Sucker bet, my (almost) bad.

[Dale10101]

Suppose two friends were simplifying 2/(7i) and one computed (2i)/(-49) and the other computed (-2i)/7, and supposing neither could find a flaw in the other's argument, the question then being which identity is correct (and one of then is):

\[\frac{{\rm{2}}}{{\left( {{\rm{7i}}} \right)}} = \frac{{(2i)}}{{( - 49)}}\] or \[\frac{{\rm{2}}}{{\left( {{\rm{7i}}} \right)}} = \frac{{(2i)}}{{( - 7)}}\]

Hey, how about pretending i is the variable x and plugging some odd value into i like 1.321 and then seeing which identity holds. I mean there are no i's raised to a power which would definitely be a problem, a symbol is a symbol, so, honestly, if you just look are the identities and imagine plugging in the same value for i on both sides of the equation ... well, Bob's your uncle, right? Sawbuck YOU'RE right?

Good thing I tried it ... made me think about unforeseen consequences when changing domains, and the limits of substitution, and in fact, my ready acceptance that this sort of error checking works with identities confined to the real numbers. (Still feeling some tremors in the after shock.)

Of course one way to check which identity is correct is to cross multiply and resolve each identity to find see which simplifies to 1 = 1, but short of changing the expression to it exponential form is there any computational way a junior algebrist could check their result? I think not, but then, I think I might have made a mistake once before (did I hear a snort somewhere).

This question is probably an under hand pitch to the plate ... but maybe it will yield a pop fly if nothing else.;-)

P.S. (-2i)/7 is correct,

\[\frac{2}{{(7i)}} = \frac{{( - 2i)}}{{(7)}} = > (2)(7) = (7i)( - 2i) = > 14 = - 14{i^2} = > 14 = - 14( - 1) = > 14 = 14 = > 1 = 1\]

 

[pka]

Dale (for my benefit!) why not simply:
-(7i 2i) / 14 = -14i^2 / 14 = -i^2 = -(-1) = 1 ?

Because that is incorrect.
You have \(\displaystyle \dfrac{1}{z} = \dfrac{{\overline z }}{{{{\left| z \right|}^2}}}\).

Thus \(\displaystyle \dfrac{2}{7i} = \dfrac{2(-7i)}{49}=\dfrac{-2i}{7}\)

 

[Dale10101]

hmmm

I am still sorting out this goofy set of circumstances but it suddenly occurs to me that the reason that the whole idea of looking at "i" as a variable for the purpose of error checking is silly/fallacious (still sort of hypnotized by the idea) is because "i" is not a variable, it is an operator ... I think? ... maybe that should be obvious ... I AM NOT putting any money on the table, period. Still fathoming the last post.

Well, it's at least a fly ball over the short stop and I am learning something, however obtuse.

 

[Dale10101]

so ...

Because that is incorrect.
You have \(\displaystyle \dfrac{1}{z} = \dfrac{{\overline z }}{{{{\left| z \right|}^2}}}\).

Thus \(\displaystyle \dfrac{2}{7i} = \dfrac{2(-7i)}{49}=\dfrac{-2i}{7}\)

You seem to be saying M. PKA that the "useful formula connecting a complex number "z", its conjugate "z bar" and it's absolute value |z|", namely:

\[(z)(\bar z) = |z{|^2}\]

cannot be transformed to:

\[\frac{1}{z} = \frac{{\bar z}}{{|{z^2}|}}\]

by the usual algebraic means of dividing both sides of the equation by the same number (a complex number in this case) ?

I think that is that you are saying ... but enough for today.

 

[daon2]

You cannot just replace i with another number to verify the identity is correct. Example, consider the simple expression \(\displaystyle i\) and the choices A) \(\displaystyle -\dfrac{1}{i}\) and B) \(\displaystyle 2i-1\)

Neither choice can be "verified" by any particular substitution for \(\displaystyle i\), but the incorrect answer (B) has the same value as the original expression when replacing \(\displaystyle i\) with \(\displaystyle 1\). So you might conclude A) is the wrong one, when it is in fact the right one.

 

[Dale10101]

Yes

You cannot just replace i with another number to verify the identity is correct. Example, consider the simple expression \(\displaystyle i\) and the choices A) \(\displaystyle -\dfrac{1}{i}\) and B) \(\displaystyle 2i-1\)

Neither choice can be "verified" by any particular substitution for \(\displaystyle i\), but the incorrect answer (B) has the same value as the original expression when replacing \(\displaystyle i\) with \(\displaystyle 1\). So you might conclude A) is the wrong one, when it is in fact the right one.

Quite right, and that is what I discovered when I tried that substitution on those original two identities ... but it was so tempting of face ... I was briefly seduced but, alas, sobered by the results.

The question then became "why doesn't it work?". My conclusion at this point is ... because "i" is essentially an operator that modifies the meaning of the number it is attached to and hence replacing "i" with a variable would be like replacing a + sign with a variable ... an error that I am sure that I made while still in an awe struck state ... but you know, maybe I was just dumbstruck. :|

 

[JeffM]

Quite right, and that is what I discovered when I tried that substitution on those original two identities ... but it was so tempting of face ... I was briefly seduced but, alas, sobered by the results.

The question then became "why doesn't it work?". My conclusion at this point is ... because "i" is essentially an operator that modifies the meaning of the number it is attached to and hence replacing "i" with a variable would be like replacing a + sign with a variable ... an error that I am sure that I made while still in an awe struck state ... but you know, maybe I was just dumbstruck. :|

I think you have made this way more complex than it need be.

\(\displaystyle \dfrac{2 * 3 + 1}{7} = 1\ but\ \dfrac{2 * 4 + 1}{7} \ne 1.\)

3 is just a symbol, but the number 3 cannot necessarily be replaced by any other number such as the number 4.

If i is a number, it has certain properties that distinguish it from every other number. If you are testing whether i has a unique property, you obviously cannot do so by using a different number.

\(\displaystyle i + 2i = 3i\) indicates a property that i shares with all the other complex numbers.

\(\displaystyle i * i = - 1\) indicates a property that is unique to i.

I think the root of your confusion is that symbols are admittedly arbitrary. Just as we use a Greek letter, pi, to represent the ratio of circumference of a circle to its diameter, we could just as well have used a Greek letter, say lambda, to represent the number that when squared equals minus 1. But numbers are not arbitrary. Once we have decided that i represents the number that when squared equals minus 1, we can't replace i with some other number and expect a valid result.

 

[Dale10101]

Yikes! you can jump down to the OMG moment if you like.

I think you have made this way more complex than it need be.

\(\displaystyle \dfrac{2 * 3 + 1}{7} = 1\ but\ \dfrac{2 * 4 + 1}{7} \ne 1.\)

3 is just a symbol, but the number 3 cannot necessarily be replaced by any other number such as the number 4.

If i is a number, it has certain properties that distinguish it from every other number. If you are testing whether i has a unique property, you obviously cannot do so by using a different number.

\(\displaystyle i + 2i = 3i\) indicates a property that i shares with all the other complex numbers.

\(\displaystyle i * i = - 1\) indicates a property that is unique to i.

I think the root of your confusion is that symbols are admittedly arbitrary. Just as we use a Greek letter, pi, to represent the ratio of circumference of a circle to its diameter, we could just as well have used a Greek letter, say lambda, to represent the number that when squared equals minus 1. But numbers are not arbitrary. Once we have decided that i represents the number that when squared equals minus 1, we can't replace i with some other number and expect a valid result.

YOU CAN JUMP DOWN TO THE "OMG" MOMENT IF YOU LIKE.

This is getting out of hand. In my original post I should have made clear that my original (bad) idea was to replace this:

\[\frac{{\rm{2}}}{{\left( {{\rm{7i}}} \right)}} = \frac{{(2i)}}{{( - 49)}}\] or \[\frac{{\rm{2}}}{{\left( {{\rm{7i}}} \right)}} = \frac{{(2i)}}{{( - 7)}}\]

with this:

\[\frac{{\rm{2}}}{{\left( {{\rm{7x}}} \right)}} = \frac{{(2x)}}{{( - 49)}}\] or \[\frac{{\rm{2}}}{{\left( {{\rm{7x}}} \right)}} = \frac{{(2x)}}{{( - 7)}}\]

and yes, the misunderstanding is my fault because I said,

"Hey, how about pretending i is the variable x and plugging some odd value into i like 1.321, instead of,

"Hey, how about pretending i is the variable x and plugging some odd value into x like 1.321, (and I actually switched the wording back and forth thinking I was emphasizing the thrust of the idea I was trying to convey, alas.)

The idea was to try and find a way to verify which identity was correct, as it turned out neither of the statement were identities (but instead were equations with a limited number of true values). So you see I was not thinking of substituting one number for another ...

OMG ... I see your point, there was no variable to begin with ... nuts, it began with the idea of punching numbers into a calculator to see if two different forms of a real valued fraction that contained radicals were equal, but that was not possible with complex numbers included ... so it was a sunny day, I turned left at the intersection, maybe I was talking on my cell phone ... blah, blah ...

well, better to be fooled for a day then a life time .... me falls down, me picks meself up ....Onward!

Thanks for your insight.