can you do an inductive proof that the sum of the first n positive integers is equal to n(n + 1) /2?

I thought what you were doing was using a,b,c,d to represent specific numbers. and in each set they would represent the same numbers. I do know what mathematical induction is though.

 

[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:
$$\sqrt(-3)^2=<br /> \sqrt(9)=+/-3$$

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.



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

$$\sqrt{(-3)^2}=<br /> \sqrt{(9)}=\pm3$$

 

[Sorry!]

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

$$\sqrt{(-3)^2}=<br /> \sqrt{(9)}=\pm3$$

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.