Can anyone spot the error in this fallacy

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Discussion Overview

The discussion revolves around a mathematical fallacy involving the manipulation of equations to conclude that all numbers are the same. Participants analyze the steps taken in the proof, particularly focusing on the implications of taking square roots and the conditions under which the equations hold true.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants point out that the error occurs when taking the square root in the transition from the sixth to the seventh line, emphasizing the need to consider both positive and negative roots.
  • Others argue that the mathematical steps seem valid but lead to inconsistencies when specific values are substituted, suggesting that the proof does not hold for all cases.
  • A participant asserts that the proof incorrectly concludes that a and b must be equal, highlighting that the equations only imply a = b under certain conditions.
  • Some participants express confusion over the calculations, questioning the validity of the results obtained from the fifth and sixth lines when specific numbers are used.
  • A later reply proposes a corrected version of the proof, indicating that a can equal b or a can equal t - b, thus allowing for the possibility that the numbers are not necessarily the same.
  • Another participant mentions that the original proof resembles other logical fallacies that incorrectly assert equivalences, such as 1 = 2.

Areas of Agreement / Disagreement

Participants generally disagree on the validity of the proof and the implications of the mathematical manipulations. There is no consensus on whether the original argument is sound, as multiple viewpoints and interpretations of the steps are presented.

Contextual Notes

Limitations include the potential misunderstanding of square root properties and the assumptions made about the values of a, b, and t. The discussion highlights the importance of careful mathematical reasoning and the conditions under which certain equations hold.

Who May Find This Useful

This discussion may be of interest to those studying mathematics, particularly in understanding common fallacies and the importance of rigor in mathematical proofs.

mjordan2nd
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<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!
 
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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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  • #10
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.
 
  • #11
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.
 
  • #12
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.
 
  • #13
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?
 
  • #14
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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  • #15
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.
 
  • #16
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.
 
  • #17
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!
 
  • #18
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 :smile: Did you get hit with a fish yet?
 
  • #19
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.
 
  • #20
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.
 
  • #21
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
 
  • #22
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!
 
  • #23
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.
 
  • #24
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:


<br /> (a - t/2)^2<br />
Our answer will be
<br /> a-t/2 OR -a+t/2
 
  • #25
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.
 
  • #26
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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  • #27
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.
 
  • #28
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.
 
  • #29
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.
 
  • #30
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.
 

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