Algebra (Mathematics)

Algebra (from Arabic aljebr meaning reunion of broken parts1) is one of the broad parts of mathematics, together with number theory, geometry and analysis. In its most general form algebra is the study of symbols and the rules for manipulating symbols2 and is a unifying thread of all of mathematics.3As such, it includes everything from elementary equation solving to the study of abstractions such as groups, rings, and fields. The more basic parts of algebra are called elementary algebra, the more abstract parts are called abstract algebra or modern algebra. Elementary algebra is essential for any study of mathematics, science, or engineering, as well as such applications as medicine and economics. Abstract algebra is a major area in advanced mathematics, studied primarily by professional mathematicians. Much early work in algebra, as the Arabic origin of its name suggests, was done in the Near East, by such mathematicians as Omar Khayyam (10501123). Elementary algebra differs from arithmetic in the use of abstractions, such as using letters to stand for numbers that are either unknown or allowed to take on many values.4 For example, in x + 2 = 5 the letter x is unknown, but the law of inverses can be used to discover its value: x=3. In E=mc2, the letters E and m are variables, and the letter c is a constant. Algebra gives methods for solving equations and expressing formulas that are much easier (for those who know how to use them) than the older method of writing everything out in words. The word algebra is also used in certain specialized ways. A special kind of mathematical object in abstract algebra is called an algebra, and the word is used, for example, in the phrases linear algebra and algebraic topology (see below). A mathematician who does research in algebra is called an algebraist.

1 There exists a polynomial P of degree 5 with the following property: if z is a complex number such that z5+ 2004z = 1, then P (z2) = 0 Find all such polynomials P

2 Let N denote the set of positive integers Find all functions f : N → N such that

holds for all real numbers x and y Prove that no very convex function exists.

First Solution: Fix n ≥ 1 For each integer i, define

∆i = f

µ

i + 1 n

¶

− f

µ

i n

¶

.

The given inequality with x = (i + 2)/n and y = i/n implies

f¡i+2 n ¢+ f¡n i¢

µ

i + 1 n

¶+ 2

¶

− f

µ

i + 1 n

¶

≥ f

µ

i + 1 n

¶

− f

µ

i n

¶+ 4

n ,

that is, ∆i+1 ≥ ∆ i + 4/n Combining this for n consecutive values of i gives ∆ i+n ≥ ∆ i + 4 Summing this inequality for i = 0 to i = n − 1 and cancelling terms yields

f (2) − f (1) ≥ f (1) − f (0) + 4n.

This cannot hold for all n ≥ 1 Hence there are no very convex functions.

Second Solution: We show by induction that the given inequality implies

for all nonnegative integers n This will yield a contradiction, because for fixed x and y the right side

gets arbitrarily large, while the left side remains fixed

We are given the base case n = 0 Now if the inequality holds for a given n, then for a, b real,

and the induction is complete

Third Solution: Rewrite the condition as

For any positive integer n,

Hence there is no very convex functions

4 Let a1, a2, , a n (n > 3) be real numbers such that

a1+ a2+ · · · + a n ≥ n and a21+ a22+ · · · + a2n ≥ n2.

Prove that max(a1, a2, , a n ) ≥ 2.

Solution: Let b i = 2 − a i , and let S = Pb i and T = Pb2

i Then the given conditions arethat

(2 − a1) + · · · + (2 − a n ) ≥ n

and

(4 − 4b1+ b21) + · · · + (4 − 4b n + b2n ) ≥ n2,

which is to say S ≤ n and T ≥ n2− 4n + 4S.

From these inequalities, we obtain

i ≤ n2 (with the equality case just mentioned), by noticing that replacing a pair

a i , a j with 2, a i + a j − 2 increases the sum of squares.

5 Let { a n } n≥0 be a sequence of real numbers such that a n+1 ≥ a2n + 1

5 for all n ≥ 0 Prove that

√

a n+5 ≥ a2

n−5 for all n ≥ 5.

First Solution: (by Alison Miller) For any k, we have the following inequality:

implying the desired result

6 Prove that the average of the numbers n sin n ◦ (n = 2, 4, 6, , 180) is cot 1 ◦

Solution: All arguments of trigonometric functions will be in degrees We need to prove

2 sin 2 + 4 sin 4 + · · · + 178 sin 178 = 90 cot 1, (∗)

which is equivalent to

2 sin 2 · sin 1 + 2(2 sin 4 · sin 1) + · · · + 89(2 sin 178 · sin 1) = 90 cos 1.

Using the identity 2 sin a · sin b = cos(a − b) − cos(a + b), we find

2 sin 2 · sin 1 + 2(2 sin 4 · sin 1) + · · · + 89(2 sin 178 · sin 1)

= (cos 1 − cos 3) + 2(cos 3 − cos 5) + · · · + 89(cos 177 − cos 179)

= cos 1 + cos 3 + cos 5 + · · · + cos 175 + cos 177 − 89 cos 179

= cos 1 + (cos 3 + cos 177) + · · · + (cos 89 + cos 91) − 89 cos 179

= cos 1 + 89 cos 1 = 90 cos 1,

sin a sin(a − b) sin(a − c) sin(a + b) sin(a + c)

= sin c(sin2a − sin2b)(sin2a − sin2c)

and its analogous forms Therefore, it suffices to prove that

x(x2− y2)(x2− z2) + y(y2− z2)(y2− x2) + z(z2− x2)(z2− y2) ≥ 0, where x = sin a, y = sin b, and z = sin c (hence x, y, z > 0) Since the last inequality is symmetric with respect to x, y, z, we may assume that x ≥ y ≥ z > 0 It suffices to prove that

x(y2− x2)(z2− x2) + z(z2− x2)(z2− y2) ≥ y(z2− y2)(y2− x2),

which is evident as

x(y2− x2)(z2− x2) ≥ 0

and

z(z2− x2)(z2− y2) ≥ z(y2− x2)(z2− y2) ≥ y(z2− y2)(y2− x2).

Note: The key step of the proof is an instance of Schur’s Inequality with r = 12

8 Let ABC be a triangle Prove that

First Solution: Let α = A2, β = B2, γ = C2 Then 0◦ < α, β, γ < 90 ◦ and α + β + γ = 90 ◦

By the Difference to Product formulas, we have

sin3C

A − B

2 = −2 sin(γ − α) sin(γ − β).

Hence it suffices to prove that

sin(α − β) sin(α − γ) + sin(β − α) sin(β − γ) + sin(γ − α) sin(γ − β) ≥ 0.

Note that this inequality is symmetric with respect to α, β, γ, so we can assume without loss of

generality that 0◦ < α < β < γ < 90 ◦ Then regrouping the terms on the left-hand side gives

sin(α − β) sin(α − γ) + sin(γ − β)[sin(γ − α) − sin(β − α)], which is positive because the function y = sin x is increasing for 0 ◦ < x < 90 ◦

Note: This proof is similar to that of the Schur’s Inequality.

Second Solution: We keep the notation of the first solution By the Addition formulas,

we have

sin 3α = sin α cos 2α + sin 2α cos α;

cos(β − α) = sin(2α + γ) = sin 2α cos γ + sin γ cos 2α;

cos(β − γ) = sin(2γ + α) = sin 2γ cos α + sin α cos 2γ;

sin 3γ = sin γ cos 2γ + sin 2γ cos γ.

By the Difference to Product formulas, it follows that

sin 3α + sin 3γ − cos(β − α) − cos(β − γ)

= (sin α − sin γ)(cos 2α − cos 2γ) +(cos α − cos γ)(sin 2α − sin 2γ)

= (sin α − sin γ)(cos 2α − cos 2γ) +2(cos α − cos γ) cos(α + γ) sin(α − γ).

Note that sin x is increasing, cos x and cos(2x) are decreasing for 0 < x < 90 ◦ Since 0 < α, γ, α+γ <

90◦, each of the two products in the last addition is less than or equal to 0 Hence

sin 3α + sin 3γ − cos(β − α) − cos(β − γ) ≤ 0.

In exactly the same way, we can show that

sin 3β + sin 3α − cos(γ − β) − cos(γ − α) ≤ 0

and

sin 3γ + sin 3β − cos(α − γ) − cos(α − β) ≤ 0.

Adding the last three inequalities gives the desired result

9 Let a, b, c be positive real numbers Prove that

It suffices to prove that f (a) + f (b) + f (c) ≤ 8 Note that

as desired, with equality if and only if a = b = c.

Second Solution: (By Liang Qin) Setting x = a + b, y = b + c, z = c + a gives 2a + b + c = x + z, hence 2a = x + z − y and their analogous forms The desired inequality becomes

Setting x = a and y = b + c yields

(2a + b + c)2+ 2(a − b − c)2= 3(2a2+ (b + c)2).

and its analogous forms Thus, the desired inequality is equivalent to

(a − b − c)2+ (b − a − c)2+ (c − a − b)2 ≥ a2+ b2+ c2.

Multiplying this out, the left-hand side of the last inequality becomes 3(a2+ b2+ c2) − 2(ab + bc + ca) Therefore the last inequality is equivalent to 2[a2+ b2+ c2 − (ab + bc + ca)] ≥ 0, which is evident

because

2[a2+ b2+ c2− (ab + bc + ca)] = (a − b)2+ (b − c)2+ (c − a)2.

Equalities hold if and only if (b + c)2 = 2(b2+ c2) and (c + a)2 = 2(c2+ a2), that is, a = b = c.

Fourth Solution: We first convert the inequality into

2a(a + 2b + 2c) 2a2+ (b + c)2 + 2b(b + 2c + 2a)

(2a − x)(10a2− 3ax − x2) = (2a − x)2(5a + x) ≥ 0,

which is evident We proved that

4a2− 12a(b + c) + 5(b + c)2

3[2a2+ (b + c)2] ≥ −

4(2a − b − c) 3(a + b + c) , hence (1) follows Equality holds if and only if 2a = b + c, 2b = c + a, 2c = a + b, i.e., when a = b = c.

Fifth Solution: Given a function f of n variables, we define the symmetric sum

where σ runs over all permutations of 1, , n (for a total of n! terms) For example, if n = 3, and

and it suffices to show the the expression in (3) is always greater or equal to 0 By the Weighted

AM-GM Inequality, we have 4a6+ b6+ c6 ≥ 6a4bc, 3a5b + 3a5c + b5a + c5a ≥ 8a4bc, and their

analogous forms Adding those inequalities yields

Note: While the last two methods seem inefficient for this problem, they hold the keys to provingthe following inequality:

where a, b, c are positive real numbers.

10 Let a, b, c be nonnegative real numbers Prove that

a + b + c

3

√ abc ≤ max{( √ a − √ b)2, ( √ b − √ c)2, ( √ c − √ a)2}.

First Solution: We prove the stronger inequality

√

¶2+

µ1

√

¶2+

µ1

¶ µ1

The first of these is the Schur’s inequality with x = a 1/3 , y = b 1/3 , z = c 1/3, while the secondfollows from the AM-GM inequality.

Third Solution: Without loss of generality, assume that b is between a and c The desired

inequality reads

a + b + c − 3 √3

As a function of b, the right side minus the left side is concave (its second derivative is −(2/3)(ac) 1/3 b −5/3),

so its minimum value in the range [a, c] occurs at one of the endpoints Thus, without loss of ality, we may assume a = b Moreover, we may rescale the variables to get a = b = 1 Now the claim

gener-reads

2c + 3c 1/3+ 1

1/2

This is an instance of weighted AM-GM inequality

Note: More generally, for nonnegative real numbers a1, a2, , a n, we have

This follows from the AM-GM inequality

We may also replace m by

m 0 = min

1≤k≤n {( √ a k − √ a k+1)2}

in (2) to obtain, in a way, a sharper lower bound A similar proof works We leave it to the reader

as an exercise

Even more generally, one can ask for a comparison between the difference between the arithmetic and

geometric means of a set of n nonnegative real numbers, and the maximum (or average) difference between the arithmetic and geometric means over all k-element subsets The authors do not know

what the correct inequalities should look like or how they may be proved

11 Let n be a positive integer Prove that

¶−1+

¶−1+

13 Prove that, for all positive real numbers a, b, c,

(a3+ b3+ abc) −1 + (b3+ c3+ abc) −1 + (c3+ a3+ abc) −1 ≤ (abc) −1

Solution: The inequality (a − b)(a2− b2) ≥ 0 implies a3+ b3 ≥ ab(a + b), so

Solution: (by David Shin) Perform the substitutions x = 1/a, y = 1/b, and z = 1/c It

suf-fices to prove that at least two of the inequalities

2x + 3y + 6z ≤ 6, 2y + 3z + 6x ≤ 6, 2z + 3x + 6y ≤ 6

are true, where

x, y, z > 0 and xy + yz + zx ≥ 1.

Assume, for the sake of contradiction, that at least two of the given inequalities are false Without

loss of generality, we may assume that 2x + 3y + 6z < 6 and 2y + 3z + 6x < 6 Then

144 > [(2x + 3y + 6z) + (2y + 3z + 6x)]2

= (8x + 5y + 9z)2

= 64x2+ 80xy + 25y2+ 81z2+ 90yz + 144zx

= 64x2− 64xy + 16y2+ 9y2− 54yz + 81z2

16 Let a0, a1, · · · , a n be numbers from the interval (0, π/2) such that

tan a0tan a1· · · tan a n ≥ n n+1

Solution: Let b k = tan(a k − π/4), k = 0, 1, , n It follows from the hypothesis that for each k,

l=0

(1 − b l)n

!1/n

,

and hence that

´

= tan a k ,

the conclusion follows

17 Let R be the set of real numbers Determine all functions f : R → R such that

f (x2− y2) = xf (x) − yf (y) for all pairs of real numbers x and y.

Solution: Setting x = y = 0 in the given condition yields f (0) = 0 Since

−xf (−x) − yf (y) = f ([−x]2− y2) = f (x2− y2)

= xf (x) − yf (y),

we have f (−x) = −f (x) for x 6= 0 Hence f (x) is odd From now on, we assume x, y ≥ 0.

Setting y = 0 in the given condition yields f (x2) = xf (x) Hence f (x2− y2) = f (x2) − f (y2), or,

f (x2) = f (x2− y2) + f (y2) Since for x ≥ 0 there is a unique t ≥ 0 such that t2 = x, it follows that

for t ≥ 0 Recall that f (x) is odd; we conclude that f (−t) = −f (t) = −tf (1) for t ≥ 0 Hence

f (x) = kx for all x, where k = f (1) is a constant It is not difficult to see that all such functions

indeed satisfy the conditions of the problem

18 Let a, b, and c be nonnegative real numbers such that

of the triples (2, 0, 0), (0, 2, 0), and (0, 0, 2).

Now we prove the upper bound Let us note that at least two of the three numbers a, b, and c are

both greater than or equal to 1 or less than or equal to 1 Without loss of generality, we assume that

the numbers with this property are b and c Then we have

for all nonnegative numbers x, y, z Observe that if z ≤ 1, then both f and g are unbounded, increasing functions of x and y.

Assume that f (a, b, c) = 4 and, without loss of generality, that a ≥ b ≥ c ≥ 0 Then c ≤ 1.

Let a 0 = (a + b)/2 Because a + b = a 0 + a 0 and ab ≤¡a−b

2

¢2

+ ab = a 02 , we have

f (a 0 , a 0 , c) ≤ f (a, b, c) = 4 and g(a 0 , a 0 , c) ≥ g(a, b, c).

Now increase a 0 to e ≥ 0 such that f (e, e, c) = 4 Note that g(e, e, c) ≥ g(a 0 , a 0 , c) It suffices to prove

that g(e, e, c) ≤ 2.

Since f (e, e, c) = 2e2+ c2+ e2c = 4, e2= (4 − c2)/(2 + c) = 2 − c We obtain that

g(e, e, c) = 2ec + (1 − c)e2 ≤ e2+ c2+ (1 − c)e2

Solution: Squaring each equation and subtracting the product of the other two yields

a2− bc = x(x3+ y3+ z3− 3xyz),

b2− ca = y(x3+ y3+ z3− 3xyz),

c2− ab = z(x3+ y3+ z3− 3xyz).

Let k = x3+ y3+ z3− 3xyz Then

(a2− bc)2− (b2− ca)(c2− ab) = k2(x2− yz) = k2a.

The same computation that produced the system above shows tha the expression on the left is

Fix some positive integer n and define a0 = n, and a k = f (a k−1 0 for k ≥ 1 The sequence {a k } k≥0

satisfies the recursive relation

a k = c1+ 2k c2cos(120◦ n) + 2 k c3sin(120◦ n) + 667k

with c1, c2, c3 some real constants

if c2> 0, then a 3(2m+1) will be negative for large m and if c2< 0, then a 6mwill be negative for large

c3 = 0 It follows that a k = c1+ 667k So the first term of the sequence determine all the others Since a0 = n, c1 = n, and hence a k = n + 667k, for all k In particular a1 = f (n) = n + 667, and

hence this is the only possible solution

21 Let a1, , a n and b1, , b n be two sequences of distinct numbers such that a i + b j 6= 0 for all i, j.

then the sum of all n2 numbers c jk is (a1+ · · · + a n ) + (b1+ · · · + b n)

Solution: Let r j =Pn k=1 c j,k Then

Then R(x) = P (x)/Q(x) where Q(x) = (x + b1)(x + b2) · · · (x + b n ) and P (x) has degree at most

n − 1 By (1), R(a1) = R(a2) = · · · = R(a n) = 1, so if we write

Q(x) ,

then S(x) is a monic polynomial of degree n and S(a1) = S(a2) = · · · = S(a n) = 0 Hence

S(x) = (x − a1)(x − a2) · · · (x − a n ).

Consider the coefficient of x n−1 in P (x) = Q(x) − S(x) Form (2), this coefficient is r1+ r2+ · · · + r n

On the other hand,

Q(x) = (x + b1)(x + b2) · · · (x + b n ) and S(x) = (x − a1)(x − a2) · · · (x − a n ).

Applying the Vieta’s theorem, this coefficient is (a1+ a2+ · · · + a n ) + (b1+ b2+ · · · + b n) Hence

we have our desired result

22 Suppose the positive integers have been expressed as a disjoint union of arithmetic progressions

as each power of x appears with coefficient 1 on both sides of the equations Multiplying both sides

But this is just the same as my first sum Indeed, for any denominator d, the number of times 1/d occurs in the first sum is the number of representations d = a b (a, b ≥ 2), while the number of times 1/d occurs in the second sum is the number of representations d = n m (n ∈ S, m ≥ 1) If we write

number of divisors of q So the sums really are the same.

24 Prove that the product of any k consecutive Fibonacci numbers is divisible by the product of the first k Fibonacci numbers.

Solution: Let [k]! = F1F2· · · F k for k ≥ 1 and [0]! = 1, and let R a,b = [a + n]!/([a]![b]!) for

which in turn follows easily from (1) The recursive relation (2), combined with the initial conditions

R a,0 = R 0,b = 1 for all a, b ≥ 0, guarantees that R a,b is always expressible as a polynomial in α and

β with integer coefficients.

Now since β = 1 − α, R a,b can be expressed as a polynomial in α with integer coefficients quently, iteratively applying α n = α n−1 + α n−2 for n ≥ 2, we can write R a,b in the form r + sα, where

integer

25 Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree n with real coefficients is the average of two monic polynomials of degree n with n real roots.

First Solution: (Tiankai Liu) Let us begin with the following lemma

Lemma For any set of n ordered pairs of real numbers {(x i , y i )} n

i=1 , with x i 6= x j for all i 6= j,

i ∈ [1, n].

We present two proofs of the lemma

• First proof By the Lagrange Interpolation Formula, there exists a unique real polynomial

i for all integers i = 1, 2, , n Then we can, and must, have P (x) = Q(x) + x n

• Second proof By the Lagrange Interpolation Formula, there exists a unique real polynomial

and must, have P (x) = Q(x) +Qn i=1 (x − x i)