Statements, for the most part, can be true or false. But they can’t be both – at least not in the common applications of logical reasoning. Much like a question, as a matter of fact. Is a question true or false? It cant be either. Nor can general exclamations. I assert that any statement which is self-contradictory simply does not have a truth value, and it is logically meaningless.  Any other statement which leads to a contradiction is false.

There is fundamental difference between proofs by contradiction and Arthur Priors perspective. In indirect proofs, a statement does have a truth value… but the value is often times unknown. By assumption (of the negation of the actual truth value) we can arrive at a contradiction, which thus proves the assumption false.

But the assumed proposition (and the concluded proposition) in indirect proofs cannot be both true and false simultaneously. The actual logical statement must be either true or false. Only the false assumption can be self-contradictory, but that false statement is not true and can never be true, for it would lead to contradiction otherwise. For a statement to actually be true and false is nonsense.

Let me rephrase what I mean. In the case of the indirect proofs, a false statement is posited as a hypothetical truth. This produces a contradiction, yes, but this contradiction is hypothetical. The implication is that a single non-contradictory statement emerges from the mess.  Dual possibilities converge to a single possibility.

This is not what is happening in the Liars Paradox. Here, the statement is self-contradictory, but not hypothetical. Its actual. Arthur Prior would have you believe that a self-contradictory statement can exist in actuality without dilemma. Furthermore, he essentially purports that “self-contradiction” and “false” are equivalent states for a single proposition. Arthur’s conclusion is that the falsity of the statement can be asserted, but with no regard that, through the statement itself, false implies true in turn.

Simply put, self-contradictory statements and entities cannot exist in actuality… only hypothetically. When in the course of hypothetical reasoning we arrive at a contradiction, we must necessarily trace our logic back to whatever assumed proposition was made, and negate it – this is the entire point of indirect proof and proof by contradiction. You cannot negate something that is, only assumptions about what might be. Arthur Priors treatment of a self-contradictory statement as actual instead of as hypothetical is nonsense to me.

If his reasoning is indeed valid then proof by contradiction is no longer a valid technique, as it may be assumed from here on that contradiction is not grounds for calling absurdity on an argument.

In proof by contradictions, the contradictory statements exists hypothetically… but then it is necessarily debunked and replaced by a new line of reasoning that does not lead to contradiction. The contradiction is removed from possibility.  However, Arthur Prior allows the continued existence of the contradictory statement.  The contradiction persists and imposes a truth value.

Arthur Prior is right about one thing. That all logical statements assert themselves as true implicitly. I agree for the most part. But bare in mind that this is true for only logical statements, not illogical or logically devoid ones.   Statements must also provide meaningful information. If the sole purpose of a statement is to declare itself false then nothing informative is contributed.  Nothing is added to an argument.  The statement is no more true or false than had you not said anything at all… which, obviously by the nature of remaining silent, has no truth value. Such statements undermine (worse, they contradict) the entire point of using language in the first place.

Undefined Sequence Terms

This is the most trivial of divergence tests.

Given any sequence term, , where (and so is of concern, it is included in the series’ sum)…

If the value is undefined, the series cannot converge because the partial sums at and after that point are undefined.

Thus, as trivial as it may seem, all terms included in a series (all terms of consequence) must be defined and finite in value.

Its unfortunate to see this fact overlooked as frequently as I do.

• If the sequence is defined everywhere of consequence then nothing can be said about the sum, or whether or not it converges.
• If there are undefined terms, including divergences to infinity coinciding with an integer , then the series is divergent.

Finite Series

If the series passes the first test and is defined everywhere of consequence then the second question to ask is: “Is the series a finite series?”

• If the series is finite then it converges.  All finite series’ that are defined at every term are convergent.
• If the series is an infinite series then nothing can be said about its convergence or its divergence, nor about the sum.

This fact is overlooked far too often, also.  Yes, I have seen students perform integral tests on a finite series.

Shifting the Series

If you have two integers, and such that , and the series is defined everywhere of consequence, then:

Either and both converge, or they both diverge.

Simply put, the finite segment, or interval , has by itself a finite sum.  That is, is a finite series and is therefore convergent.  This interval has no affect on the convergence or the divergence of the whole.

Realizing this can simplify alternative approaches later.  Students are often times too concerned for what happens near the beginning of the series that they fail to recognize that the only thing that matters is what happens to the series past any arbitrary finite point.

Yes, I have seen students dumbfounded by the fact that a series does not start at the default zero or one index.

Sequence Limit Test, or the “n^th Term Test”

Generally speaking though, the first three “tests” are trivial and hardly worthy of note.  The real issues arise with infinite, everywhere-defined series’, with an arbitrary initial series index – which is what this this test is about.

This test is very easy, probably the easiest, and yet I see it completely ignored all the time in favor of other, more difficult, less conclusive tests.  It checks for divergence but cannot verify convergence.

Simply take the limit at infinity of the sequence.

• If diverges infinitely or diverges due to failure to converge, then the series diverges.
• If – the limit converges to a non-zero value – then the series diverges.
• If , converges to zero, then nothing can be said about the series’ convergence or divergence, nor about the sum.
  • There is no convergence verification check using this test

This test is ignored far too frequently.   I have seen people check with, say, the ratio test or the roots test, falsely determine “convergence”, or wasted effort concluding that the test was “inconclusive”, even though this simple check would have easily demonstrated divergence.

11 – Alternatively add and subtract the digits of an integer, from one end to the other.  If the resultant is a multiple of 11 then the whole is a multiple of 11.  For example, take the value 365167484.  We add and subtract its digits alternatively: .  The sum is a multiple of 11 (Zero is a multiple of eleven!).  Because the resultant is a multiple of eleven, so too is the whole.  The process is reiterative.

12 – The number twelve can be factored into 3 and 4.  A number is divisible by 12 only if it is divisible by both 3 and 4, simultaneously.  Refer to those rules.

13 – The method for thirteen is not unlike the first method for seven.  So similar in fact, I will actually copy and paste it here and perform only minor edits. Compare:

“Suppose the integer s is in the form , where m and n are both integers as well, but restrict n to a single digit.  This is saying, take n to be the ones digit and take m to be everything to the left of the ones digit as its own integer.  For example, , therefore and .  The process is relatively simple.  Multiply n by nine and subtract it from m: .  In this example we take 1697 and we subtract 72 for a difference of 1625.  Now, 16978 is divisible by 13 only if 1625 is divisible by 13.  That is to say, more generally, is divisibe by 13 if is divisible by 13.  The process is reiterative.”

13 – An alternative method from the one mentioned above is such that: .  Everything else is otherwise the same.

14 – Fourteen can be factored into 2 and 7.  Thus, a number is divisible by 14 only if it is divisible by both 2 and 7. It must be divisible by 7 and be even.

15 – Fifteen can be factored into 3 and 5. Thus, a number is divisible by 15 only if it is divisible by both 3 and 5.  That is, divisible by 3 while ending in a 0 or a 5.

16 – The divisibility rules for 16 follow the same pattern as for 4 and 8.  There are two approaches.  A number must be divisible by 2 four distinct times. Or, the four least significant digits must be divisible by 16 by themselves.

2^k – To generalize the rule of 2, 4, 8, 16 and the like – If you are testing if a number, s, is divisible by n, where , for some positive integer k (such is the case with 2, 4, 8, 16, etc.), then the two tests are as follows: The number s must be divisible by 2 in total k distinct times; or the least significant k digits of s must be divisible by n.

This method I have found useful and effective on multiple instances.  It isnt always effective, however – it depends heavily on the polynomial itself and you might waste some time applying it ineffectually.  But if it does works… you simplify the problem and save yourself a lot of work. Its one of those tricks that you might use when you run out of options otherwise.

I am a bit proud of myself for it, though.  I have no idea what this trick might be called, and Im sure it already exists, but I independently derived this technique myself. So, Im a bit pleased by it, especially.  Its nothing particularly special, just a natural extension of basic algebraic reasoning.  And yet, as trivial as it might be, I have never been formally taught this and I have never before seen it, neither online nor in a book – and believe me, I have done quite a bit of research into root-finding techniques.  If anyone knows the name or a reference, or has any mathematics to contribute to it, please do tell.

Suppose you have some polynomial for which you suspect that there might be complex roots.  Suppose you have exhausted your alternative techniques at factoring. It seems irreducible, though perhaps not. Some arbitrary nth degree polynomial with arbitrary coefficients.  As it turns out, this technique applies to any polynomial, even ones with irrational and complex coefficients.

For the sake of simplicity, let us use the trivial cubic polynomial as an example:

We make the assumption that a complex root zi exists, where z is a complex number unto itself, being multiplied by the imaginary unit. Nothing is wrong with this, since z is an arbitrary value to begin with of the form , the product zi is yet another arbitrary complex number of the form .  We only need assume that some complex number zi is a root of the polynomial.

Let us plug this arbitrary value in and simplify:

Group the real terms and the imaginary terms:

In order for zi to be a root, P(zi) must equal zero.  Therefore, it stands to reason that the real component and the imaginary component must also be zero.  There is an assumption here which I will explain later.  Let us assume for now that the real and the imaginary components must both equal zero independently of one another.

We arrive then at two simultaneous equations:

And factoring out a zero root on the right equation.

The great thing about this technique (if the coefficients of the polynomial are real) is that it allows you to split one polynomial into two simultaneous polynomials, each with reduced degree. In addition, one will have even exponents and the other odd exponents.

All that need be done now is find the roots of each of these.  Find the roots that make the real component zero, and the roots that make the imaginary component zero.

We want to find those values of z which make both the real and imaginary components equal to zero. So, taking the intersection of these two sets produces the following set:

This set is the set of all z-values such that zi is a root of the original polynomial.  If z is real then zi is purely imaginary, and you have purely imaginary roots.  If z is purely imaginary then zi will indeed be real roots.  And if z is a complex number, as it was defined to be, then zi will be a complex number.

The technique does work, generally speaking, for all polynomials with arbitrary coefficients.  However, it likely will be less effective and more difficult if those coefficients are irrational or complex in themselves.  The imaginary and real components of P(zi), for example, wont necessarily produce two reduced polynomials.

As mentioned earlier, the technique doesnt always work as an effective and practical tool… and for the same reason as the assumption I also eluded to earlier.  On occasion – more frequently than not, actually – the technique will not account for all possible complex roots.  The assumption that I made earlier was that both the imaginary and the real components must both equal zero independently of one another. That independence is only the simple case, but the assumption is essential to the technique.

In truth, the real components and the imaginary components, when you plug in the arbitrary complex z value, may themselves be arbitrary complex numbers instead of zero.  Thus the true imaginary and real components after evaluation at z may not reflect the imaginary and real components of the general equation in terms of z.  If that makes any sense. In another way of saying, the sum of the real component and the imaginary component does equal zero, but they dont equal zero independently of one another. They equal zero dependently of one another.

Just so long as the real component evaluated at complex z equals the negative of the imaginary component evaluated at complex z, the equation will still balance to zero and zi will still be a root.  It just wont show up in the intersection of the two sets on account of the fact that we assumed each component of the general equation was individually zero, and not the sum after evaluation.

I dont want to kick a dead horse – I hope I have explained this point sufficiently well.  Feel free to ask for elaboration.

In conclusion, the reasoning is sound and the technique is valid, it just isnt affective for all arbitrary polynomials in a practical sense.  I use it when I am out of alternative options… and I have found it beneficial in more than a few cases.  It has allowed me to find and factor out complex roots when I still hadnt accounted for all real roots, and when I wasnt even able to factor terms satisfactorily using more well known algebraic methods. I adamently believe that being aware of this technique, having it in your arsenal, no matter how trivial or ineffectual it might be, and being aware of its limitations and hold-ups, will only add to the mathematicians root-finding skills.  You never know when it might just work.

Feel free to give me feedback on this post.  I wouldnt mind a thorough discussion of the method, its shortcomings and its strengths.

0 – No number is ever divisible by zero.  This is undefined.

1 – All numbers are divisible by one. Nothing is achieved by doing so. Completely trivial.

2 – Only even numbers are divisible by two.  That is, any number that ends in a 0, 2, 4, 6, or an 8.

3 – If the sum of the digits of the integer is divisible by three, then so is the integer itself.  This is iterative – you may find the sum of the digits of the sum of the digits of the sum of the digits, so on so forth, until you are left with a single digit number: 0, 3, 6, or 9, (or any larger number that is also divisible by 3).

4 – If the number is divisible by 2 twice then it is divisible by 4.  That is, if after factoring out a 2 you can factor out another 2 from the remaining factor then the original integer is a multiple of 4.

4 – Alternatively, take the last two digits (the least significant digits – the ones and the tens place).  If these two digits are themselves divisible by 4 then the whole integer is divisible by 4.

5 – All numbers ending in a 0 or a 5 in the ones place are divisible by 5.

6 – The prime factors of 6 are 2 and 3.  Thus, any number that is divisible by both 2 and 3, simultaneously, is also divisible by 6.  It must pass both tests – refer to the tests for divisibility by 2 and by 3.  If the integer’s digits add to a multiple of 3 and the integer is even, then its a multiple of six.

7 – Suppose the integer s is in the form , where m and n are both integers as well, but restrict n to a single digit.  This is saying, take n to be the ones digit and take m to be everything to the left of the ones digit as its own integer.  For example, , therefore and .  The process is relatively simple.  Double n and subtract it from m: .  In this example we take 36 and we subtract 8 for a difference of 28.  Now, 364 is divisiby by 7 only if 28 is divisible by 7.  That is to say, more generally, is divisiby by 7 if is divisible by 7.  The process is reiterative.

7 – An alternative method exists.  Memorize the set .  What we want to do is pair each digit of your number with a digit from the set of six values, above.  The ones digit gets paired with the first value, the tens digit gets paired with the second value, the hundreds digit gets paired with the third value, etcetera.  When you have reached the end of the set, cycle around to the beginning.  Continue pairing digits of the number with a value from the set until all digits have been paired.  Multiply these pairings and add their products.  For example, take the number .  We add the products of the associated pairs:

.

The value 2037 is divisible by 7 only if 28 is divisible by 7.  The algorithm is reiterative.

8 – This divisibility test for 8 is not unlike the divisibility test for 4.  A number must be divisible by 2 three distinct times.  . That is, after dividing by 2 evenly, the remaining factor must still be divisibly by 4.

8 – The alternative divisibility test for 8 is not unlike the alternative divisibility test for 4.  We take the three least significant digits (the ones, tens and hundreds place).  As its own unique integer, if it is divisible by 8 then the whole is divisible by 8.

9 – The divisibility test for 9 is not unlike the divisibility test for 3.  We must add up all the digits of the integer, re-iteratively if you wish.  The sum must ultimately be divisible by 9.

9 – Alternatively, not unlike the divisibility test for four, 9 is the square of 3.  So a number is divisible by 9 if it is divisible by 3, and the remaining factor is also divisible by 3.

10 – If the integer ends in a zero then it is divisible by 10.  This is the same as passing the divisibility tests for 2 and 5 simultaneously.  For 5, the number must end in a 0 or 5, while for 2 the number must be even.

For divisibility rules for integers larger than ten, refer to this followup post.

Whenever you check a root by polynomial long division or synthetic division, these tests apply and you’ve already done all the work!

Suppose you wanted to check a root, r, not by plugging it directly into the polynomial, P(x), but by dividing the term out.  Evaluating P(r) would tell you if r is a root or not, but it doesn’t simplify the polynomial by reducing its degree, nor does it provide you with any additional information.  Unless its easy arithmetic, its not really practical in the grand scheme of root-finding.  It is recommended to synthetically divide r out of P(x), or to use polynomial long division to divide the binomial (x-r) out of the polynomial.

Your polynomial is arbitrary, but it must have real coefficients.  If its coefficients have common imaginary factors then they must be factored out.

Pick some value you want to test, it doesn’t have to be a root or even so much as a suspected root.  Call this value t, a test value.  When you test this value, when you divide, however you chose to do it, you will arrive at a new polynomial of reduced order.

Notice the rational expression.  This is zero (because c=0) whenever t is a root.  As a general rule, c = P(t).

The important aspect here is the coefficients of each term.  The coefficients of the polynomial and the coefficient, c, of the rational expression.

Whether you use polynomial long division or synthetic division, the point of the procedure which is fundamental to this test is to make note of the coefficients of the resultant polynomial and rational expression.

List the coefficients of the new function in order.

Bare in mind, it is important to include all terms, even those with a zero coefficient.  If a term, say for example the quadratic term, does not exist in the new quotient polynomial, the implication is that it does exist with a coefficient of zero.  All coefficients from the rational expression, to the constant term, on up to the n-1^st degree term must be accounted for.

Nothing up until this point is new to you.  All we have done thus far was done division of a binomial into a polynomial, by one means or another, as you likely do all the time anyway whenever testing a root.  And if you don’t, you should.

Now, the hard part is over.  All you have to do now is take a look at the quotients coefficients, as per the list generated above.

Upper Bounds Test

Suppose the test value, t, was a positive number. This is a condition for the upper bounds test to work.  If, then, the list of coefficients of the new quotient polynomial are all non-negative then we can say that t is an upper bound.

That is to say, two conditions must be met.  One, t must be positive.  Two, the new polynomials coefficients (including the numerator of the rational expression) must all be non-negative.  The phrase “non-negative” implies that positive values and zero are both permissible.

If those conditions are met then t is an upper bound on the real roots.  No root can be greater than t.

Lower Bounds Test

This test is similar to the upper bounds test, with two variations.

First of all, the test value t must be negative.

Secondly, the new coefficients must follow a different pattern entirely.  The new coefficients must alternate between non-negative and non-positive.  It doesn’t matter if it starts out negative or positive, just so long as it alternates thereafter according to the pattern.  Again, it need not alternate positive and negative, but it does need to alternate non-negative and non-positive.  In other words, zeros count as something.

If these two conditions are met, if the testing value t is negative and the quotient polynomial alternates non-negative and non-positive, then we can say that t is a lower bound on the real numbers.  No root can be less than t.

Conclusion

These tests help to restrict the real domain significantly.  It can eliminate possible rational roots derived from the Rational Roots Theorem so that they dont need to be tested.  It can narrow the real domain far more restrictively than the Single Bounds Theorem.  Even if the roots end up being irrational and difficult to find, at least you know whereabouts (in what interval) to look when you employ alternative methods.

I have noted that some people are confused about the existence of irrational roots that dont have conjugate pairs.

Complex Conjugates

To start, let us first discuss complex conjugates.

If a polynomial has only real-valued coefficients (irrational or transcendental, it doesnt matter), then the rule of complex conjugates applies.

For any root or zero that is complex, and therefore having the form where , there is a second root at .  Where a and b are real.

That is the jist of the complex conjugate rule, and it always applies to all complex roots, no matter the values of a or b, just so long as b is non-zero and the polynomial in question has real coefficients.

Irrational Conjugates

In the case of irrational conjugates, there is a similar rule.  It stipulates that the coefficients of the polynomial must all be rational and real.  Just so long as the coefficients of the polynomial are real and rational, the irrational conjugates rule applies.

The rule states that any irrational root of the form has an irrational conjugate of the form .  Where a, b, and c are all rational numbers.

Note that b is just a constant value that can be factored out of or into the radical, so for any irrational root of the form there is another root at . Where a and b are rational.  This is algebraically equivalent.

Equivalently, and without loss of ambiguity, if a root takes the form then we know there exists a root .  In addition, if you allow then the irrational conjugate is equal to the negative of the irrational.

The Existence of Irrational Roots Without Conjugate Pairs

The rule is very specific about the form an irrational conjugate may take.  It excludes all other irrational roots having different forms.

People get confused by this for some reason.  It is very possible for irrational roots to exist that do not take the form mentioned that the rule applies to.  For example, a root of the form would not meet the very specific criteria (namely the form ), and therefore we cannot make any general statements about any conjugate existing.

Irrational conjugates exist only if the polynomial has real and rational coefficients… AND… an irrational root takes the form .  Under no other condition does the irrational conjugate rule apply.

Thus, if you know that real irrational roots exist but have also determined that it doesnt have a conjugate pair, then you immediately know that it doesnt take the form of a rational number added to the square root of another rational number.

Irrational and Complex Conjugation

Notice that irrational roots can take the form , but now consider the possibility that b is a negative value.  Suddenly, the complex conjugate rule emerges from the irrational conjugate rule.

Recognizing the fact that in the complex number the value given to b can be irrational in itself, in the form of a square root of a rational number, and that i, the imaginary unit, is nothing more than the square root of negative one.

By letting the irrational conjugate contain a negative value in the square root operator, the irrational conjugate rule is equivalent to the complex conjugate rule.  By letting the imaginary value of the complex conjugates contain a square root operation, the complex conjugate is equivalent to the irrational conjugate.

To start, you must set up your polynomial appropriately.  If your coefficients are irrational or complex, you must factor our any common coefficient, in particular the irrational and complex components, in order to make all coefficients rational and real.  Multiplying the polynomial on whole by an arbitrary irrational/complex constant is ultimately irrelevant.  But in order to apply the theorem, all coefficients of the polynomial you are finding the roots to must be rational and real.  This theorem cannot be applied if your coefficients cannot be rationalized.

In fact, the coefficients must all be integers in order to apply the theorem.  So if your coefficients are non-integers but rational then multiply through by a the common denominator.  Multiplying the polynomial by a constant in no way affects the roots themselves, but it may be necessary to make all coefficients integers.

We should ultimately end up with a polynomial of the form:

Where all are integers and is an arbitrary constant (irrational or complex is fine).

Notice that the roots of this polynomial occur when it is set equal to zero.

Thus the constant C drops out and is irrelevant.

The theorem also assumes that there is a constant term. That is to say, .  If it is equal to zero then is a root and is a factor.  It should be noted and dropped out to produce a new polynomial with a decremented degree having a constant term.

The theorem is simple from here on out.  All you need to focus on is the lead coefficient and the constant term

Applying the Theorem

You must be able to find the factors of a constant.  Not the prime factors exclusively, but all factors.  All integers, composite or prime, that can divide into a constant.

Create a set of numbers which form the integer factors of your lead coefficient.  Assume for a moment that , then the list of integer factors are .

Create a second set of numbers which form the integer factors of the constant term.  Assuming then the set of integer factors is .

The next step is to take all rational numbers that can be constructed from each possible combination, in which the second set forms the numerator and the first set forms the denominator.

That is, each number from the set constructed of the factors of the constant term divided by each number from the set constructed of the factors of the lead coefficient.

Taking each combination:

This sets components can be reduced and the set consolidated.

This is pretty simple so far, eh?

Now we take the positive and negative of each of those values:

Or simply:

This set contains all of the possible rational roots that the polynomial can have.  The polynomial cannot have rational roots outside of this set.

Any generic roots that the polynomial may have that is not accounted for by the Rational Roots Theorem fall in the realm of non-rational numbers.  Roots that exist outside of the set produced by the theorem are either complex or real-irrational.

Given any polynomial P(x), written in order such that exponents are decreasing from largest to smallest, lead term to constant term…

If some term then disregard it. It has no affect and doesnt need to be listed.  All it means is that a coefficient was zero, which is neither positive nor negative, and can count as both or neither, and therefore has no baring on the outcome of the test.

If the constant term then there is no constant term, and is a root and is a factor.  This should be removed and factored out of the polynomial before moving on with these tests.

Applying the Rule

What we do is we count the number of sign changes on the coefficients.  Using the sign on as the initial sign, the reference, we count the number of sign changes from to .

Sign changes, of course, refers to the change from a positive to a negative value, or from a negative to a positive value.

An example would be the polynomial

Note the coefficients, which form the ordered set:

The lead coefficient is a positive 3 and has a positive sign. It then changes once to become a negative (-2 in this case) on the second term.  It changes a second time to become a positive (+2) on the third term.  On the fourth term it is still positive (5) and therefore doesnt change.  But on the sixth and final term it changes to negative again (-6), for a total of three sign changes.

Thus, there are three sign changes in this example polynomial.  Descartes’ Rule of Signs tells us that there can be no more than three positive real roots.

In general, Descartes’ Rule of Signs tells us that if a polynomial has sign changes then there are no more than s positive real roots.  The value s of the number of sign changes is the maximum number of positive real roots that there can be.

Fewer Than Maximum

“Maximum” is the key word there.  The actual number of positive real roots may not in actuality be precisely s, the number of sign changes. It can be fewer, but only by multiples of 2.

That is to say that if the number of sign changes is s then the actual number of positive real roots could be: etcetera, to the smallest non-negative number, 0 or 1.  Another way of saying all of this is the maximum number of sign changes may be decremented by multiples of two to some integer value in the interval to produce some value that is the actual number of positive real roots.

As an equation: where , and p is the actual number of positive real roots, s is the number of sign changes, and n is some non-negative integer.

In the case of the example polynomial above, in which the number of sign changes was 3, we can say with absolute certainty that there is either 1 or 3 positive real roots.  It cannot be more than 3, because the number of sign changes indicates a maximum. And it cannot be fewer than 0, because no countable quantity may be negative.  And it cannot be either 0 or 2, because neither is fewer than 3 by an even integer.

Negative Real Roots

There is no direct method for finding the number of negative real roots from the polynomial P(x).  Instead, we must get creative by constructing a new polynomial.

What we have to do is reflect the polynomial across the y-axis by replacing x with negative x.  The polynomial becomes a new polynomial which is a perfect reflection of the original.  Positive x-values become negative, negative x-values become positive.  All positive roots become negative roots and all negative roots become positive roots.

Simplify the polynomial so that the coefficicients absorb a negative where applicable.  In the case of a general polynomial,

For terms with even exponents, the coefficient doesnt change. For terms with odd exponents, the coefficients become negated. In our example polynomial, the entire expression simplifies into:

At this point we may treat the new reflected polynomial as its own unique polynomial.  Let us apply Descartes’ Rule of Signs to determine the number of positive real roots just like we did before.

In our example polynomial I count exactly 1 sign change between the third and fourth terms.  This count, 1, is a maximum and it cannot be decremented by even integers without becoming negative.  So, the exact number of positive real roots for P(-x) is exactly 1.

However, P(-x) is a reflection of the original P(x) about the y-axis, and what are positive real roots for P(-x) are in fact negative real roots for P(x).

Thus, we can conclude that in our example polynomial, there are 1 or 3 positive real roots and 1 negative real root.

Considering the Complex

Descartes’ Rule of Signs determine a set of possibilities for the number of positive and the number of negative real roots.  It does not account for complex roots.  However…

A corollary to the Fundamental Theorem of Algebra says that for any n-degree polynomial, there are exactly n roots.  If the roots are not real then they exist as complex.  Whatever roots are not accounted for by Descartes’ Rule of Signs must be unreal… since positive and negative real numbers together encompass all the real numbers (and the zero, but zero roots were accounted for at the beginning)

In the case of our example polynomial above, P(x) was a fourth degree polynomial, a quartic. We know that there are exactly four roots in total.  These roots may not all be unique and they may not all be real.

We may take each possible pair between the number of possible positive and negative real roots. In the case of negative real roots, the only possibility was 1.  In the case of positive real roots, the only possibilities are 1 and 3.  When considering all possible real roots (positive or negative) we see that there are either 2 or 4.  The numbers 2 and 4 come from the sum between negative and positive, to produce a whole new set of possible real roots.

We know now that there are either 2 or 4 real roots.  And since we know that there are 4 roots in total, we may deduce that the difference, 2 and 0 in this case, are the number of possible complex roots.

Conclusion

Determining the quantity of possible real, possible complex, possible positive real and possible negative real roots are as simple as counting sign changes. No one should be struggling to find another positive real root when they have already accounted for all of them.

Sometimes simply knowing that at least one negative real root exists gives you all the information you need to stop searching for more positives.  Sometimes knowing that so many roots are negative is all you need to know to decide whether or not to look for complex.

One very surprising result in math is the relationship between the Fibonacci sequence and the golden ratio.

The Golden Ratio, also called Phi as we use the Greek lowercase letter to denote it, is a famed ratio.  This ratio is apparent in art and science alike, built into art even if by accident, and can be seen throughout mathematics and nature everywhere, from life to physics.  Its regarded as a “beautiful” ratio and has a certain aesthetic quality that most people will recognize as appealing even if they fail to realize that the ratio is involved.

Simply, the ratio can be computed explicitly as:

The Fibonacci sequence, on the other hand, is defined as:

where and

This produces the sequence:

The Fibonacci sequence also appears in nature and often has to do with population growth and diversity.  But it has no apparent or directly obvious relationship to the Golden Ratio.  Their relationship emerges in math as a quite astonishing result.

The explicit equation that emerges is:

Which is an exact equality.  However, also an exact expression, we may also use

Another equation, which is simpler and requires less computation can be used, but it involves the floor function (returning the next lower integer):

So if you ever need to know what a particular Fibonacci number is, it isnt too hard to find out.

Similarly, the following formula allows you to determine the index within the Fibonacci sequence of a Fibonacci number.  If you know that F is a Fibonacci Number then F is the nth Fibonacci number.: