[A Conundrum of Algebra]
Algebra is just a variety of arithmetic [A] [A] Issac Asimov, Realm of Algebra, Houghton Mifflin, 1961.
- Issac Asimov
x = 3 + 4
The dreaded “x” of algebra; not until the bottom of page 10 of the “realm of algebra” does Asimov introduce the symbol x. We see that x is equal to 7. Nothing to “dread”. But what can we make of this expression?:
7 = a + b
Assigning a = 3 and b = 4 we satisfy the relationship following the well established rules of arithmetic.
Except a could just as well be equal to 6 (a = 6) and b equal to 1 (b = 1) and still satisfy the equation 7 = a + b. Taking this idea one step further, let a = 242 and b = − 235 in
c = a + b
and what do we find as c? Of course c = 7!
This is an interesting observation worth pointing out more directly; in the expression
7 = a + b
there is an “infinite” number of pairs of a&b that satisfy our algebraic relationship. Restricting ourselves to “whole” numbers (or integers), we have 6&1, 5&2, 4&3, ... 242&-235 or even 456796&-456789 that we can substitute for a&b. And that’s just for letting c equal the integer 7! And the same applies to each of the “infinite” number of integers we can substitute in for our c! All this from the most fundamental axiom of algebraic arithmetic:
c = a + b
This observation may seem trivial; but hold this thought.
Let’s skip over subtraction, multiplication and division (the remaining
core arithmetic operations of algebra) and move right on to raising a
number to a power. Raising a number to a power is simply multiplying the
number by itself a “number” - or “power” of “times” and we can say that
multiplication is really adding one number to itself repeatedly a
number of “times”. In the case of raising a number c to a power
(and again we’ll stick with integers - no need to unnecessarily
complicate matters more than we may already have...) the power is the
number of copies we multiply together; for example multiplying 3 by
itself we have
3⋅3 = 9
and if we again multiply by 3,
3⋅3⋅3 = 27
We first “squared” 3, raising it to the second power and then “cubed” it
by raising to the third power. We can express these operations as 3^{2} = 9 and 3^{3} = 27. Using symbols of algebra we can write the operation of raising a number to a power as
c^{n}
where c is our integer number and n is the power to which it is raised. For example, for c = 3 and n = 4
3^{4} = 3⋅3⋅3⋅3 = 81
Simple enough. Of course we are not introduced to squares and cubes
until page 72 of the “realm of algebra” so we’ve left a few details out.
This is elementary algebra - but not “elementary” in the pejorative
sense. In fact, using the two basic operations we have been discussing,
we are going to run head first into a bit of a conundrum. But let’s not
digress - at least not yet.
We have another interesting observation to make.
If we raise a continuous sequence of integers to a given power we will
have a sequence of integers having integer roots - the root being the
original integer. Another way of stating this idea is to say we can
raise all integers to any power resulting in integers with integer
roots, but not all integers have integer roots! Using our general
expression of c^{n}, let’s raise the integer sequence from 1 to 5 to the 3rd power:
1^{3} = 1
2^{3} = 8
3^{3} = 27
4^{3} = 64
5^{3} = 125
Cubing (raising to the 3^{rd} power) the first 5 integers
results in 5 integers with cube roots for the first 125 integers;
meaning that of the first 125 integers, only 5 have integer cube roots!
And if we continue to 10^{3} = 1000
we will only find another 5 integers with integer cube roots! In fact,
to find all integers with roots of a given power, we need to raise all
integers to that power! Honestly, this is a bit mind boggling (exponents
and logarithms! oh my!) - especially when we consider that not all integers have integer roots for any power!
To restate, if we raise an integer to an integer power, we find an
integer having an integer root of that power. This is a tautology -
reasonably so as it is a definition - an axiom.
c^{n} = q, ^{ n}√(q) = c
We can express this more concisely
^{n}√(c^{n}) = c
And to emphasize again: to find all those integers having an integer
root of an integer power, we raise all integers to all integer powers.
(It is far easier than finding roots of all integers by taking roots!)
Ok. So having come this far we can take a deep breath - and then plunge into logarithms! Kidding - just kidding!
Let’s ask if there is any other way to find all integers - and only
those integers - having integer roots for any and all integer powers?
What might we discover by bringing our first axiom of algebraic arithmetic to the question; c = a + b? Can we substitute a + b for c in c^{n}?
c^{n} = (a + b)^{n}
Yes! Yes indeed: let’s try it out with c = 7 , a = 3, b = 4, and n = 3;
7^{3} = 3^{3} + 3(3^{2})(4) + 3(3)(4^{2}) + 4^{3} = 343
and of course ^{3}√(343) = 7 !
Here’s what we did:
(a + b)^{3} = (a + b)⋅(a + b)⋅(a + b) = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}
This is a binomial expansion. It follows quite simply from our two algebraic axioms, c = a + b and c^{n}. In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial.
expand discussion of binomial theorem
For n = 4 we have
a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4}
and for n = 5,
a^{5} + 5a^{4}b + 10a^{3}b^{2} + 10a^{2}b^{3} + 5ab^{4} + b^{5}
and so forth. And so on..
The trivial but none-the-less interesting point here is - as with c^{n} - we can find all integers - and only those integers - having integers roots for integers power using (a + b)^{n}! And... there is an “infinite” (or unbounded set) of integer pairs (integer binomials) for each and every integer c where c = a + b! Oh my!
Interesting observation: what do we get with the expression: ?
a^{n} + (a + b)^{n}
If a, b and n are integers, then this expression will also evaluate to an integer. Letting a = 3, b = 4, and n = 3 we have
3^{3} + (3 + 4)^{3} = 370
BUT notice, there is no integer cube root! ^{3}√(370) = 7.179054352
This should not be surprising to us, since we established (a + b)^{n} expresses all integers - and only those integers -having integer roots for all integer powers n, and it would seem to follow there can be no integer root for the expression: Let’s try to formulate a “conjecture”:
a^{n} + (a + b)^{n}
Conjecture^{*}: For all positive integers a, b, and n, the binomial (a + b)^{n} expresses integers - and only those integers - having an integer root a + b = c for an integer power n; therefore there can be no integer root for the expression a^{n} + (a + b)^{n} for the integer power n.
The astute algebraist will notice if a = 3, b = 1, and n = 2 we have
3^{2} + (3 + 1)^{2} = 3^{2} + 4^{2} = 5^{2}
the archetype Pythagorean triple!
And hence, despite carefully outlining both the logic and elementary
algebra of our “conjecture” we have run head long into a contradiction!
Shall we throw the baby out with the bath water?
Our discovery would have nicely demonstrated the almost four centuries old conjecture: no three positive integers a, b, and c satisfy the equation a^{n} + b^{n} = c^{n} for any integer value of n greater than 2.
If we allow for the limiting stipulation “for any integer value of n greater than 2” can we then reconcile our discovery with the older conjecture?
Let’s first see why our conjecture is contradicted by the Pythagorean equation.
We can point out that while there are an infinite number of Pythagorean
Triples, the Theorem itself simply states a fundamental relationship: the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
So while there are “many” integer triples that can be found, there are
many more combinations of a, b, and c not integer triples! So long as we
specify any two of the three variables, we can find the third; integers
or not!
Interesting point for us to recall - but this doesn’t help explain why our conjecture fails for n = 2.
Here might be a good place to give Paul Lockhart’s words some thought [B] [B] Paul Lockhart, Measurement, Belknap Press, 2012; page 60. :
“The point of doing algebra is not to solve equations; it’s to allow
us to move back and forth between several equivalent representations,
depending on the situation at hand and depending on our taste. In this
sense, all algebraic manipulation is psychological. The numbers are
making themselves known to us in various ways, and each different
representation has its own feel to it and can give us ideas that might
not occur to us otherwise.”
The Pythagorean equation is simply written as
c^{2} = a^{2} + b^{2}
We have already stumbled on one psychological algebraic manipulation where we substituted a binomial of a + d = b to obtain
c^{2} = a^{2} + (a + d)^{2}
(We should point out that we have to “reassign” our character symbols
here; a moment’s thought will avoid confusion; previously we had written
our binomial of “c” as a + b but as we begin with the Pythagorean equation, our binomial for b (rather than c) requires bringing in a fourth symbol, d.)
Another manipulation we can explore - either term on the right side of
the equality sign can be subtracted from both sides of our equation; an
algebraically “legal” maneuver;
c^{2} − a^{2} = b^{2}
Ah! what have we here? The difference of two squares identity!
(c − a)(c + a) = b^{2}
Let’s keep manipulating: let c + a = m and c − a = n:
m⋅n = b⋅b
If we substitute our classic 3-4-5 Pythagorean integer triple into our “manipulated” equation, we get
8⋅2 = 4⋅4
We have rearranged a “second” order equation into a linear relationship where it is apparent we can find integers m, n, and b that satisfy the equation!
Can we similarly rearrange c^{3} = a^{3} + b^{3} ?
c^{3} − a^{3} = b^{3}
Factoring the difference of two cubes, the best we can do is
(c − a)(c^{2} + ca + a^{2}) = b^{3}
A cursory examination suggests this is a bit impenetrable algebraically.
If we let c = a + d , and substitute we have, after expanding the binomial
a^{3} + 3a^{2}d + 3ad^{2} + d^{3} − a^{3} = b^{3}
and subtracting terms
3a^{2}d + 3ad^{2} + d^{3} = b^{3}
For b^{3} to have an integer root, our expanded binomial in terms of integers a & d would need to be expressed as a^{3} + 3a^{2}d + 3ad^{2} + d^{3}
as we established previously, anything else will NOT have an integer
root. And of course the same argument applies for all powers of n > 2.
Leaving our equation in the original form
c^{3} = a^{3} + b^{3}
and substituting b = a + d
c^{3} = a^{3} + a^{3} + 3a^{2}d + 3ad^{2} + d^{3} = 2a^{3} + 3a^{2}d + 3ad^{2} + d^{3}
And again here for c^{3} to have an integer root our expanded binomial in terms of integers a & d would need to be expressed as a^{3} + 3a^{2}d + 3ad^{2} + d^{3}. We can make the observation, by reference to the proof of the binomial theorem;
there is no power to which a binomial may be raised for which the first
or last term of the n+1 terms of the polynomial (the expanded binomial)
has a coefficient other than one.
Given this observation, we rephrase our conjecture to a theorem?
Theorem: For all positive integers a, b, and n, the binomial (a + b)^{n} equates to only those integers having an integer root a + b = c for an integer power n; therefore there can be no integer root for the expression a^{n} + (a + b)^{n} for integer powers n > 2. [C] [C] Restated: for all positive integers a, b and n: (a + b)^{n} ∈ ℤ^{n}. It follows that for n greater than two a^{n} + (a + b)^{n}∉ℤ^{n}.
Actually, we can express our theorem more concisely as a simple inequality
(a + b)^{n} ≠ a^{n} + (a + b)^{n}
but doing so introduces a “cognitive illusion [D] [D] You might find this example interesting.”. This inequality is always true for positive integers a, b, and n (even n = 2) BUT one may tend to discard the argument as it relates to integers - thinking it may be possible for a^{n} + (a + b)^{n} = (c + d)^{n}. We have shown this cannot be true EXCEPT for n = 2; courtesy of the difference of two squares identity - where the difference may also be an integer square.
Have we demonstrated - and further reconciled - our conjecture? Is it a “proof”?
Again, Paul Lockhart [E] [E] Paul Lockhart, Measurement, Belknap Press, 2012; page 12.
“This is an example of a mathematical argument, otherwise known as a
proof. A proof is simply a story. The characters are the elements of the
problem, and the plot is up to you. The goal, as in any literary
fiction, is to write a story that is compelling as a narrative. In the
case of mathematics, this means that the plot not only has to make
logical sense but also be simple and elegant. No one likes a meandering,
complicated quagmire of a proof. We want to follow along rationally to
be sure, but we also want to be charmed and swept off our feet
aesthetically.”
Whether or not our demonstration is a “proof” is up to you, the reader,
to decide. Considering the [aura] surrounding the Last Theorem, one must
be reluctant to make any claim - especially in light of the body of
literature that has accumulated over the last almost four centuries
since the original “conjecture” was first posited. And while the Theorem
was famously “proved” by Andrew Wiles
in May 1995, it most certainly is an imposing story and one that few
can follow. Not to deprecate the accomplishment of Wiles, or the efforts
of countless mathematicians, but we may wish to invoke Occam's razor here. Once “seen”, our explanation is almost trivial. And in this respect it may be unsettling. It can be written down on less than a sheet of paper (avoiding the “explanatory” quagmire we may have put forth here...).
It would most certainly vindicate Fermat - who after discovering it - may have considered it trivial and uninteresting.