No irrationals
Poak...@msn.com
consequences, if any, of someone proving that all numbers are rational,
that is, that there are no irrational numbers? Thank you.
Peter
Gene Ward Smith
If you proved sqrt(2) was rational, you would prove arithmetic was
contradictory, since you can also prove it is not rational.
Dave L. Renfro
Assuming by "numbers" you mean the set of real numbers,
they would have proved that mathematics, as an axiomatic
theory in logic, is inconsistent.
However, if we don't assume as much as you're asking us to,
and only assume that one or more people claimed to have
proved all real numbers are rational, then there's also
the possibility that they made a false statement (perhaps
deliberately, perhaps not).
Dave L. Renfro
William Hughes
This only makes sense in the form "those things we call real numbers
are only numbers if they are rational".
Forcing people to use rational numbers only in calculations would have
little practical effect. (In practice rational approximations to real
numbers
have to be used).
Mathematically, there would be little change, all we have is a change
in names. There would be no number x such that x^2 - 2 = 0, but we
would have sequences of numbers such that |x^2 -2| was as small as we
please (alternately we could divide our numbers into two sets,
those for which x^2-2 <0 and those for which x^2 -2 >0). We could not
calls these entities numbers (l'academie mathematique would forbid
this)
but these things would still exist and they would be interesting to
study.
So in the end we still have the real numbers we just call them
something
else.
-William Hughes
MoeBlee
Proving from what axioms and in what system of logic?
If there is a proof from the axioms of Z set theory, then one
consequence is that Z set theory is inconsistent.
MoeBlee
Bart Goddard
Dave L. Renfro
> Forcing people to use rational numbers only in calculations
> would have little practical effect. (In practice rational
> approximations to real numbers have to be used).
I don't know about you, but I perform calculations with
irrational numbers all the time. For example, to rewrite
{ [1 / sqrt(2)] + [1 / sqrt(3)] } ^ (-1)
as
3*sqrt(2) - 2*sqrt(3)
requires computation with irrational numbers, or to
check that sqrt(5) - 2 is a zero of x^3 - 17x + 4 is
a straightforward computation with irrational numbers.
In fact, I had to do some computation with irrational
numbers just to come up with this last example -- I cubed
sqrt(5) - 2 to see where it'd get me, then I subtracted
an appropriate multiple of sqrt(5) - 2.
Dave L. Renfro
Dave L. Renfro
[computations with irrational numbers]
Speaking of computations with irrational numbers, here's
something I came across the other day (Amer. Math. Monthly,
Volume 57, p. 74).
"Proof" that sqrt(52) = 2*sqrt(13):
sqrt(13) = sqrt(4 + 9) = sqrt(4) + sqrt(9) = 2 + 3 = 5
sqrt(52) = sqrt(16 + 36) = sqrt(16) + sqrt(36) = 4 + 6 = 10
Therefore, sqrt(52) = 10 = 2*5 = 2*sqrt(13).
Question for sci.math:
If you were grading a college algebra test and saw the
above work for a 5-point problem that asked the student
to rewrite sqrt(52) in simplest radical form, how much
credit would you give? I'd probably give 1 to 2 points,
assuming of course that the write up was as logical as
what I gave and the only mistake was assuming sqrt is
distributive over addition.
Dave L. Renfro
P.S. Newbies may want to watch how fast this thread grows now!
porky_...@my-deja.com
Yes, 1 or 2 points, since besides the incorrect assumption, the
reasoning is quite logical. Incidently, the fact that the answer is
correct is pure incidental and shouldn't affect the final grade (IMHO).
Dave L. Renfro
>> "Proof" that sqrt(52) = 2*sqrt(13):
>>
>> sqrt(13) = sqrt(4 + 9) = sqrt(4) + sqrt(9) = 2 + 3 = 5
>>
>> sqrt(52) = sqrt(16 + 36) = sqrt(16) + sqrt(36) = 4 + 6 = 10
>>
>> Therefore, sqrt(52) = 10 = 2*5 = 2*sqrt(13).
>>
>> Question for sci.math:
>>
>> If you were grading a college algebra test and saw the
>> above work for a 5-point problem that asked the student
>> to rewrite sqrt(52) in simplest radical form, how much
>> credit would you give? I'd probably give 1 to 2 points,
>> assuming of course that the write up was as logical as
>> what I gave and the only mistake was assuming sqrt is
>> distributive over addition.
porky_...@my-deja.com wrote:
> Yes, 1 or 2 points, since besides the incorrect assumption,
> the reasoning is quite logical. Incidently, the fact that
> the answer is correct is pure incidental and shouldn't
> affect the final grade (IMHO).
Actually, this is more of an ethical dilemma than I first
realized. For a problem like this, you ordinarily don't
expect to see work, although you might award partial credit
for something relevant that you happen to see if they get
the wrong answer. However, in this case, only awarding
1 or 2 points seems "wrong", because someone might be thinking
the same thing and yet not show you their work like this
person did. In the rare cases that I've seen something like
this (correct answer to something not really requiring work,
but they show incorrect work that still leads to the correct
answer), I just bit the bullet and awarded them full credit.
Of course, I wrote an explanation on their paper about the
situation. For something like this, where you aren't really
grading any work by the student and it's just a result they
could easily do in their head, we're only talking about
a few percentage points anyway, and it hardly ever happens.
(Except to new teachers, who might put something like "Rewrite
x^2 * x^2 as x to some power" on a test!)
Incidentally, the grading of ap-calculus tests follows a rather
hard line. If the student writes down something incorrect,
regardless of whether it's relevant to the problem or not,
one or more points are deducted. (I think the way it's
phrased when you're grading ap-calculus tests is that one
or more points will not be awarded.)
Dave L. Renfro
jank...@hotmail.com
Poak...@msn.com skrev:
> Hi! Please, could somebody tell me what would be the mathematical
> consequences, if any, of someone proving that all numbers are rational,
> that is, that there are no irrational numbers? Thank you.
Then they could team up with James Harris and have all mathematicians
fired.
---
J K Haugland
http://home.no.net/zamunda
Jesse F. Hughes
If this was a test regarding algebraic manipulation of radicals, I
think I'd give zero points.
--
Jesse F. Hughes
"Things are pretty mixed up, but I think the worst is over."
-- A LaTeX error message or a psychic forecast?
porky_...@my-deja.com
> Actually, this is more of an ethical dilemma than I first
> realized. For a problem like this, you ordinarily don't
> expect to see work.
Why? Last May I was working with one kid from high school (as a
volunteer) to teach him exactly the same problem: to simplify square
roots. I told him to break down the number under the root into the
prime factors, explicitly, and then simplify if possible. Unless the
factorization is so simple it can be done in one's head. So you *may*
or *may not* see the work.
But suppose it did happen. I would argue, the example from the journal
is a concoted one. Normally not knowing the distributive laws (or
mis-interpreting them) would lead to the wrong answer. So I would have
safely assumed that if the answer was correct then the logic was
correct, and if the answer was wrong, then the logic was wrong --- if
no intermediate steps were shown.
porky_...@my-deja.com
> William Hughes wrote:
> Mathematically, there would be little change, all we have is a change
> in names. There would be no number x such that x^2 - 2 = 0, but we
> would have sequences of numbers such that |x^2 -2| was as small as we
> please (alternately we could divide our numbers into two sets,
> those for which x^2-2 <0 and those for which x^2 -2 >0).
Yeah. Dedekind cuts. But then the rational numbers are Dedekind cuts as
well.
fishfry
"Poak...@msn.com" <Poak...@msn.com> wrote:
You could be walking down the number line and fall down a hole. Like the
one at sqrt(2).
fishfry
They must have been making a joke. It would take more cleverness to do
what they did, than it would to just do the correct simplification.
Proginoskes
Actually, the number line would be almost entirely holes; you'd have to
jump from one "number" to the next one instead.
--- Christopher Heckman
Dave L. Renfro
> They must have been making a joke. It would take more
> cleverness to do what they did, than it would to just
> do the correct simplification.
Actually, it was an example in an article on college
teaching in the American Mathematical Monthly (brief
cite given in my other post), something that was claimed
to have actually come up on a student's test, either by
a student of the author's or a student of someone else
who told the author about it. (I don't have that particular
Amer. Math. Monthly volume with me right now and I don't
remember any more than this.) The author did say the
student was quite bright, and I think the author also
said the student hadn't had math in several years or
had a very weak background from high school (I don't
remember which).
Dave L. Renfro
Dave L. Renfro
>> Actually, this is more of an ethical dilemma than I first
>> realized. For a problem like this, you ordinarily don't
>> expect to see work.
porky_...@my-deja.com wrote:
> Why? Last May I was working with one kid from high school
> (as a volunteer) to teach him exactly the same problem:
> to simplify square roots. I told him to break down the
> number under the root into the prime factors, explicitly,
> and then simplify if possible. Unless the factorization
> is so simple it can be done in one's head. So you *may*
> or *may not* see the work.
Sure, but in the case of something like this on a test,
sqrt(52), work usually isn't required. So if they show
work that's wrong but leads to the correct answer, they
could legitimately challenge points being taken off
when someone else has no points taken off for no work
shown. That is, unless the instructions *specifically*
say that work is required on that problem (and if you
do this, you had better tell them in advance what "work"
means for a problem like this, such as actually writing
52 = 4*13, giving the complete sequence sqrt(52) =
sqrt(4*13) = sqrt(4)*sqrt(13) = 2*sqrt(13), or whatever)
or the instructions *specifically* say that points may
be deducted for incorrect statements not identified as
such (i.e. crossing out something). Actually, this last
cravat is probably a good idea. Note that you're not
saying you _will_ count off, only that you might.
> But suppose it did happen. I would argue, the example
> from the journal is a concoted one. Normally not knowing
> the distributive laws (or mis-interpreting them) would
> lead to the wrong answer. So I would have safely assumed
> that if the answer was correct then the logic was correct,
> and if the answer was wrong, then the logic was wrong ---
> if no intermediate steps were shown.
Sure, if no work was given, I'd assume the student did
it correctly (or they saw it on someone's test next to
him/her) if they had the correct answer. My concern is not
with what I believe (even if very strongly), but with what
I can successfully defend _to_the_student_ in an airtight
way. I'm sure most department chairs would back me up on
something like this if a student didn't like my taking
off for incorrect work, when no work was requested and
no mention was made that incorrect work for correct answers
could lead to a loss of points, and the student went to the
department chair about it (I've been this situation before,
although I don't think it was for something precisely
like this), but you'll save yourself and others a lot of
time by avoiding such a situation from coming up in the
first place.
By the way, the example in the journal supposedly really
did happen. See my most recent post in this thread.
Dave L. Renfro
C6...@shaw.ca
What is the rational number next to 1/3 on the larger side?
R.G. Vickson
>
> --- Christopher Heckman
Jonathan Hoyle
> consequences, if any, of someone proving that all numbers are rational,
> that is, that there are no irrational numbers? Thank you.
It would be approximately the same as someone proving that 2+2=5.
Your welcome.
The Qurqirish Dragon
It may not be a joke- although that is likely.
Consider this method for simplifying square-roots:
Step 0: if the radicand is a perfect square, then simplify and you are
done.
Step 1: see if the is a perfect square in the radicand.
Step 2: if no, go to step 3. If yes, pull the square outside the
radical and return to step 1.
Step 3: combine the values outside the radical and multiply by what's
left.
Now, this is a very poorly worded algorithm for the task, but if you
know the proper method, then you can understand the description.
Let's say you don't recall how this works, but remember these steps.
We sant to simlpify sqrt(52)
By step 1, since 1,4,9,16,25,36,49 are less than 52, they are perfect
squares in the radicand. Let's see if anything works nicely. We have
seen through factoring problems that it is often easier to try several
pairs of factors, and seeing if any one of them is easy, rather than
immediately going through a full factorization.
we check:
52 = 1 + 51, but 51 is not a square
52 = 4 + 48, but 48 is not a square
52 = 9 + 43, but 43 is not a square
52 = 16 + 36- AHA! 16 and 36 are both squares!.
We pull out 16 and get: sqrt(52) = 4 + sqrt(36)
Now we pull out the 36: ...= 4 + 6 = 10
Similarly with sqrt (13):
13 = 1 + 12, but 12 isn't a square
13 = 4 + 9. This works, and gets the result of 5 as above. thus
sqrt(52)/sqrt(13)=10/5=2, which rewritten gives the desired
sqrt(52)=2*sqrt(13) (which is the statement that needed to be proved).
Of course, this works because each addend for 52 is 4 times the
corresponding addend for 13. (16=4*4; 36=4*9) Any perfect square would
work similarly.
eg: sqrt(45)=3*sqrt(5), since 45 = 36+9=9*(4+1) and 5 = 4+1 would work
out similarly with the flawed procedure.
Poak...@msn.com
If the sum of an infinite number of rational numbers is rational,
sqrt(2) is rational.
Peter
Dave L. Renfro
> If the sum of an infinite number of rational numbers
> is rational, sqrt(2) is rational.
Huh?? An infinite sum of rational numbers can be rational
or it can be irrational (when the sum exists). I don't see
how this has any bearing on the irrationality of sqrt(2).
Also, it's not clear whether you mean an infinite
sum of rational numbers *can be* rational or *must be*
rational. The second is false, by the way:
1/4 + 1/8 + 1/16 + ... = 1/2.
Dave L. Renfro
Dave L. Renfro
> Also, it's not clear whether you mean an infinite
> sum of rational numbers *can be* rational or *must be*
> rational. The second is false, by the way:
>
> 1/4 + 1/8 + 1/16 + ... = 1/2.
Ooops, wrong example!
2 + 1/2! + 1/3! + 1/4! + ... = e,
which is irrational (transcendental, in fact).
Dave L. Renfro
porky_...@my-deja.com
it out.
Poak...@msn.com
Your example is very good. Does it mean the sum of an infinite number
of rational numbers can be irrational? How come? That is
counterintuitive. What is the proof?
Peter
Poak...@msn.com
I said "if."
Peter
Gerry Myerson
"C6...@shaw.ca" <C6...@shaw.ca> wrote:
That's easy: .33333333333.... with a 4 at the end.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Gerry Myerson
"Poak...@msn.com" <Poak...@msn.com> wrote:
> Your example is very good. Does it mean the sum of an infinite number
> of rational numbers can be irrational? How come? That is
> counterintuitive. What is the proof?
Take your favorite irrational number. Look at its decimal expansion.
Write that expansion as a + (b / 10) + (c / 100) + (d / 1000) + ...,
where a, b, c, d, ... are integers, and b, c, d, ... are between 0
and 9, inclusive. Voila!
G.E. Ivey
James Dolan
poak...@msn.com <poak...@msn.com> wrote:
|Hi! Please, could somebody tell me what would be the mathematical
|consequences, if any, of someone proving that all numbers are
|rational, that is, that there are no irrational numbers? Thank you.
it's easier to list the non-consequences:
--
Dik T. Winter
If you are very small yourself. I think I can stand on the number line
even if all non-integers are missing. (Shoe size 45, for those who
understand such sizes. And for those who do not, it is about 10 in
Mexico, about 11 in the English speaking world and about 29.5 in Japan.)
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Jonathan Hoyle
> of rational numbers can be irrational? How come? That is
> counterintuitive. What is the proof?
Pick any irrational number, say pi.
Note that pi = 3 + 0.1 + 0.04 + 0.001 + 0.0005 + 0.00009 + ...
Any irrational can be composed as a sum rationals in this manner.
(Duh.)
Robert Israel
Hmm: this error is more sophisticated than it seemed. To simplify
sqrt(N), you first look for ways of writing N = x^2 + y^2 (which
will exist if the squarefree part of N has no prime factors == 3 mod
4), and take that pair [x,y] whose gcd is maximum. If gcd(x,y) = b,
then N = b^2 (u^2 + v^2) where u = x/b and v = y/b. Then the
rest of the calculation goes:
sqrt(N) = sqrt(x^2 + y^2) "=" x + y = b (u + v)
"=" b sqrt(u^2 + v^2)
with a perfectly correct conclusion despite the incorrect = signs
(but we all know students often write = when they don't really
mean =).
Maybe it is worth 2 points after all...
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Proginoskes
"Next" is in the sense of what number you jump to, not the next larger
rational.
--- Christopher Heckman
Aatu Koskensilta
> Hi! Please, could somebody tell me what would be the mathematical
> consequences, if any, of someone proving that all numbers are rational,
> that is, that there are no irrational numbers? Thank you.
Everyone would scream in horror as the universe destroyed itself.
Mathematicians would weep powerlessly as every open problem was
instantly solved, in at least two ways to boot.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Poak...@msn.com
O. K., but shouldn't the sum of either a finite or infinite number of
rational numbers be rational? Why it is not?
Peter
Poak...@msn.com
Please, is there any other proof of the existence of irrational numbers
than sqrt(2) = a/b . . .?
Peter
Dave L. Renfro
> Please, is there any other proof of the existence of
> irrational numbers than sqrt(2) = a/b . . .?
The Math Forum: Ask Dr. Math: "The Irrationality of e".
http://mathforum.org/library/drmath/view/53910.html
Xavier Gourdon and Pascal Sebah, "Irrationality proofs",
Numbers, Constants and Computation web pages.
http://tinyurl.com/4c62w
Here are two proofs, [A] and [B], that show irrational
numbers exist without trying to target any specific
irrational number, such as sqrt(2), e, pi, etc.
[A] The rational numbers are countable [1] and the real
numbers are uncountable [2]. Hence, there must exist
a real number that isn't rational, since no set can
be both countable and uncountable. [If no irrational
numbers existed, the set of rational numbers would
be equal to the set of real numbers.]
[1] http://groups.google.com/group/sci.math/msg/bb6bc7787ad9255b
[2] http://www.answers.com/topic/cantor-s-diagonal-argument
[B] Arrange the rational numbers in a list [1]. Denote this
list by r_1, r_2, r_3, ... . For each positive integer n,
let I_n be the open interval (r_n - 1/2^n, r_n + 2^n)
of length 1/2^n + 1/2^n with midpoint r_n. Consider the
union of all these open intervals. Every rational number
belongs to this union, but not every real number can
belong to this union, because the sum of the lengths
of these intervals is 1 + 1/2 + 1/4 + 1/8 + ... = 2.
This last statement actually takes some proof. One
way to take care of it is to consider those intervals
that intersect [0,3], which gives an open convering
of [0,3], from which a finite subcovering can be
extracted. Now it's a matter of showing that a finite
union of open intervals, whose individual lengths add
to less than 2, must omit some points in [0,3]. (This
last statement actually isn't all that easy to nail
down in an airtight way, unless I'm forgetting
something.) Those omitted points don't belong to
the original union of open intervals (easy, but it
does need to be taken care of), and hence none of
the omitted points can be rational. This method is
essentially due to Borel (his 1894 Ph.D. Dissertation,
published in 1895). Incidentally, Borel's original
proof of the finite covering property used transfinite
induction. See my comments in [3]. A link to Borel's
1895 published version is given in [4], and Borel's
proof using transfinite induction is on pp. 51-52.
[3] http://groups.google.com/group/sci.math/msg/1a243c7fd0d64bac
[4] http://groups.google.com/group/sci.math/msg/cf41572445cd404a
Dave L. Renfro
Randy Poe
The sum of a finite number of rationals is rational. This
follows from the fact that the sum of two rationals is
rational. In fact, the concept of a sum of n rationals must
be defined inductively from the concept of the sum of 2
rationals.
> Why it is not?
An "infinite sum" has no a priori definition, especially not
in terms of the sum of two rationals. It is a limit of a
sequence, the limit point of a set of points. All of those
points are rational, but there's no requirement that the
limit point of a point set be a member of that set. There
are many simple counterexamples.
I think you'll find your intuition is being led astray because
you don't have a firm notion of what is meant by "infinite
sum" and that it actually doesn't mean "adding up infinitely
many terms".
- Randy
Dave L. Renfro
> An "infinite sum" has no a priori definition, especially not
> in terms of the sum of two rationals. It is a limit of a
> sequence, the limit point of a set of points.
Nit-pick ...
The limit of a sequence is not always a limit point
of the set of terms of that sequence. Consider an
eventually constant sequence, for example, which
you will get if the sum had a tail all of whose
terms are the rational number 0.
Dave L. Renfro
Han.de...@dto.tudelft.nl
> Hi! Please, could somebody tell me what would be the mathematical
> consequences, if any, of someone proving that all numbers are rational,
> that is, that there are no irrational numbers? Thank you.
OK. Let's start with the naturals and define a (real) number (a) to the
n-th power as the product of (n) times an (a):
a^n = a.a.a.a ... a
With mathematical induction, the following properties can be proved for
any natural (m) and (n):
a^(m+n) = a^m.a^n
(a^m)^n = a^(m.n)
Other theorems are (if we denote the k-th root from x as kVx):
a^(m-n) = a^m/a^n provided that m > n
nV(a^m) = a^(m/n) provided that m divisible by n
Time to generalize. Assume that a > 0 in the sequel. Then we _define_:
a^(m-n) := a^m/a^n for _all_ natural (m,n)
a^(m/n) := nV(a^m) for _all_ natural (m,n)
Special cases are: a^0 = 1 and a^1 = a , a^(1/2) = sqrt(a) .
In this way, any _rational_ exponent (r) in a^r can be defined. Right?
But ... HOW ABOUT IRRATIONAL EXPONENTS ? (No logarithms please)
Han de Bruijn
cbr...@cbrownsystems.com
If some real r is the limit of a sequence of rationals (q_1, q_2, q_3,
..., q_n, ...), why not define a^r as lim n->oo (a^q_n)?
Cheers - Chas
Han.de...@dto.tudelft.nl
> But ... HOW ABOUT IRRATIONAL EXPONENTS ? (No logarithms please)
While thinking about a partial answer to this question ...
program poster;
{
It is demonstrated in this program how arbitrary real powers
of a real number can be approximated, with nothing else than
elementary arithmetic operations and a square root function.
}
function macht(x,a : double) : double;
{
Power x^a
}
var
m,k : integer;
g,p,q,y : double;
begin
m := Trunc(a);
y := x;
p := 1;
while m > 0 do begin
if (m and 1) > 0 then p := p * y;
m := m div 2; { m := m shr 1 }
y := y * y;
end;
g := a - Trunc(a);
y := x;
q := 1;
for k := 0 to 50 do
begin
g := g * 2;
y := sqrt(y); { Square Root HERE }
if (Trunc(g) and 1) > 0 then q := q * y;
end;
macht := p * q;
end;
procedure Test;
{
Just a Test
}
begin
Writeln(macht(4,0.5));
Writeln(macht(9,0.5));
Writeln(macht(81,1/4));
Writeln(macht(27,1/3));
Writeln(macht(exp(1),7),' = ',exp(7));
Writeln(macht(exp(1),0.7),' = ',exp(0.7));
Writeln(macht(exp(1),7.7),' = ',exp(7.7));
Writeln(macht(exp(1),Pi),' = ',exp(Pi));
end;
begin
Test;
end.
Han de Bruijn