Definitions of algebraic and transcendental numbers

Motivation of the definition of the term algebraic number . The numbers 7/5 , 2^(1/3)+1/5 , and 1/2+sqrt(1-sqrt(2))/2 are, in one sense, quite different:

  • the first is a rational number
  • the second is a real irrational number, and
  • the third is (clearly) a complex number (with non-zero 'imaginary' part)

> evalc(1/2 + sqrt(1 - sqrt(2)));


> I^2; # in Maple 'I' is sqrt(-1)



But in another sense they are not so different: all three are solutions of equations of the same kind :

  • 7/5 is a solution of the equation a*x+b = 0 , where a and b are integers: (a, b) = (5, -7)
  • 2^(1/3)-1/5 is a solution of the equation a*x^3+b*x^2+c*x+d = 0 , where a , b , c and d are integers: (a, b, c, d) = (125, 75, 15, -249)
  • 1/2+sqrt(1-sqrt(2))/2 is a solution of the equation a*x^4+b*x^3+c*x^2+d*x+e = 0 , where a , b, c , d and e are integers: (a, b, c, d, e) = (8, -16, 8, 0, -1)

> solve(8*x^4 - 16*x^3 + 8*x^2 - 1 = 0);

1/2-1/2*sqrt(1+sqrt(2)), 1/2+1/2*sqrt(1+sqrt(2)), 1...

> evalc(1/2 + sqrt(1-sqrt(2))/2);



DEFINITION . A real or complex number alpha is said to be algebraic ('over the integers ') if x = alpha is a solution of some polynomial equation a[0]*x^n+a[1]*x^(n-1)+`...`+a[n] = 0 , where the a[0], a[1], `...`, a[n] are integers, a[0] <> 0 .

REMARK on that DEFINITION. You can sometimes see an algebraic number defined as follows: a real or complex number alpha is said to be algebraic ('over the rationals ') if x = alpha is a solution of some polynomial equation a[0]*x^n+a[1]*x^(n-1)+`...`+a[n] = 0 , where the a[0], a[1], `...`, a[n] are rationals, a[0] <> 0 .

One should appreciate that it doesn't matter which of these definitions is adopted as the basis for defining the term algebraic , for - trivially - a number that is algebraic according to one of these definitions is automatically algebraic according to the other definition, and vice versa. (For example, x = alpha = 8/3-2*sqrt(21)/3 is a solution of the equation 3*x^2/4-4*x-5/3 = 0 , and so alpha is algebraic according to the latter definition. But clearly that same alpha is a solution of the equation 9*x^2-48*x-20 = 0 (the earlier one multiplied by the product of the denominators of the rational coefficients) :

> solve(3/4*x^2 - 4*x - 5/3 = 0);

8/3+2/3*sqrt(21), 8/3-2/3*sqrt(21)

> solve(9*x^2 - 48*x - 20 = 0);

8/3+2/3*sqrt(21), 8/3-2/3*sqrt(21)


DEFINITION . An algebraic number alpha is said to be of degree n if x = alpha is a solution of some polynomial equation a[0]*x^n+a[1]*x^(n-1)+`...`+a[n] = 0 , where the a[0], a[1], `...`, a[n] are integers ( a[0] <> 0 ), but x = alpha is not a solution of any polynomial equation (with integer coefficients, not all zero) of smaller degree.

Example 1. 7/5 is clearly an algebraic number of degree 1.

Example 2. (sqrt(7)+3)/5 is clearly an algebraic number (set alpha = (sqrt(7)+3)/5 , then 5*alpha-3 = sqrt(7) , and thus

(5*alpha-3)^2 = sqrt(7)^2 , 25*alpha^2-30*alpha+9 = 7 , and so x = alpha is a solution of the equation 25*x^2-30*x+2 = 0 .

Also, (sqrt(7)+3)/5 , being irrational, cannot be of the first degree, and thus is of degree 2.

Example 3. sqrt(2)+sqrt(3) is clearly an algebraic number (but do not jump to the hasty conclusion that it is of degree 2 on the flimsy grounds that sqrt(2) and sqrt(3) are both algebraic, and of the second degree): set alpha = sqrt(2)+sqrt(3) , then

alpha^2 = (sqrt(2)+sqrt(3))^2 = 5+2*sqrt(6) , (alpha^2-5)^2 = (2*sqrt(6))^2 = 24, giving alpha^4-10*alpha^2+25 = 24 , and finally

alpha^4-10*alpha^2+1 = 0 . Thus x = sqrt(2)+sqrt(3) is a solution of the fourth degree equation x^4-10*x^2+1 = 0 .

However, it is not that sqrt(2)+sqrt(3) is of degree 4 simply because of the fact that the equation is of degree 4. After all, if one set alpha = sqrt(5) , then alpha^4 = sqrt(5)^4 = 25, giving alpha^4-25 = 0 , and so x = alpha is a solution of the fourth degree equation x^4-25 = 0 . But, the polynomial x^4-25 happens to factor as the product of two polynomials with integer coefficients, of smaller degree: x^4-25 = (x^2-5)*(x^2+5) , and it's the x^2-5 that is the fundamental polynomial associated with sqrt(5) .

Without going any further, let me simply remark that x^4-10*x^2+1 doesn't factor as the product of two polynomials in a similar fashion (and thus sqrt(2)+sqrt(3) is of degree 4):

> factor(x^4 - 25);


> factor(x^4 - 10*x^2 + 1);



Elementary properties of algebraic numbers (which are easily proved).

1. If alpha and beta are algebraic numbers then so too are alpha+beta, alpha-beta, alpha*beta and alpha/beta ( beta <> 0 )

2. If alpha is algebraic then so also is every one of alpha^(1/2), alpha^(1/3), alpha^(1/4), `...`


Concerning Hardy & Wright 's It is not immediately obvious that there are any transcendental numbers ...

Why was it natural at one time to wonder about the possibility that every real (or complex) number is algebraic? (Of course one has to put out of one's mind the world view after Cantor, with his clarification that there are only a countable number of algebraic numbers, but an uncountable number of non-algebraic (=transcendental) ones. The interested reader might wish to consult at my web site my 2nd year BA summary course notes on The Real Numbers and Cantorian Set Theory .)

So, imagine starting with a completely empty complex plane, and imagine it being filled up, bit by bit, with every kind of number one can think of... every time one thinks of a new number or numbers (especially an infinite number of numbers) a light comes on at each of those numbers. The plane begins to be lit up with those tiny point specks of light... slowly at first... The integers, even all of them, hardly make any impression, but certainly the rationals fill up a lot of the real line (itself, of course, swamped by the vastness of the entire complex plane), though not, of course, the entire real line: there are all those real irrational numbers... all those ones that are algebraic of the second degree, but they don't fill the entire real line, there are all those of the third, fourth, fifth, sixth, ... degrees (Do they - perhaps - fill the entire real line?). And then there are all those complex numbers with non-zero imaginary parts... all those ones of the form ( r[1]+r[2]*i ), where r[1] and r[2] vary in all possible ways over the rational numbers... Every one of those is a solution of a quadratic equation with integer coefficients (and so is of degree 2). They fill up a lot of the complex plane, but not, of course, all of it. But then we imagine switching on lights at all the complex solutions of third, fourth, fifth, sixth, ... degree algebraic numbers. Have we now filled up the entire complex plane; or, at the very least, earlier, the entire real line?

Suppose, just for the moment , that we hadn't filled the entire complex plane, and let us return to the above definition of the actual term algebraic , with its reference to 'integer' (or 'rational') coefficients. Let's now make the observation that those coefficients are merely (with hindsight) algebraic numbers of the first degree, and, in the light of that, let us now speculate that we might - perhaps - get our hands on a real or complex number that isn't an algebraic number (according to our earlier definition) by doing attempting like this: let's try to find a polynomial equation that doesn't have an algebraic solution by choosing some (or possibly all) of its coefficients not to be integers (or rational numbers)...

Note that every such polynomial equation has solutions (as many, in fact, as the degree of the polynomial, thought some may be repeated), every one of which is a complex number: that's because of the so-called Fundamental Theorem of Algebra (remarkable itself in its day, and still so...), namely that every polynomial equation with real/complex coefficients has exclusively complex solutions (some of which may be real).

Thus (we wonder) could we find/construct a polynomial, all of whose coefficients are algebraic numbers (but where some/all of them are of degree greater than 1), such that the corresponding equation (all of whose solutions we know from the Fundamental Theorem of Algebra must be complex numbers) possesses at least one solution that is not an algebraic number? For example, consider the following equation in which all of the coefficients are algebraic numbers (not all rational):

sqrt(7)*x^2-9*x/11+3/4-sqrt(5/3) = 0 ... (E)

with coefficients sqrt(7) (2nd degree), -9/11 (1st degree), and 3/4-sqrt(5/3) (2nd degree algebraic).

The solutions of that equation are:

> solve(sqrt(7)*x^2 - 9*x/11 + 3/4 - sqrt(5/3) = 0);



A beginner might not immediately notice, but those two solutions are algebraic numbers according to the standard definition of being algebraic ' over the integers '.

Incidentally, it is an entirely elementary matter to prove - independent of actual calculation - that the two solutions of the above equation are algebraic numbers (my 2nd year BA students do that sort of thing). In fact, it is an entirely elementary standard result (this is not the place in which to prove it) that no matter what polynomial equation one takes of whatever degree ( m ), in which all of the coefficients are algebraic numbers of whatever degrees, then not only does that equation have m complex solutions (some of which may be real numbers), possibly with some 'repeated' solutions, but every one of those solutions is an algebraic number (according to the standard definition). In other words, if

b[0]*x^m+b[1]*x^(m-1)+`...`+b[m] = 0 ... (B)

is a polynomial equation of degree m in which every coefficient b[0], b[1], `...`, b[m] is an algebraic number (according to the standard definition, and b[0] <> 0 ) then every solution of (B) is a solution of:

a[0]*x^n+a[1]*x^(n-1)+`...`+a[n] = 0 ... (A)

in which every coefficient a[0], a[1], `...`, a[n] is an integer, and a[0] <> 0 . (The standard way of expressing the above succinctly is to say that the field of algebraic numbers is algebraically closed .)

In short, any hope of creating/finding/... a non-algebraic number by merely tinkering with the coefficients (in the manner suggested above) is doomed to failure .

The question then, in Euler's time (or possibly earlier than Euler?), was:

are there any (real or complex) numbers that are not algebraic?

Euler (or someone earlier?) defined a transcendental number to be a (real or complex) number that is not algebraic; it "transcendended the power of the algebraic".

Of course if there were any, then there would automatically be infinitely many. That's a matter that a novice might like to think about.

I return to Euler in another subsection.


A final word about computer computation/demonstration. I'd like to return briefly to the example (E) above, just in case a novice (using Maple or whatever) attempts to vary the coefficients, and wonders about why sometimes things don't appear to work out nicely... Consider the example:

sqrt(7)*x^5-9*x/11+3/4-sqrt(5/3) = 0 ... (E')

in which I have merely changed in (E) the x^2 to x^5 , but otherwise kept everything else unaltered. Now try 'solving', and you get:

> solve(sqrt(7)*x^5 - 9*x/11 + 3/4 - sqrt(5/3) = 0);



The software appears to fail, and with good mathematical reason... Here a beginner needs to know that one has now come up against a barrier: the classical result (Abel and Galois) concerning the non-existence of a solution 'using radicals' for general polynomial equations of degree at least 5... But that is "another day's work"...

You can of course do things like:

> plot(sqrt(7)*x^5 - 9*x/11 + 3/4 - sqrt(5/3), x = -2..2);

[Maple Plot]

> plot(sqrt(7)*x^5 - 9*x/11 + 3/4 - sqrt(5/3), x = -1..1);

[Maple Plot]


> fsolve(sqrt(7)*x^5 - 9*x/11 + 3/4 - sqrt(5/3) = 0);



Enough of this.

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After August 31st 2007 please use the following Gmail address: jbcosgrave at

This page was last updated 18 February 2005 15:09:55 -0000