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Quadratic Mystery

If a "discovery" counts as an "invention"...
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Given the number (c), and told that it equals (a)*(b), where all three are different odd numbers, then when attempting to develop an algorithm to find (a) and (b), one of the very common things that appears in the mathematics is the Quadratic Formula. To refresh your memory about that, see the first link.

An example line of reasoning, that leads to needing the quadratic formula, goes like this:
I know that (a)*(b) = (c)
Suppose I knew what (a)+(b) added to equal? Call it (d), to represent it, even though I don't actually know it.
If (a)+(b)=(d), then it logically follows that (b) = (d)-(a)
I can plug that into the multiplication equation, like this:
(a)*[(d)-(a)] = (c)
That can be algebraically manipulated to become this:
[(a)*(d)] - [(a)*(a)] = c --and then:
0 = c + [(a)*(a)] - [(d)*(a)] --and then:
[(a)*(a)] - [(d)*(a)] + c = 0

That last equation is suitable for applying the quadratic formula. Please don't be confused by the way certain variables are used here, and the way they are used in the formula. There, the (a) and (b) and (c) of the formula are "coefficients" of a descending series of "powers" of (x) --including x-raised-to- the-zero-power (an invisible 1 that (c) multiplies). Here I'm using (a) instead of (x), and the relevant/respective coefficients of powers of (a) are an invisible multiplying (1) for the second power of (a), a (-d) for the first power of (a), and (c) for the zeroth power of (a). Thus we will end up with this:
(a) = [(d) ± sqrt([(d)*(d)] - [(4)*(c)])] / 2

Now we could make guesses regarding (d), and see what value of (a) is computed --which should be one of the factors of (c). Indeed, because of that ± symbol (add or subtract), the quadratic formula can always compute two results, and in this case both results will be factors of (c).

Feel free to experiment with some known values. For example, since we know that 5*7=35, we could say (a)=5, (b)=7, (c)=35, and (d)=5+7=12. So plug the 12 and the 35 into the formula, and see it produce 5 and 7.

Now here's the thing. During a number of years of playing with algebra and the problem of finding the factors of (c), every time I found an application of the quadratic formula to compute one of the factors of (c), the formula would also always be able to compute the other factor of (c).

Until just a day or two ago, as I write this. That's the Quadratic Mystery! Why in this new formula (below) did it only compute one of the factors of (c)? I think the answer has something to do with "rigging" the logic....

Here's the initial line of reasoning:
If I assume that (b) is larger than (a) --see the start of this text, where it states that (a), (b), and (c) are all different from each other-- then I could imagine creating an equation to compute (b) from (a), like this:
If (a)=5 and (b)=13, then (a)+8 = 13= (b), or [(a)*2] + 3 = 13 = (b)

I chose to use the second possibility, because very often (b) is greater than (a) by several times, and I didn't want the number I added (8 or 3 above) to get really large --it will never need to be larger than (a) if I use the multiplier version. So, I need two variables, one for the unknown multiplier --call it (d)-- and one for the unknown amount to add --call it (e). Therefore:
[(a)*(d)] + (e) = (b)
I can plug THAT into the original equation (a)*(b) = (c), and get:
(a)*{[(a)*(d)] + (e)} = (c) --which can become:
[(a)*(a)*(d)] + [(a)*(e)] = (c) --which can become:
[(d)*(a)*(a)] + [(e)*(a)] - (c) = 0 --which is now ready for the quadratic formula. We still want to compute (a), but now the relevant/respective coefficients of the powers of (a) are (d), (e), and (-c):

(a) = [(-e) ± sqrt([(e)*(e)] + [4*(d)*(c)])] / [2*(d)]

When known values are plugged into the formula to test it, it can compute two different values for (a) --one will be a positive number and the other will be a negative number- - and when either of those values is plugged into the original equation:
(a)*{[(a)*(d)] + (e)} = (c)
both result in a valid calculation of (c). But only one of them is actually a factor of (c)! That value is the positive number (replace the ± symbol with a simple +), which also is (a), the **smaller** factor of (c). (But then see above about "rigging" the logic....)

Now it just so happens that a slight variation of the formula is able to compute the other factor of (c), which is (b):

(b) = [(e) + sqrt([(e)*(e)] + [4*(d)*(c)])] / 2

You don't have to take my word for it; you can take the equation
(a)*(b)=(c)
and replace both (a) and (b) with the complicated formulas, and it works --a mess of algebra can ultimately yield (c)=(c) --which technically counts as proof the second version of the quadratic formula is as valid as the first (with the ± symbol replaced by +), each for computing ONE factor of (c).

Is all this mathematical playing around actually useful for finding the factors of (c) when it is a really big number (like a thousand digits long)? I think the answer to that is "NO!". But I also thought this particular unusual thing was interesting enough to post here at the HalfBakery. Hope you liked it!

Vernon, Jul 30 2016

Quadratic Formula http://www.purplema...odules/quadform.htm
As mentioned in the main text. [Vernon, Jul 30 2016]

Mathematical Powers http://www.purplema...odules/exponent.htm
Another item for refreshing of the memory. [Vernon, Jul 31 2016]

[link]






       Nope, you’re good as is.
Ian Tindale, Jul 30 2016
  

       [Ian Tindale], someone might be able to ONCE get away with posting an Idea with "Mystery" in the title, and leaving the main text blank, like I accidentally did by pressing "Enter" at the wrong time, starting the process of adding this Idea, but I'm pretty sure it would not be accepted more than once. (And now that I've described it, maybe not even once. But I think I'll try it and see....)
Vernon, Jul 30 2016
  

       Did you work at Bletchley Park?
whatrock, Jul 31 2016
  

       Hi Vernon, can you explain the second round of substitution you're doing (i.e. what's e all about?) and, can you give some data for what values fall into the "working" category, and what values fall into the "non-working" category?
zen_tom, Aug 03 2016
  

       I did explain what (e) was about. You can convert any number into a larger number by multiplying it by something (even if only "1"), and then adding something. (e) is the number that gets added. If you divided your smaller number into the larger, (e) would be the remainder of the division.
Vernon, Aug 03 2016
  
      
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