Polynomials of an odd degree have at least one real root

Today, I present a proof taken from Edwards & Penney's Calculus and Analytic Geometry and PlanetMath.org.

I show that any polynomial of odd degree must have at least one root that is real.

Lemma 1:

lim (x → ∞) c/x = 0

where c is a nonzero constant

NOTE: This means that for any positive number ε, we can find a number δ such that:

for all abs(x) ≥ δ, abs(c/x) is less than ε

[See Definition 1, here for definition of mathematical limits]

Proof:

(1) Let ε be any positive number

(2) Assume c is positive.

(3) Let δ = 1 + c/ε

(4) So, then δ is greater than c/ε

(5) Which means that 1/δ is less than ε/c

(6) Which means that c/δ is less than ε

(7) For all n ≥ δ, it follows that:

c/n ≤ c/δ [which is less than ε from step #5 above]

(8) -c/δ is greater than -ε [this follows directly from step #5 from multiplying -1 to both sides]

(9) For all n ≤ -δ, it follows that:

c/n ≥ -c/δ [which is greater than -ε from step #7 above]

(10) Assume that c is negative

(11) Let δ = c/ε - 1

(12) So δ is less than c/ε

(13) So 1/δ is greater than ε/c

(14) Since c is negative, c/δ is less than ε [from multiplying a negative number to both sides]

(15) So for all n ≥ δ, it follows that:

c/n ≤ c/δ [which is less than ε from step #14 above]

(16) -c/δ is greater than -ε [this follows directly from step #14 from multiplying -1 to both sides]

(17) For all n ≤ -δ, it follows that:

c/n ≥ -c/δ [which is greater than -ε from step #16 above]

QED

Corollary 1.1:

limit (c/xⁱ) = 0

where c is a constant and i is a positive integer

Proof:

(1) For i=1, it is true from Lemma 1 above

(2) Let δ be the same δ from Lemma 1 above which depends on the sign of c.

(3) For all n ≥ δ, we have:

c/nⁱ ≤ c/n ≤ c/δ which is less than ε

(4) For all -n ≤ -δ, we have:

c/nⁱ ≥ c/n ≥ -c/δ which is greater than -ε

QED

Corollary 1.2:

Let g(x) = a₁/x + a₂/x² + ... + a_n-1/xⁿ

limit (x → ∞) g(x) = 0

Proof:

(1) For each a_i/xⁱ, the limit is 0. [From Corollary 1.1 above]

(2) Since g(x) is the sum of these values, we apply the Addition Rule [see Corollary 8.1, here] to get:

limit (x → ∞) g(x) = 0

QED

Lemma 2:

If f(x) is a polynomial of odd degree, there exists values a,b such that:

f(a) is less than 0 and f(b) is greater than 0

Proof:

(1) Let f(x) be a polynomial of an odd degree n such that:

f(x) = a₀xⁿ + a₁x^n-1 + ... + a_n-1x + a_n = 0

(2) We can make f(x) monic (where xⁿ does not have a coefficient) by dividing both sides by a₀.

So, we can assume the following monic form:

xⁿ + a₁x^n-1 + ... + a_n-1x + a_n = 0

(3) Let g(x) = (1/xⁿ)[a₁x^n-1 + ... + a_n-1x + a_n] =

= a₁/x + a₂/x² + ... + a_n-1/x^n-1 + a_n/xⁿ

(4) Then:

f(x) = xⁿ[1 + g(x)]

(5) Using Corollary 1.2 above, we know that there exists δ such that:

for all abs(x) ≥ δ, abs(g(x)) is less than 1 [where ε = 1]

(6) So, it follows that 1 + g(x) is greater than 0.

(7) Let b be any positive number greater than δ.

(8) It follows that f(b) is greater than 0 since bⁿ*[1 + g(x)] is greater than 0.

(9) Let a be any negative number where a is less than -δ.

(10) It follows that f(a) is less than 0 since aⁿ*[1 + g(x)] is less than 0.

QED


Theorem 3: A polynomial of odd degree has at least one real root

Proof:

(1) Let f(x) be a polynomial of odd degree.

(2) All polynomials are continuous [see Corollary 1.1, here], so f(x) is continuous.

(3) There exists a number a such that f(a) is less than 0. [see Lemma 2 above]

(4) There exists a number b such that f(b) is greater than 0. [see Lemma 2 above]

(5) Therefore, there exists at least one real root [see Weierstrass Intermediate Value Theorem, here]

QED

References

• "polynomial equation of odd degree", PlanetMath.Org
• Edwards, Jr. C. H. & Penney, David E., Calculus and Analytic Geometry, Prentice Hall, 1990