Register to reply 
Can anyone spot the error in this fallacy 
Share this thread: 
#1
Dec209, 02:50 PM

P: 118

<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! 


#2
Dec209, 03:00 PM

P: 922

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



#3
Dec209, 09:03 PM

P: 118

Could you explain why that is erroneous?



#4
Dec209, 09:53 PM

P: 17

Can anyone spot the error in this fallacy
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 dont know why is that. but i would like to know 


#5
Dec209, 10:03 PM

P: 4,513

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



#6
Dec209, 11:37 PM

P: 571




#7
Dec209, 11:47 PM

P: 17

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. 


#8
Dec209, 11:50 PM

Mentor
P: 7,315

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. 


#9
Dec209, 11:52 PM

P: 571

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. 


#10
Dec309, 12:22 AM

P: 571

I don't think you are doing your math correctly anyways (from what I read up above) You do remember BEDMAS correct?
[tex](1\frac{3}{2})^2=(2\frac{3}{2})^2[/tex] 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: [tex]1\frac{3}{2}[/tex] which = 0.5 or we have [tex]1+\frac{3}{2}[/tex] 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
Dec309, 12:25 AM

P: 17




#12
Dec309, 12:27 AM

P: 571




#13
Dec309, 12:32 AM

P: 17

5th line 1^2 3(1) + 3^2/4 = 2^2  3(2) + 3^2/4 0.25 = 0.25 6th line (13/4)^2 = (23/4)^2 0.5625 = 1.5625 how is that wrong math? 


#14
Dec309, 12:42 AM

P: 571

[tex]a^2  ta + (t^2)/4 = b^2  tb + (t^2)/4[/tex] OR [tex]13+1.5=46+1.5[/tex] Both sides equal 0.5 6th line: [tex](a  t/2)^2 = (b  t/2)^2[/tex] OR [tex](11.5)^2=(21.5)^2[/tex] [tex](0.5)^2=(0.5)^2[/tex] 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 (13/4)^2 does not equal 0.5625 


#15
Dec309, 12:47 AM

P: 17




#16
Dec309, 12:51 AM

P: 571




#17
Dec309, 01:03 AM

P: 17




#18
Dec309, 01:47 AM

P: 571

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


Register to reply 
Related Discussions  
Can you spot the fallacy?  Calculus  10  
Where's the fallacy  General Math  5  
Spot the error in my code. (in C).  Programming & Computer Science  11  
Find the fallacy  Calculus & Beyond Homework  2  
Another fallacy  General Math  2 