# Can anyone spot the error in this fallacy

• mjordan2nd
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.
mjordan2nd
<prepared>a + b = t
<prepared>(a + b)(a - b) = t(a - b)
<prepared>a^2 - b^2 = ta - tb
<prepared>a^2 - ta = b^2 - tb
<prepared>a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
<prepared>(a - t/2)^2 = (b - t/2)^2
<prepared>a - t/2 = b - t/2
<prepared>a = b

Therefore all numbers are the same!

When you take the square root from the 6th to 7th line.

Could you explain why that is erroneous?

the formulas seem legit, but if you put numbers in each, as i have done a =1 b =2 t =3

the 6th line shows inconsistency. BOTH 5th lines would appear to be -1.4375=-1.4375 and the next line you would expect the same, but one appears to be 0.5625 = 1.5625

i still don't know why is that. but i would like to know

I'm perfectly happy with this. All numbers are equal.

mjordan2nd said:
Could you explain why that is erroneous?

You need to take +/- when you take a square root unless it's an absolute value.

Sorry! said:
You need to take +/- when you take a square root unless it's an absolute value.

lets just argue for sakes here that it is absolute value because it really doesn't matter, if you put the -/+ in front of the numbers no way will it still equal each other. the magnitude is still different.

ok might be a bit confusing, assume absolute for our case. the magnitudes should equal in theory but by placing numbers its not. but there is nothing wrong with the square root.

mjordan2nd said:
<prepared>a + b = t
<prepared>(a + b)(a - b) = t(a - b)
<prepared>a^2 - b^2 = ta - tb
<prepared>a^2 - ta = b^2 - tb
<prepared>a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
<prepared>(a - t/2)^2 = (b - t/2)^2
<prepared>a - t/2 = b - t/2
<prepared>a = b

Therefore all numbers are the same!

Why in the world would think that?

This does not say anything about a and b beyond what you have done. To satisfy the process you have defined you must have a = b . It does not say anything in general about anything.

Perau said:
lets just argue for sakes here that it is absolute value because it really doesn't matter, if you put the -/+ in front of the numbers no way will it still equal each other. the magnitude is still different.

ok might be a bit confusing, assume absolute for our case. the magnitudes should equal in theory but by placing numbers its not. but there is nothing wrong with the square root.

EDIT: I think you misread what I wrote. The solutions to a square root must be +/- or you just give ABSOLUTE VALUE.

I think my wording may have confused you. What I mean is that if you give an absolute value of 3 it can be +/-3. So I was saying that the solutions have to be +/- unless you give your answer as an absolute (which we are not doing in this with numbers.)

All that this proves as of now with letters is that a=b, if you assume that all letters must be different numbers then YES the square root is the error.

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I don't think you are doing your math correctly anyways (from what I read up above) You do remember BEDMAS correct?

$$(1-\frac{3}{2})^2=(2-\frac{3}{2})^2$$

We end up with this correct? Now what we have is (-0.5)^2=(0.5)^2 which is true.

But if we take the square root in the previous step we end up with:

$$1-\frac{3}{2}$$ which = -0.5 or we have $$-1+\frac{3}{2}$$ which gives us 0.5.

Same for the other side. If you end up with -0.5=0.5 Then you are wrong in your taking of the square root and this is not a valid solution.

Sorry! said:
I don't think you are doing your math correctly anyways (from what I read up above) You do remember BEDMAS correct?

$$(1-\frac{3}{2})^2=(2-\frac{3}{2})^2$$

We end up with this correct? Now what we have is (-0.5)^2=(0.5)^2 which is true.

But if we take the square root in the previous set we end up with:

$$1-\frac{3}{2}$$ which = -0.5 or we have $$-1+\frac{3}{2}$$

Same for the other side. If you end up with -0.5=0.5 Then you are wrong in your taking of the square root and this is not a valid solution.

oh okay, i see your point, but this still goes back to my first post on this, the calculation i done from 5th to 6th line goes haywire, where squareroot hasnt been applied yet.

Perau said:
oh okay, i see your point, but this still goes back to my first post on this, the calculation i done from 5th to 6th line goes haywire, where squareroot hasnt been applied yet.

You're not doing your math correctly.

Sorry! said:
You're not doing your math correctly.

a = 1 b = 2 t = 3

5th line
1^2 -3(1) + 3^2/4 = 2^2 - 3(2) + 3^2/4
0.25 = 0.25

6th line
(1-3/4)^2 = (2-3/4)^2
0.5625 = 1.5625

how is that wrong math?

Perau said:
a = 1 b = 2 t = 3

5th line
1^2 -3(1) + 3^2/4 = 2^2 - 3(2) + 3^2/4
0.25 = 0.25

6th line
(1-3/4)^2 = (2-3/4)^2
0.5625 = 1.5625

how is that wrong math?

5th line:
$$a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4$$

OR

$$1-3+1.5=4-6+1.5$$

Both sides equal -0.5

6th line:
$$(a - t/2)^2 = (b - t/2)^2$$

OR

$$(1-1.5)^2=(2-1.5)^2$$
$$(-0.5)^2=(0.5)^2$$

Both sides equal 0.25... This is way off what you got. If you post your steps maybe I can help you?

EDIT: I noticed you have it as 3/4... why? As well I'm quite certain (1-3/4)^2 does not equal 0.5625

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Sorry! said:
5th line:
$$a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4$$

OR

$$1-3+1.5=4-6+1.5$$

Both sides equal -0.5

6th line:
$$(a - t/2)^2 = (b - t/2)^2$$

OR

$$(1-1.5)^2=(2-1.5)^2$$
$$(-0.5)^2=(0.5)^2$$

Both sides equal 0.25... This is way off what you got. If you post your steps maybe I can help you?

OH, i left my t at t/4 istead of t/2. my bad.

Perau said:
OH, i left my t at t/4 istead of t/2. my bad.

Well even so your answer for line 5 is still wrong and so is the one half of line 6... I'll help you out with that if you wanted.

Sorry! said:
Well even so your answer for line 5 is still wrong and so is the one half of line 6... I'll help you out with that if you wanted.

haha cheers, but how i showed you is exactly how i calculated. should have brought a calculator. but i think i seem to know why it's wrong. thanks anyway!

Perau said:
haha cheers, but how i showed you is exactly how i calculated. should have brought a calculator. but i think i seem to know why it's wrong. thanks anyway!

Heh sure, I'm not trying to make you look 'stupid' or anything just if you need help with how to calculate it I'd be glad to show you. (I'm not sure how old you are but I'm assuming still in high school? No offense.)

Welcome to the forums by the way Did you get hit with a fish yet?

Square root issues aside I do not see where you are saying anything other than a and b are the same number and t is any number which is the sum of those numbers.

Without giving too much away,

a+b = t.

For two terms

a - t/2 and b - t/2,

either both terms are equal and a=b=t/2,

or one of 'a or 'b is less than t/2.

sandbanana said:
Square root issues aside I do not see where you are saying anything other than a and b are the same number and t is any number which is the sum of those numbers.

Basically, if what is being stated were true, then for any numbers that satisfy a+b=t, a=b. So, we can define a and b arbitrarily, and set the value of t to a+b, and a should equal b, which is clearly wrong. For instance, I could pick a = 24 and b = 68, and then set t = 92. And by this "proof", 24 = 68. Of course, the proof isn't valid thanks to the square root issues, so obviously it isn't true. But that's the claim, which is akin to other logical fallacies which prove 1 = 2 or 1 = 0 or other impossibilities.

DaveE

Here is the corrected version:
<prepared>a + b = t
<prepared>(a + b)(a - b) = t(a - b)
<prepared>a^2 - b^2 = ta - tb
<prepared>a^2 - ta = b^2 - tb
<prepared>a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
<prepared>(a - t/2)^2 = (b - t/2)^2
<prepared>a - t/2 = $$\pm$$(b - t/2)
<prepared>a = b or a = t - b

Therefore all numbers are the same or they are different!

Norman said:
When you take the square root from the 6th to 7th line.

mjordan2nd said:
Could you explain why that is erroneous?

You cannot take the square root of real negative numbers.

mugaliens said:
You cannot take the square root of real negative numbers.

Well first this isn't true and second it's not why the square root is the error in this.

The reason the square root is the error has already been explained:

When you take the square root your answer must be +/-. The OP does not do this and so the wrong square root value ended up being chosen to continue to the final answer.

When we take the square root of:

$$(a - t/2)^2$$
$$a-t/2$$ OR $$-a+t/2$$

Sorry! said:
Well first this isn't true...

I shoudl have completed the thought: "You cannot take the square root of real negative numbers and wind up with a single real number."

As you explained, you wind up with an either/or, and there the equality fails.

Even though the proof is flawed, the conclusion is true. Let A = {a, b, c, ..., d} be a set of numbers. Then a = b= c= ... = d as I prove by induction on the number of elements in A. First of all notice that for all singletons, A = {a}, the statement is vacuously true. Next assume that it is true when the set contains n elements {a1, a2, ..., an} and let A = {a1, a2, ..., a(n+1)}. Then the set B= {a1, a2, ..., an} has n elements and so they are all the same. Likewise the set C = {a2, a3, ..., a(n+1)} has n elements and so all of them are the same. Thus a1 = a2 = ... = an as they are all in B, and an = a(n+1) as they are both in C. So all the elements of A are the same. Hence all numbers are the same.

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mugaliens said:
I shoudl have completed the thought: "You cannot take the square root of real negative numbers and wind up with a single real number."

As you explained, you wind up with an either/or, and there the equality fails.

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

jimmysnyder said:
Even though the proof is flawed, the conclusion is true. Let A = {a, b, c, ..., d} be a set of numbers. Then a = b= c= ... = d as I prove by induction on the number of elements in A. First of all notice that for all singletons, A = {a}, the statement is vacuously true. Next assume that it is true when the set contains n elements {a1, a2, ..., an} and let A = {a1, a2, ..., a(n+1)}. Then the set B= {a1, a2, ..., an} has n elements and so they are all the same. Likewise the set C = {a2, a3, ..., a(n+1)} has n elements and so all of them are the same. Thus a1 = a2 = ... = an as they are all in B, and an = a(n+1) as they are both in C. So all the elements of A are the same. Hence all numbers are the same.

How do you go from 'I defined A,B,C... as these sets a,b,c,d...' to mean that all numbers are the same.

Sorry! said:
How do you go from 'I defined A,B,C... as these sets a,b,c,d...' to mean that all numbers are the same.
A is a set of numbers a, b, c, ...
For instance, if A is the set { a, b, c, d }, a set with 4 numbers a, b, c, and d, then B is the set { a, b, c}, a set with 3 numbers, and C is the set { b, c, d} also a set with 3 numbers. If it is the case that for all sets with 3 numbers, the numbers are all the same, then it is also the case that for all sets with 4 numbers, all the numbers are the same. Since for all sets with 1 element, all elements are obviously the same (there is only one element) the induction is complete and all numbers are the same.

jimmysnyder said:
A is a set of numbers a, b, c, ...
For instance, if A is the set { a, b, c, d }, a set with 4 numbers a, b, c, and d, then B is the set { a, b, c}, a set with 3 numbers, and C is the set { b, c, d} also a set with 3 numbers. If it is the case that for all sets with 3 numbers, the numbers are all the same, then it is also the case that for all sets with 4 numbers, all the numbers are the same. Since for all sets with 1 element, all elements are obviously the same (there is only one element) the induction is complete and all numbers are the same.

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

Sorry! said:
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.

jimmysnyder said:
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}

Sorry! said:
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?

Sorry! said:
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.

jimmysnyder said:
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.

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