Does 1= -1?
bob Slotz
PROOF:
1. 1 = 1
2. 1 = sqrt(1)
3. 1 = sqrt( (-1)(-1) )
4. 1 = sqrt( -1) * sqrt( -1)
5. but i = sqrt(-1) by definition.
6. Therefore: 1 = i * i
7. but i^2 = -1
8. Therefore: 1 = -1
Explain where and what is logically "wrong" here.
IF ANYONE CAN HELP ME, I WOULD BE VERY HAPPY.
-**** Posted from remarQ, Discussions Start Here(tm) ****-
http://www.remarq.com/ - Host to the the World's Discussions & Usenet
Randy Yates
I'll give you a clue bob: If 4 = 2 * 2, and 4 = -2 * -2, then
does sqrt(4) equal +2 or -2??? Look up square roots in your
textbook for more information.
--
% Randy Yates % "So now it's getting late,
%% DIGITAL SOUND LABS % and those who hesitate
%%% Digital Audio Sig. Proc. % got no one..."
%%%% <ya...@ieee.org> % 'Waterfall', *Face The Music*, ELO
http://www.shadow.net/~yates
Roger Andersson
bob Slotz <Schmy...@hotmail.com> wrote in
article<Wlxq2.5475$IJ3.44928090@WReNphoon2>...
> My Teacher Handed out this problem which I am trying to solve.
>
> PROOF:
> 1. 1 = 1
> 2. 1 = sqrt(1)
> 3. 1 = sqrt( (-1)(-1) )
> 4. 1 = sqrt( -1) * sqrt( -1)
> 5. but i = sqrt(-1) by definition.
> 6. Therefore: 1 = i * i
> 7. but i^2 = -1
> 8. Therefore: 1 = -1
>
> Explain where and what is logically "wrong" here.
>
> IF ANYONE CAN HELP ME, I WOULD BE VERY HAPPY.
>
>
>
> -**** Posted from remarQ, Discussions Start Here(tm) ****-
> http://www.remarq.com/ - Host to the the World's Discussions & Usenet
>
The logic seems to be alright.
There's just one fatal error ; the definition of "i" is not i = sqrt(-1)
which I know is commonly used.
"i" is defined as the number that equals -1 if squared.
That is i^2 = -1.
Using this definition we can't go wrong as in step 5 above.
I think the reasoning above (steps 1 - 8 ) shows the extreme
importance of exact definitions in mathematics.
oz
at step two. See something wrong there? --------- Oz
bob Slotz wrote in message ...
Stan Brown
In the next-to-last year of the millennium, bob Slotz
(Schmy...@hotmail.com) wrote in newsgroup alt.algebra.help, article
<Wlxq2.5475$IJ3.44928090@WReNphoon2>:
> 2. 1 = sqrt(1)
> 3. 1 = sqrt( (-1)(-1) )
This depends on sqrt(ab) = sqrt(a)*sqrt(b). But that is valid only if a
and b are both positive.
--
Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
http://www.concentric.net/%7eBrownsta/
Can anyone suggest a good national (U.S.) ISP with TCP/IP, good
Usenet, and a UNIX shell? After five months of disconnects at
Concentric, I'm ready for a change.
Stan Brown
newsgroup alt.algebra.help, article <F62qn...@news2.new-york.net>:
>bob Slotz wrote in message ...
>> 2. 1 = sqrt(1)
>at step two. See something wrong there? --------- Oz
No, step 2 is fine. It depends on substitution, not on taking the sqrt of
just one side of the equation.
The problem lies in step 3 of the original.
Virgil Hancher
> >My Teacher Handed out this problem which I am trying to solve.
> >
> > PROOF:
> > 1. 1 = 1
> > 2. 1 = sqrt(1)
> > 3. 1 = sqrt( (-1)(-1) )
> > 4. 1 = sqrt( -1) * sqrt( -1)
> > 5. but i = sqrt(-1) by definition.
> > 6. Therefore: 1 = i * i
> > 7. but i^2 = -1
> > 8. Therefore: 1 = -1
> >
> > Explain where and what is logically "wrong" here.
In going from equation 2 to equation 3, you are using the rule that
sqrt(a*b) = sqrt(a)*sqrt(b), but this rule only applies when a and
b are both non-negative. It does NOT hold in this situation!
David Brownridge
> In math, what ever is done to one side must be done to the other side....
Actually that's not so (assuming you're talking about the two sides of
an equation). One may do something to one side only of an equation that
does not change it's value. Eg, suppose we have
Eq.1 3x + 2 = y
and we also know
Eq.2 x = y + 1
Then we substitute Eq.2 into the left hand side of Eq.1 to get
3(y + 1) + 2 = y
and proceed to work with that -- but we haven't done anything to "the
opposite side".
I suspect that this is what's happening at step 2 of the original
"proof".
> Look at step two. See something wrong there? --------- Oz
> bob Slotz wrote in message ...
> >My Teacher Handed out this problem which I am trying to solve.
> >
> > PROOF:
> > 1. 1 = 1
> > 2. 1 = sqrt(1)
> > 3. 1 = sqrt( (-1)(-1) )
> > 4. 1 = sqrt( -1) * sqrt( -1)
> > 5. but i = sqrt(-1) by definition.
> > 6. Therefore: 1 = i * i
> > 7. but i^2 = -1
> > 8. Therefore: 1 = -1
> >
> > Explain where and what is logically "wrong" here.
--
rgds
DVD
(David Brownridge) <mailto:d...@melbpc.org.au>
David Brownridge
> > >
> > > PROOF:
> > > 1. 1 = 1
> > > 2. 1 = sqrt(1)
> > > 3. 1 = sqrt( (-1)(-1) )
> > > 4. 1 = sqrt( -1) * sqrt( -1)
> > > 5. but i = sqrt(-1) by definition.
> > > 6. Therefore: 1 = i * i
> > > 7. but i^2 = -1
> > > 8. Therefore: 1 = -1
> > >
> > > Explain where and what is logically "wrong" here.
>
> sqrt(a*b) = sqrt(a)*sqrt(b), but this rule only applies when a and
> b are both non-negative. It does NOT hold in this situation!
Surely you mean going from eq.3 to eq.4 ?
From eq.2 to eq.3 is just substituting -1*-1=1
--
rgds
DVD
(David Brownridge) <mailto:d...@melbpc.org.au>
(mailed & posted)
David Brownridge
>
> [posted and e-mailed]
>
> In the next-to-last year of the millennium, bob Slotz
> (Schmy...@hotmail.com) wrote in newsgroup alt.algebra.help, article
> <Wlxq2.5475$IJ3.44928090@WReNphoon2>:
> > 2. 1 = sqrt(1)
> > 3. 1 = sqrt( (-1)(-1) )
>
> and b are both positive.
Umm -- methinks you refer to the step:
> 3. 1 = sqrt( (-1)(-1) )
> 4. 1 = sqrt( -1) * sqrt( -1)
I wonder if there's a problem with some writers referring to step
numbers and others referring to line/equation numbers, where step n
_might_ be the step from eq.n to eq.n+1, or it might be the step
resulting in eq.n ??
--
rgds
DVD
(David Brownridge) <mailto:d...@melbpc.org.au>
(posted & mailed)
Gypsogecky
Usenet, and a UNIX shell? After five months of disconnects at
Concentric, I'm ready for a change.
>>
You could try Earthlink. I don't know if it has what you're looking for, but I
do know that it gives me great service. =) it's $19.95/mo I believe. =) hope
this helps!! check out their website at http://www.earthlink.net. See ya!!
Maggie
***Need some advice? Feel free to e-mail me at laure...@earthlink.net ...
I'll get back to you within one day! =D***
Stan Brown
(d...@melbpc.org.au) wrote in newsgroup alt.algebra.help, article
<36AD0E...@melbpc.org.au>:
>> > 2. 1 = sqrt(1)
>> > 3. 1 = sqrt( (-1)(-1) )
>>
>> This depends on sqrt(ab) = sqrt(a)*sqrt(b). But that is valid only if a
>> and b are both positive.
>
>Umm -- methinks you refer to the step:
>
>> 3. 1 = sqrt( (-1)(-1) )
>> 4. 1 = sqrt( -1) * sqrt( -1)
>
You're right, I quoted the wrong two lines; I had meant to quote the ones
you quoted.
--
Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
http://www.concentric.net/%7eBrownsta/
Darrell Ryan
>
> oz wrote:
>
> > In math, what ever is done to one side must be done to the other side....
>
> Actually that's not so (assuming you're talking about the two sides of
> an equation). One may do something to one side only of an equation that
> does not change it's value. Eg, suppose we have
>
> Eq.1 3x + 2 = y
>
> and we also know
>
> Eq.2 x = y + 1
>
> Then we substitute Eq.2 into the left hand side of Eq.1 to get
>
> 3(y + 1) + 2 = y
>
> and proceed to work with that -- but we haven't done anything to "the
> opposite side".
Nor have we really "done" anything to the lhs either. Just replaced x
with an equivalent value. Say we have the equation:
1 = 1
...and since 1 = 3/3 we do:
1 = 3/3
...or for that matter:
3/3 = 3/3
I don't think this is what is meant when most people think of "doing"
things to both sides of an equation in the sense that adding,
subtracting, multiplying, dividing, exponentiating, or taking roots,
etc. are "doing" things to an equation.
>
> I suspect that this is what's happening at step 2 of the original
> "proof".
>
> > Look at step two. See something wrong there? --------- Oz
Look closely at step 4 and how it was obtained from step 3. The
property sqrt(ab)=sqrt(a)sqrt(b) only works when a and b are not
*both* negative.
>
> > bob Slotz wrote in message ...
> > >My Teacher Handed out this problem which I am trying to solve.
> > >
> > > PROOF:
> > > 1. 1 = 1
> > > 2. 1 = sqrt(1)
> > > 3. 1 = sqrt( (-1)(-1) )
> > > 4. 1 = sqrt( -1) * sqrt( -1)
> > > 5. but i = sqrt(-1) by definition.
> > > 6. Therefore: 1 = i * i
> > > 7. but i^2 = -1
> > > 8. Therefore: 1 = -1
> > >
> > > Explain where and what is logically "wrong" here.
--
Darrell
Darrell Ryan
> No, step 2 is fine. It depends on substitution, not on taking the sqrt of
> just one side of the equation.
>
> The problem lies in step 3 of the original.
1. 1 = 1
2. 1 = sqrt(1)
3. 1 = sqrt((-1)(-1))
4. 1 = sqrt(-1)sqrt(-1)
Sqrt((-1)(-1)) = sqrt(1). Step 3 is just a substitution of an equal
value just like step 2 was. the problem is going from step 3 to step
4. Sqrt(ab)<>sqrt(a)sqrt(b) when a and b are both negative.
--
Darrell
Mark Olson
taken, the result will be denoted as (plus or minus). And when looking at the
final result, one or both of your answers might have to be thrown out.
Proof: 1 = 1
1 = (-1) (-1)
(1)^1/2 = {plus or minus}[(-1) (-1)]^1/2
1 = {plus or minus} (-1)^1/2 (-1)^1/2
1 = {plus or minus} i * i
1 = {plus or minus} i^2
1 = {plus or minus} (-1)
1 = {minus} (-1) Plus is thrown out.
1 = 1
Mark Olson
Darrell Ryan wrote:
> > > > 1. 1 = 1
> > > > 2. 1 = sqrt(1)
Darrell Ryan
>
> The problem in your proof starts in step 3. When the square root of a square is
> taken, the result will be denoted as (plus or minus). And when looking at the
> final result, one or both of your answers might have to be thrown out.
>
> Proof: 1 = 1
> 1 = (-1) (-1)
> (1)^1/2 = {plus or minus}[(-1) (-1)]^1/2
> 1 = {plus or minus} (-1)^1/2 (-1)^1/2
> 1 = {plus or minus} i * i
> 1 = {plus or minus} i^2
> 1 = {plus or minus} (-1)
> 1 = {minus} (-1) Plus is thrown out.
> 1 = 1
Timeout :-)
You are assuming that someone took the square root of both sides.
That is not necessarily so. The statement 1=1 expresses the equality
of two expressions (in this case the same expression.) The statement
1=sqrt(1) also expresses the equality of two expressions, although
they are not identical expressions. BUT...the expression sqrt(1)
represents the exact same number as the expression 1. No plus or
minus.
The radical (sqrt) by convention implies only the principal root (not
the other root) hence 1=sqrt(1) and 1=1 are equivalent equations
(every solution to one is also a solution to the other.) Recall when
taking sqrt of both sides not every solution to one is a solution to
the other because as you know we often have to throw out a proposed
solution (implying the equations are not necessarily equivalent)).
If I have:
1 = 1
and I then write:
1 = sqrt(1)
I am not necessarily implying that I took the square root of both
sides of the equation. All I did was substitute an equally valid
expression on one side and one side only. Just like if I had:
1 = 1
and then wrote:
1 = 3/3
1 and 3/3 represent the exact same number just like 1 and sqrt(1)
represent the exact same number. Your "proof" above uses this very
act of substitution when going from:
1 = 1
...to
1 = (-1)(-1)
Of course you did not perform any one manipluation to both sides of
the equation. All you did was substitute the expression (-1)(-1) for
the expression 1 on the rhs only. That's OK, there is nothing wrong
with that.
In the next step, you make it clear that you are taking the square
root of both sides and properly indicate both the + and - roots.
Nothing wrong there.
In the next step, you make use of a property that says
sqrt(ab)=sqrt(a)sqrt(b) and this you *cannot* do when a and b are
*both* negative.
Counterexample:
sqrt[(-1)(-1)] = sqrt(1) = 1
...while
sqrt(-1)sqrt(-1) = i^2 = -1
Although technically an *illegal* move, the +/- resulting from your
explicit taking the square root of both sides makes it appear the
method is correct because you just conveniently throw out the one that
doesn't fit.
But...
1 = 1
and
1 = sqrt(1)
does not imply taking roots. We just substituted an equivalent
expression for 1 on the right side only. Nothing wrong with that.
In case anyone is wondering, the property sqrt(ab)=sqrt(a)sqrt(b)
holds when a *or* b is negative. Just as long as they are not *both*
negative.
Example:
sqrt[(-2)(3)] = sqrt(-6) = i*sqrt(6)
...and
sqrt(-2)sqrt(3) = (i*sqrt(2)) * sqrt(3) = i*sqrt(2*3) = i*sqrt(6)
--
Darrell
Stan Brown
>Stan Brown wrote:
>
>> No, step 2 is fine. It depends on substitution, not on taking the sqrt of
>> just one side of the equation.
>>
>> The problem lies in step 3 of the original.
>
>2. 1 = sqrt(1)
>4. 1 = sqrt(-1)sqrt(-1)
>
>Sqrt((-1)(-1)) = sqrt(1). Step 3 is just a substitution of an equal
>value just like step 2 was. the problem is going from step 3 to step
I agree -- as I posted back on January 26 when someone pointed out that I
had meant to capture steps 3-4 instead of 2-3.
--
Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
http://www.concentric.net/%7eBrownsta/
The above e-mail address and URL are valid for only the next few days.
Randy Yates
>
> In the next-to-last year of the millennium, Darrell Ryan (ry...@edge.net)
> wrote in newsgroup alt.algebra.help, article
> <36B0066A...@edge.net>:
> >Stan Brown wrote:
> >
> >> No, step 2 is fine. It depends on substitution, not on taking the sqrt of
> >> just one side of the equation.
> >>
> >> The problem lies in step 3 of the original.
> >
> >1. 1 = 1
> >2. 1 = sqrt(1)
> >3. 1 = sqrt((-1)(-1))
> >4. 1 = sqrt(-1)sqrt(-1)
> >
> >Sqrt((-1)(-1)) = sqrt(1). Step 3 is just a substitution of an equal
> >value just like step 2 was. the problem is going from step 3 to step
> >4. Sqrt(ab)<>sqrt(a)sqrt(b) when a and b are both negative.
>
> I agree -- as I posted back on January 26 when someone pointed out that I
> had meant to capture steps 3-4 instead of 2-3.
I disagree with you both.
The problem is that the "sqrt()" operation is being
applied improperly. BY DEFINITION, sqrt(1)=+1 *OR* -1.
Or, more precisely, sqrt(a*a) = +/-|a|. That is to
say, you cannot specify whether sqrt(1) is -1 or +1 until
you have more information about the problem or the
way in which the argument is factored. Therefore, the
problem in the "proof" above is in going from step
1 to step 2 since one cannot dictate that sqrt(1) = +1.
Alternately, if you do choose the positive square root (in,
step 2) you are implying that we are choosing the positive
factors, and then the problem is in going from 2 to 3.
However, in either case, the problem occurs prior to
step 4.
--
% Randy Yates % "Midnight, on the water...
%% DIGITAL SOUND LABS % I saw... the ocean's daughter."
%%% Digital Audio Sig. Proc. % 'Can't Get It Out Of My Head'
%%%% <ya...@ieee.org> % *El Dorado*, Electric Light Orchestra
http://www.shadow.net/~yates
Randy Yates
>
> In the next-to-last year of the millennium, Randy Yates
> (ya...@shadow.net) wrote in newsgroup alt.algebra.help, article
> <36B27767...@shadow.net>:
> >The problem is that the "sqrt()" operation is being
> >applied improperly. BY DEFINITION, sqrt(1)=+1 *OR* -1.
> >Or, more precisely, sqrt(a*a) = +/-|a|.
>
That's a good question. I have to answer no at the moment because
I'm at home and all my textbooks are at work. I'll check next
time I'm there.
I do find the question of authority somewhat interesting myself,
but I must say that even if I do not find it in any of my books,
I will continue to operate the way I have discussed because my
experience has shown this method to be fruitful.
> Your "definition" works only if one gives up the idea of sqrt() as a
> function.
Precisely.
> Furthermore, if sqrt(x) meant either of two values, why would there be a
> +- sign in the conventional quadratic formula? If sqrt() had the meaning
> you wish to give it, then the quadratic formula would be written with
> just a + sign (or just a - sign) before the radical. But I have *never*
> seen it written that way, and I daresay you haven't either.
No, of course I've never seen it that way, Stan. :) It is just
simpler to convey the idea by writing it that way.
> Conclusion: the square-root sign signifies the positive square root; and
> since sqrt() in Usenet stands for that sign, sqrt() is also the positive
> square root.
What about the other half of the answer??? The symbol "sqrt(b)" is
asking the question, "What value, when squared, gives b?" If we only
give the positive square root, we are missing half of the answer.
Also, who gives a hoot what "sqrt()" stands for in Usenet??? The original
poster was ostensibly quoting a problem from class.
Now I'll put you on the hot seat, Stan (and Darrell). You stated that
"Sqrt(ab)<>sqrt(a)sqrt(b) when a and b are both negative" right? From
what authority do you take this? Isn't the exponent rule (a*b)^z=(a^z)*(b^z)
valid in all cases?
--
% Randy Yates % "Bird, on the wing,
%% DIGITAL SOUND LABS % goes floating by
%%% Digital Audio Sig. Proc. % but there's a teardrop in his eye..."
%%%% <ya...@ieee.org> % 'One Summer Dream', *Face The Music*, ELO
http://www.shadow.net/~yates
Randy Yates
>
> Randy Yates wrote:
>
> >
> > I disagree with you both.
> >
> > applied improperly. BY DEFINITION, sqrt(1)=+1 *OR* -1.
>
> is +1 (pardon the puns on this one).
>
> 1 = sqrt(1) is a perfectly valid equation. It is a true statement.
> Sqrt(1) is just another name for the exact same number that 1 is.
This is the gist of the issue, and we disagree on this point. I see
no other discussion to be had between you and I, Darrell, because
we simply disagree.
Let me reiterate, just to be clear. "sqrt(1)" is either "+1" or
"-1" but the determination cannot be made until more information
on the problem or context in which the sqrt() is used is known.
That is precisely why you cannot simply replace "1" with "sqrt(1)"
when going from step 1 to step 2 in the "proof" that started this
thread.
Let me add that I present this information
on the sqrt() operation not simply by a making a series of deductions
in my head while sitting in front of the computer posting this
usenet news article nor by reading it out of a book, but rather
by having been bitten multiple times in the past when attempting
to solve problems using the (incorrect) interpretion of sqrt(a*a) as
"+|a|" (as Darrell does) instead of carrying through the "+/-" as
I suggest.
--
% Randy Yates % "...the answer lies within your soul
%% DIGITAL SOUND LABS % 'cause no one knows which side
%%% Digital Audio Sig. Proc. % the coin will fall."
%%%% <ya...@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
http://www.shadow.net/~yates
Stan Brown
(ya...@shadow.net) wrote in newsgroup alt.algebra.help, article
<36B27767...@shadow.net>:
>applied improperly. BY DEFINITION, sqrt(1)=+1 *OR* -1.
Can you quote any authority for this rather surprising statement?
Your "definition" works only if one gives up the idea of sqrt() as a
function.
Furthermore, if sqrt(x) meant either of two values, why would there be a
+- sign in the conventional quadratic formula? If sqrt() had the meaning
you wish to give it, then the quadratic formula would be written with
just a + sign (or just a - sign) before the radical. But I have *never*
seen it written that way, and I daresay you haven't either.
Conclusion: the square-root sign signifies the positive square root; and
since sqrt() in Usenet stands for that sign, sqrt() is also the positive
square root.
--
Albert Y.C. Lai
--text follows this line--
> Let me reiterate, just to be clear. "sqrt(1)" is either "+1" or
> "-1" but the determination cannot be made until more information
If you say in human language "find the square root(s) of 1", then the
answer is obviously {1, -1}.
However, the solution 1 is of a special importance and use: it is the
principal root. We want some way to name the function that maps a
number to its principal square root (or principal n-th root in
general). Most mathematicians have chosen the surd symbol, and most
mathematical users of computers have chosen to translate the surd
symbol to the function name "sqrt".
As such, "square root" and "sqrt" do not mean the same thing. "Square
root" gives both 1 and -1, "sqrt" gives 1.
Similarly, the "cube roots of 1" are 1, cos(2pi/3) + i sin(2p/3), and
cos(2pi/3) - i sin(2pi/3). But when we write out in symbols "n surd
1", we mean 1, and just that.
If you don't like this convention, too bad; but you still need some
way to name the function that maps a number to its principal square
root.
Indeed, the need for a concise symbol for the principal n-th root (or
at least the non-negative real n-th root) of a number is apparent in
the statement of de Moivre's theorem. Moreover, we do not need
another concise symbol for all the n-th roots of a number, thanks to
de Moivre's theorem, which already gives a compact expression for all
of the n-th roots in closed form.
[...]
> by having been bitten multiple times in the past when attempting
> to solve problems using the (incorrect) interpretion of sqrt(a*a) as
> "+|a|" (as Darrell does) instead of carrying through the "+/-" as
> I suggest.
Show us one example of that. I will show you the real cause of the
bite, i.e., that the correct, complete solution can still be obtained
using "sqrt(a*a) = |a|", thus demonstrating that the flaw lies
somewhere else.
--
[If you post a response, no need to cc me; if you cc me, please say so.]
Accounting is no more accurate than the input data.
Theory is no less accurate than the user.
Darrell Ryan
>
> Stan Brown wrote:
> > Conclusion: the square-root sign signifies the positive square root; and
> > since sqrt() in Usenet stands for that sign, sqrt() is also the positive
> > square root.
>
> asking the question, "What value, when squared, gives b?" If we only
> give the positive square root, we are missing half of the answer.
This, I submit, is the real gist of the disagreement. The symbol
"sqrt(b)" is not asking that question. It represents only the
principal root. This is common convention in every text I have seen.
If we want/need to consider both roots then it is written +/-sqrt(b).
This is so commonly understood I can'tr believe we are even arguing
that. :-)
>
> Also, who gives a hoot what "sqrt()" stands for in Usenet??? The original
> poster was ostensibly quoting a problem from class.
Now we are _really_ getting off on a tangent. Of course, if effective
communication is to be made we must be using the same language here
just like we would use the same language on a chalkboard or when
writing or reading a textbook. If you look at the FAQ you will see
that on this forum sqrt() denotes precisely what Stan and I have been
trying to tell you, which BTW is in no way different from the radical
symbol with an index of 2 on more conventional media (books,
chalkboards, etc.)
>
> Now I'll put you on the hot seat, Stan (and Darrell). You stated that
> "Sqrt(ab)<>sqrt(a)sqrt(b) when a and b are both negative" right?
Yes, speaking for myself of course.
> From
> what authority do you take this?
That is a fair question. Although I know I have seen this in other
texts (not to mention the several professors that I have addressed
this to in the past) the text that I have at hand right now is the 6th
Ed. of College Algebra by Lial, et al. ISBN 0-673-46752-X on pg. 92:
"The rule sqrt(c)*sqrt(d) = sqrt(cd) is valid only when c and d are
not both negative. For example,
sqrt[(-4)(-9)] = sqrt(36) = 6
while
sqrt(-4)sqrt(-9) = 2i(3i) = 6i^2 = -6
so that
sqrt[(-4)(-9)] is not equal to sqrt(-4)sqrt(-9)"
The example speaks for itself. Of course, the validity of this
example relies on the convention that the symbol sqrt() denotes only
the principal root. That, Randy, seems to be the gist of the
disagreement. Surely you won't argue the validity of the above
counterexample if one accepts that convention. If you want an
authority on that convention of sqrt() being only the principal root,
you can pretty much find that easily enough by consulting with just
about any algebra text.
Again, I find it strange you disagree with that since it has been
driven in all our heads ever since we were first taught to put +/- on
one side when we take the sqrt of both sides of an equation.
> Isn't the exponent rule (a*b)^z=(a^z)*(b^z)
> valid in all cases?
Let's see:
[(-4)(-9)]^1/2 ?= (-4)^1/2 * (-9)^(1/2)
(36)^1/2 ?= 2i * 3i
6 ?= 6i^2
6 <> -6
The gist of this argument is really the same as the gist of the
radical argument. The expression a^1/2 is equivalent, by definition,
to sqrt(a). It is not equivalent to -sqrt(a).
--
Darrell
Darrell Ryan
>
> Darrell Ryan wrote:
> >
> > Randy Yates wrote:
> >
> > >
> > > I disagree with you both.
> > >
> > > applied improperly. BY DEFINITION, sqrt(1)=+1 *OR* -1.
> >
> > is +1 (pardon the puns on this one).
> >
> > 1 = sqrt(1) is a perfectly valid equation. It is a true statement.
> > Sqrt(1) is just another name for the exact same number that 1 is.
>
> This is the gist of the issue, and we disagree on this point. I see
> no other discussion to be had between you and I, Darrell, because
> we simply disagree.
>
> "-1" but the determination cannot be made until more information
> That is precisely why you cannot simply replace "1" with "sqrt(1)"
> when going from step 1 to step 2 in the "proof" that started this
> thread.
No. It is well known and accepted convention that sqrt() denotes only
the principal root. Since 1 obviously does not equal -1, then 1 sure
doesn't equal -sqrt(1) (which *is* -1). That is all the context one
needs to equate 1 with sqrt(1).
Let me reiterate to be clear one more time that _if_ we are taking the
square root of both sides of an equation, then yes you must at least
consider _both_ roots:
1 = 1
sqrt(1) +/- sqrt(1)
CONVENTION stipulates that sqrt() is the principal root. That's why
we write +/- on one side!!!!!
..so we have the two equations:
sqrt(1) = sqrt(1) sqrt(1) = -sqrt(1)
...which is the same as:
1 = 1 1 = -1
One equation is a contradiction, so we have:
1 = 1
which is equivalently written as:
1 = sqrt(1)
...and we are right back to where we started.
BUT....
We never took the sqrt of both sides in the original procedure. All
we did was equate 1 and sqrt(1). This is *just* a valid equation as 1
= 3/3. To say that 1 = -sqrt(1) is just as much a contradiction as to
say 1 = -3/3.
>
> Let me add that I present this information
> on the sqrt() operation not simply by a making a series of deductions
> in my head while sitting in front of the computer posting this
> usenet news article nor by reading it out of a book, but rather
> by having been bitten multiple times in the past when attempting
> to solve problems using the (incorrect) interpretion of sqrt(a*a) as
> "+|a|" (as Darrell does) instead of carrying through the "+/-" as
> I suggest.
_No where_ have you heard me say that nor have you seen me use that
argument in showing why 1<>-1 in this thread. In fact, since you
mention it, sqrt(a^2)=|a| by definition which in fact supports even
more what Stan and I have been trying to tell you........sqrt()
implies the principal root (and *only* the principal root.) since
absolute value is _never_ negative, how do you justify that sqrt(a^2)
can _ever_ be negative?
--
Darrell
Stan Brown
(ya...@shadow.net) wrote in newsgroup alt.algebra.help, article
>What about the other half of the answer??? The symbol "sqrt(b)" is
>asking the question, "What value, when squared, gives b?"
Exactly: what VALUE, singular.
It is true that there is another value that can also yield the original
when squared, but x^(1/2) or sqrt(x) is asking for the principal value.
Similarly, x^(1/3) or cbrt(x) has one answer, not three, even though
there are three complex cube roots of any complex number except 1.
>I will continue to operate the way I have discussed because my
>experience has shown this method to be fruitful.
You are free to use whatever definitions you wish. But if you wish to
communicate, rather than merely spout off, you need to accept the common
definitions.
--
Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
http://www.mindspring.com/~brahms/
Please note new email address and Web page. After five months of
disconnects at Concentric, I finally gave up.
Randy Yates
>
> Randy Yates wrote:
> >
> > Darrell Ryan wrote:
> > >
> > > Randy Yates wrote:
> > >
> > > >
> > > > I disagree with you both.
> > > >
> > > > The problem is that the "sqrt()" operation is being
> > > > applied improperly. BY DEFINITION, sqrt(1)=+1 *OR* -1.
> > >
> > > 1^2 and (-1)^2 are both 1, but sqrt(1) is one and *only one* 1 and it
> > > is +1 (pardon the puns on this one).
> > >
> > > 1 = sqrt(1) is a perfectly valid equation. It is a true statement.
> > > Sqrt(1) is just another name for the exact same number that 1 is.
> >
> > This is the gist of the issue, and we disagree on this point. I see
> > no other discussion to be had between you and I, Darrell, because
> > we simply disagree.
> >
> > Let me reiterate, just to be clear. "sqrt(1)" is either "+1" or
> > "-1" but the determination cannot be made until more information
> > on the problem or context in which the sqrt() is used is known.
> > That is precisely why you cannot simply replace "1" with "sqrt(1)"
> > when going from step 1 to step 2 in the "proof" that started this
> > thread.
>
> No.
Yes. Yes yes yes. There, did that accomplish anything? Neither do
your "no"s, Darrell. I suggest you remove them and simply get on
with your discussion.
> It is well known and accepted convention that sqrt() denotes only
> the principal root.
It is not accepted by me. Or, as I alluded to earlier, if one does
use the positive root, then one must carry that knowledge
forward in subsequent steps (e.g., from step 2 to step 3).
Conventions are usually very good things, but I'm afraid in
this case it is not. In addition, just because something
is a convention does not make it correct. It was widely
held as the convention in Galileo's time that the sun
revolved around the earth.
Are you so doggedly tied to a convention that you cannot
consider for a moment that it may be a bad one? Please,
I respectfully request this of you, Darrell.
> Since 1 obviously does not equal -1, then 1 sure
> doesn't equal -sqrt(1) (which *is* -1). That is all the context one
> needs to equate 1 with sqrt(1).
>
> Let me reiterate to be clear one more time that _if_ we are taking the
> square root of both sides of an equation, then yes you must at least
> consider _both_ roots:
>
> 1 = 1
>
> sqrt(1) +/- sqrt(1)
What was that last line? Did you mean to write:
sqrt(1) = +/- sqrt(1)
???
>
> CONVENTION stipulates that sqrt() is the principal root. That's why
> we write +/- on one side!!!!!
>
> ..so we have the two equations:
>
> sqrt(1) = sqrt(1) sqrt(1) = -sqrt(1)
>
> ...which is the same as:
>
> 1 = 1 1 = -1
>
> One equation is a contradiction, so we have:
>
> 1 = 1
>
> which is equivalently written as:
>
> 1 = sqrt(1)
>
> ...and we are right back to where we started.
I see what you are saying, but I disagree with your
methods. Let me proceed further with my argument.
I assert that if sqrt(a) were a well-defined
function, then it would be perfectly acceptable
to take the sqrt() of both sides of an equation
WITHOUT using +/-.
If sqrt(a) is a well-defined function, as you seem to
be asserting with your definition of sqrt(a) as
the "principal root of a," then it would be the
case that
a = b ==> sqrt(a) = sqrt(b)
That is one of the criteria for a well-defined
function. Therefore, taking the square root
of both sides does NOT require the +/-, as
you have said, based on this property of
a well-defined function.
If you therefore acknowledge sqrt(a) is
not a well-defined function, then we can
also dispense with the idea that the problem
only occurs when operating on both sides of
an equal sign, because even the *expression*
"sqrt(a)" must be understood to have two
possible answers.
What you seem to be doing (and admittedly, it's
not just you, but actual textbook authors) is
trying to force a well-defined function from
something that is not. Contrast this to my
approach, where I simply acknowledge up front
that sqrt(a) is not well-defined and proceed
from there.
> BUT....
>
> We never took the sqrt of both sides in the original procedure. All
> we did was equate 1 and sqrt(1). This is *just* a valid equation as 1
> = 3/3. To say that 1 = -sqrt(1) is just as much a contradiction as to
> say 1 = -3/3.
>
> >
> > Let me add that I present this information
> > on the sqrt() operation not simply by a making a series of deductions
> > in my head while sitting in front of the computer posting this
> > usenet news article nor by reading it out of a book, but rather
> > by having been bitten multiple times in the past when attempting
> > to solve problems using the (incorrect) interpretion of sqrt(a*a) as
> > "+|a|" (as Darrell does) instead of carrying through the "+/-" as
> > I suggest.
>
> _No where_ have you heard me say that nor have you seen me use that
> argument in showing why 1<>-1 in this thread. In fact, since you
> mention it, sqrt(a^2)=|a| by definition which in fact supports even
> more what Stan and I have been trying to tell you........sqrt()
> implies the principal root (and *only* the principal root.) since
> absolute value is _never_ negative, how do you justify that sqrt(a^2)
> can _ever_ be negative?
I'm sorry Darrell but I have read, re-read, and triple-read the paragraph
above and can't figure out what you're talking about. I was simply trying
to say that I disagree with your convention that sqrt(a) is always
positive. You were saying that, right?
--
Darrell Ryan
>
> Yes. Yes yes yes. There, did that accomplish anything? Neither do
> your "no"s, Darrell. I suggest you remove them and simply get on
> with your discussion.
>
> > It is well known and accepted convention that sqrt() denotes only
> > the principal root.
>
> It is not accepted by me.
<...>
> It was widely
> held as the convention in Galileo's time that the sun
> revolved around the earth.
> > sqrt(1) +/- sqrt(1)
>
> What was that last line? Did you mean to write:
>
> sqrt(1) = +/- sqrt(1)
>
> ???
> to say that I disagree with your convention that sqrt(a) is always
> positive. You were saying that, right?
You are free to define the symbol sqrt() in any way you wish. But
when doing so, it should not surprise you when others disagree because
they (like the rest of the mathematical world) use a different
convention.
Randy, you are blowing off steam. Why don't you accept the standard
convention? If you insist on a different convention, fine and dandy,
but it shouldn't surprise you when you run into resistance. It's
kinda like you are saying pa-tay-to and the rest of the world is
saying pa-tah-to and you just won't be quiet until the rest of the
world says pa-tay-to.
FACT----it is standard convention that the symbol sqrt() denotes the
principal root and only the principal root. This in no way
contradicts the fact that there are two roots. It is just a simple
way to distinguish between the two. The negative root is denoted by
-sqrt(). This is all very standard as I have stated several times.
If you don't accept that, it is not my problem.
I give up. 1 must equal -1 if that will make this go away.
Darrell
Randy Yates
> [...]
> Why don't you accept the standard
> convention?
Because it simply did not make sense, as you so conveniently
snipped out from this most-recent response. You kept the part
that is consistent with your beliefs, and simply chose to
ignore the other part, which dealt with well-defined functions.
Well sure, you can always stick our head in the sand, I suppose.
I realized last night, after I posted, what the problem is, and
I have since seen Albert Lai's response which stated it
precisely: there are TWO "operations" we commonly want to
perform when dealing with square roots. One is the FUNCTION
"sqrt(a)", the principal square root, and the other
is the operation of finding the square roots, which I
refer to here as the "non-principal" square root (and
which is not a function).
However, you did not make that distinction, Darrell. You went
along your merry way and used the words "square root" for both
"principal" and "non-principal" square root. Further, you
muddied the water by attempting to construct some half-cocked
scheme of distinguishing between taking the non-principal
square root of both sides of an equation versus taking the
principal square root of an expression when there
should be no distinction. Both operations are possible with
both types of constructs.
It is PRECISELY these problems that kept me from agreeing
with your POVs, and I submit that these are indeed important
problems to resolve. For example, do we yet have a USENET news
standard for expressing a non-principal square root? That
issue appears to be open.
Finally, if we are really interested in answering the original
poster's question, we might want to clarify whether he meant
"principal" or "non-principal" square root, since past experience
has shown that many posters of questions can't even get
simple, basic notation right, much less something as subtle
as the meaning of "sqrt()" as "principal square root."
> I give up. 1 must equal -1 if that will make this go away.
Perhaps if you simply acknowledged the truth, Darrell, things
would straighten themselves out.
And since I'm blowing off steam, as you put it, or "spouting
off" as Stan puts it, I might as well play out the character
you have already made me and say what's really on my mind.
You people amaze me. You get up on high horses and look down
your snotty noses at anyone who would DARE to challenge the
common, established ways of thinking or persist in bringing
some inconsistency to light. You ridicule and in the process
attempt to oppress them that would challenge your world view.
I acknowledge that scrutiny of new ideas is necessary and
right, but there is absolutely no need for the holier-than-thou
attitude I have seen in this thread. A little unprejudiced
open-minded discussion would have brought out the issues much
more quickly.
--
Randy Yates % "She's sweet on Wagner-I think she'd die for Beethoven-
DIGITAL SOUND LABS % She love the way Puccini lays down a tune, and
Digital Audio Sig. Proc. % Verdi's always creepin' from her room."
<ya...@ieee.org> % "Rockaria", *A New World Record*, ELO
http://www.shadow.net/~yates
Randy Yates
> > by having been bitten multiple times in the past when attempting
> > to solve problems using the (incorrect) interpretion of sqrt(a*a) as
> > "+|a|" (as Darrell does) instead of carrying through the "+/-" as
> > I suggest.
>
> bite, i.e., that the correct, complete solution can still be obtained
> using "sqrt(a*a) = |a|", thus demonstrating that the flaw lies
> somewhere else.
"Find the negative voltage that would produce 1 watt into 1 ohm."
Solution: (?) P = V^2/R ==> V^2 = P*R
Take the "principal square root" of both sides:
V = +sqrt(P*R) = +sqrt(1*1) = +1. ???
(Where is the negative value?)
Darrell Ryan
<...>
> I realized last night, after I posted, what the problem is, and
> I have since seen Albert Lai's response which stated it
> precisely: there are TWO "operations" we commonly want to
> perform when dealing with square roots. One is the FUNCTION
> "sqrt(a)", the principal square root, and the other
> is the operation of finding the square roots, which I
> refer to here as the "non-principal" square root (and
> which is not a function).
>
> However, you did not make that distinction, Darrell. You went
> along your merry way and used the words "square root" for both
> "principal" and "non-principal" square root.
<...>
Man, c'mon. You are in denial. worse, you deny you are in denial.
This is beyond irritating. It has become humerous! Time after time
after time I have told you thqat the symbol sqrt() denotes only the
principal root. You are the one who started all this mess by
insisting it implies both roots. I thought the saying went if you
can't beat 'em join 'em. I didn't know you redefined it to if you
can't beat em' trade places with 'em!
It's funny, Randy. I'm in tears with laughter. You have gone from
completely irritating to a world class comedian as far as I'm
concerned!
Darrell
Albert Y.C. Lai
> "Find the negative voltage that would produce 1 watt into 1 ohm."
>
> Solution: (?) P = V^2/R ==> V^2 = P*R
> Take the "principal square root" of both sides:
> V = +sqrt(P*R) = +sqrt(1*1) = +1. ???
V^2 = P*R
Take the principal square root of both sides:
|V| = sqrt(P*R)
(As promised, I use sqrt(a*a) = |a|.)
Since P=1 and R=1,
|V| = sqrt(1*1) = 1
So V=1 or V=-1.
Now we use the last bit of the information that has not been used: the
desired voltage is supposed to be negative, so we keep the solution
V=-1 and discard the solution V=1.
The "keeping and discarding" is justified by the following logic
formalization of the problem and the solution:
Problem: find V s.t. V < 0 and V^2 = 1*1.
Solution:
V<0 and V^2=1*1 take principal square root
iff V<0 and |V|=1
iff V<0 and (V=1 or V=-1)
iff (V<0 and V=1) or (V<0 and V=-1)
iff False or (V<0 and V=-1)
iff V<0 and V=-1 (y and x) = (x implies y) and x
iff (V=-1 implies V<0) and V=-1
iff True and V=-1
iff V=-1
--
[If you post a response, no need to cc me; if you cc me, please say so.]
"The instrument which brings about the adjustment of differences between theory
and practice, between thought and experiment, is mathematics. It builds the
connecting bridge and continually strengthens it." - David Hilbert
(Constance Reid, _Hilbert_, pp 195)
Randy Yates
>
> Randy Yates <ya...@shadow.net> writes:
>
> > "Find the negative voltage that would produce 1 watt into 1 ohm."
> >
> > Solution: (?) P = V^2/R ==> V^2 = P*R
> > Take the "principal square root" of both sides:
> > V = +sqrt(P*R) = +sqrt(1*1) = +1. ???
>
> V^2 = P*R
>
> Take the principal square root of both sides:
>
> |V| = sqrt(P*R)
> (As promised, I use sqrt(a*a) = |a|.)
>
> Since P=1 and R=1,
>
> |V| = sqrt(1*1) = 1
>
> So V=1 or V=-1.
>
> Now we use the last bit of the information that has not been used: the
> desired voltage is supposed to be negative, so we keep the solution
> V=-1 and discard the solution V=1.
I've never looked at it that way. You've shown me a new way
of solving these types of problems, i.e., you've shown that
they can be solved using only principal square roots.
I would've normally taken the non-principal square root of
both sides:
V = +/- sqrt(P*R)
which of course is logically identical to
|V| = sqrt(P*R).
This is a very subtle difference. I have been enlightened. Thanks, Albert!
Randy Yates
I just wanted to address one more point with respect to your
recent posts on this issue:
"Albert Y.C. Lai" wrote:
>
> Randy Yates <ya...@shadow.net> writes:
>
> > "Find the negative voltage that would produce 1 watt into 1 ohm."
> >
> > Solution: (?) P = V^2/R ==> V^2 = P*R
> > Take the "principal square root" of both sides:
> > V = +sqrt(P*R) = +sqrt(1*1) = +1. ???
>
> V^2 = P*R
>
> Take the principal square root of both sides:
>
> |V| = sqrt(P*R)
> (As promised, I use sqrt(a*a) = |a|.)
>
> Since P=1 and R=1,
>
> |V| = sqrt(1*1) = 1
>
> So V=1 or V=-1.
>
> Now we use the last bit of the information that has not been used: the
> desired voltage is supposed to be negative, so we keep the solution
> V=-1 and discard the solution V=1.
As I have already responded, I agree that this problem can
be solved using only principal square roots, as you have done
here.
BUT, in another post, Albert Lai wrote:
> Show us one example of that. I will show you the real cause of the
> bite, i.e., that the correct, complete solution can still be obtained
> using "sqrt(a*a) = |a|", thus demonstrating that the flaw lies
> somewhere else.
You seem to be saying here that the problem I was referring to
is because I used non-principal square roots instead of
principal square roots. That is not true. The flaw I was
referring to is when one takes a problem like that above and
reduces it to
V = sqrt(P*R).
The problem here is that the non-principal square root (npsqrt) is used
on the left side and the principal square root on the other, but
the problem can indeed be solved correctly using only npsqrt()s.
--
% Randy Yates % "My Shangri-la has gone away, fading like
%% DIGITAL SOUND LABS % the Beatles on 'Hey Jude'"
%%% Digital Audio Sig. Proc. %
%%%% <ya...@ieee.org> % 'Shangri-La', *A New World Record*, ELO
http://www.shadow.net/~yates
Albert Y.C. Lai
> You seem to be saying here that the problem I was referring to
> is because I used non-principal square roots instead of
> principal square roots. That is not true.
Right. I did not intend to say that. I was just saying that using
principal square roots would solve the problem equally well. Of
course, both methods are fine; the key is to use them correctly.
--
[If you post a response, no need to cc me; if you cc me, please say so.]
Ancient number system: one, two, many.
Futuristic GUI number system: click, double-click, repeatedly click.
silvio
bob Slotz ha scritto nel messaggio ...
>My Teacher Handed out this problem which I am trying to solve.
>
> PROOF:
> 2. 1 = sqrt(1)
> 4. 1 = sqrt( -1) * sqrt( -1)
> 5. but i = sqrt(-1) by definition.
> 6. Therefore: 1 = i * i
> 7. but i^2 = -1
> 8. Therefore: 1 = -1
>
> Explain where and what is logically "wrong" here.
>
>
>
>
> -**** Posted from remarQ, Discussions Start Here(tm) ****-
>http://www.remarq.com/ - Host to the the World's Discussions & Usenet
silvio
bob Slotz ha scritto nel messaggio ...
>My Teacher Handed out this problem which I am trying to solve.
>
see my corrections greetings Silvio
> PROOF:
> 1. 1 = 1
> 2. +/- 1 = sqrt(1)
> 3.+/- 1 = sqrt( (-1)(-1) )
> 4. +/- 1 = sqrt( -1) * sqrt( -1)
> 5. but +/- i = sqrt(-1) by definition.
> 6. Therefore: +/- 1 =(+/- i )*(+/- i)
> 7. but(+/- i)^2 = -1
> 8. Therefore: +/-1 = +/-1 therefore +1=+1 and -1=-1