Proofs in IB Math AA SL: Comprehensive Notes with 30 Examples

Solution (Direct Proof):

1. Let the two even integers be \(2m\) and \(2n\), where \(m\) and \(n\) are integers.
2. The sum is \(2m + 2n = 2(m + n)\).
3. Since \((m+n)\) is an integer, the sum is of the form \(2(\text{integer})\), which is even.

Hence proved: The sum of any two even integers is even.

Solution (Direct Proof):

1. Let the even integer be \(2a\). Let the other integer be \(b\).
2. The product is \((2a) \cdot b = 2(ab)\).
3. Since \(ab\) is an integer, the product is a multiple of 2, thus even.

Conclusion: The product of an even integer and any integer is even.

Solution (Direct Proof):

1. Let \(n = 3k\) for some integer \(k\). Thus \(n\) is a multiple of 3.
2. Compute \(n^2 = (3k)^2 = 9k^2 = 3(3k^2)\).
3. Clearly, \(3k^2\) is an integer, so \(n^2\) is a multiple of 3.

Conclusion: If \(n\) is a multiple of 3, then \(n^2\) is also a multiple of 3.

Solution (Direct Proof in Geometry):

1. Consider triangle \(ABC\). Draw a line through \(A\) parallel to side \(BC\).
2. Let the angles at \(B\) and \(C\) be \(\angle ABC\) and \(\angle ACB\). By the alternate interior angles theorem, the angle formed at \(A\) on the parallel line corresponds to \(\angle ABC\) and \(\angle ACB\).
3. The sum of angles on a straight line is \(180^\circ\). Hence, \(\angle BAC + \angle ABC + \angle ACB = 180^\circ\).

Conclusion: The interior angles of a triangle sum to \(180^\circ\).

Solution (Direct Proof):

1. Let \(n\) be an odd integer. Then \(n = 2k + 1\) for some integer \(k\).
2. \(n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1\).
3. Since \((2k^2 + 2k)\) is an integer, \(n^2\) is of the form \((2m + 1)\), which is odd.

Hence: The square of an odd integer remains odd.

Solution (Partial Induction - Base Case):

1. Base Case: For \(n=1\), LHS \(= 1\). RHS \(= \frac{1\cdot(1+1)}{2} = 1\). So it holds for \(n=1\).
2. The full induction would then assume it is true for \(n=k\) and prove for \(n=k+1\). (Not shown here in detail.)

Note: This is the classic formula for the sum of the first \(n\) natural numbers.

Solution (Direct Checking):

1. We want all positive integers \(d\) such that \(10 = d \times k\) for some integer \(k\).
2. Check systematically: \(1 \times 10\), \(2 \times 5\). No other positive integer pairs yield 10.
3. Hence the factors are \(1, 2, 5, 10\). That’s exactly 4 factors.

Conclusion: 10 has exactly 4 positive divisors.

Solution (Algebraic Manipulation):

1. Start with the RHS: \((x - y)(x + y) = x^2 + xy - xy - y^2\).
2. This simplifies to \(x^2 - y^2\).
3. Hence \(x^2 - y^2 = (x - y)(x + y)\).

Conclusion: The identity is proven by direct expansion.

Solution (Direct Definition):

1. \(\sqrt{4}\) is defined as the positive number which, when squared, equals 4.
2. Check \(2^2 = 4\). Since 2 is positive, \(\sqrt{4} = 2\).

Conclusion: By definition of principal square root, \(\sqrt{4} = 2\).

Solution (Symmetry of Definition):

1. \(\gcd(m, n)\) is the greatest common divisor of \(m\) and \(n\).
2. By definition, it is the largest integer that divides both \(m\) and \(n\).
3. Whether we say “\(\gcd(m, n)\)” or “\(\gcd(n, m)\),” the pairs of integers are the same, so the same integer divides them.
4. Hence \(\gcd(m, n) = \gcd(n, m)\).

Conclusion: The gcd function is commutative in its arguments.

Solution (Contraposition or Direct Proof Approach):

1. We can do this via contrapositive: “If \(n\) is not even (i.e., \(n\) is odd), then \(n^2\) is not even (i.e., odd).”
2. Let \(n = 2k + 1\). Then \(n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1\) which is odd.
3. Thus \(\neg Q \implies \neg P\). Hence \(n^2\) being even (\(P\)) forces \(n\) to be even (\(Q\)).

Conclusion: If \(n^2\) is even, \(n\) must be even.

Solution (Number Theory / Euclidean Algorithm):

1. Recall that \(\gcd(a, b)\) is the largest integer that divides both \(a\) and \(b\).
2. If \(d\) divides both \(a\) and \(b\), then it also divides \((b-a)\). Likewise, if \(d\) divides \(a\) and \((b-a)\), it also divides \(b\). Thus the sets of common divisors of \((a, b)\) and \((a, b-a)\) are the same.
3. Hence \(\gcd(a, b) = \gcd(a, b-a)\). This fact underpins the Euclidean Algorithm for gcd.

Conclusion: The gcd does not change if we replace \(b\) by \(b-a\).

Solution (Combinatorial Argument):

1. Definition: \(\displaystyle \binom{n}{k} = \frac{n!}{k!(n-k)!}\).
2. Then \(\displaystyle \binom{n}{n-k} = \frac{n!}{(n-k)!(n-(n-k))!} = \frac{n!}{(n-k)!k!}\).
3. Observe that \(\frac{n!}{k!(n-k)!} = \frac{n!}{(n-k)!k!}\). Hence they are equal.

Alternatively: You can argue it by choosing \(k\) objects out of \(n\) is equivalent to excluding \((n-k)\) objects from \(n\).

Solution (Contradiction Outline):

1. Assume there are three distinct solutions: \(x_1, x_2, x_3\).
2. A quadratic polynomial can be factored (if solutions exist) as \(a(x - x_1)(x - x_2)\). This leaves no room for a third distinct root \(x_3\), because a quadratic is degree 2.
3. Hence having three solutions leads to contradiction. So the equation can have at most two solutions.

Conclusion: No quadratic with real coefficients has more than two distinct real solutions.

Solution (Mathematical Induction):

1. Base Case (n=1): LHS = 1, RHS = \(\frac{1\cdot(1+1)}{2} = 1\). Holds.
2. Inductive Step: Assume it holds for \(n=k\), i.e., \(1 + 2 + \cdots + k = \frac{k(k+1)}{2}\). We need to show it holds for \(n=k+1\).
3. LHS for \(k+1\) is \((1 + 2 + \cdots + k) + (k+1)\). By induction hypothesis, that is \(\frac{k(k+1)}{2} + (k+1)\).
4. Factor out \((k+1)\): \(\frac{k(k+1)}{2} + (k+1) = (k+1)\left(\frac{k}{2} + 1\right) = \frac{(k+1)(k+2)}{2}.\)
5. Thus the formula holds for \(k+1\). By induction, it is true for all \(n\).

Solution (Proof by Induction - Outline):

1. Base Case: For \(n=1\), \(2^1 = 2\) which is greater than \(1\). So it holds.
2. Inductive Step: Assume \(2^k > k\) for some \(k\ge 1\). Then for \(k+1\): \[ 2^{k+1} = 2 \cdot 2^k > 2k \ge k+1 \quad \text{(for }k\ge 1\text{)}. \]
3. Thus \(2^{k+1} > k+1\). By induction, \(2^n > n\) for all \(n\ge 1\).

Solution (Classical Geometry):

1. Let triangle \(ABC\) have \(AB = AC\). We want to show \(\angle ABC = \angle ACB\).
2. Draw the angle bisector or altitude from \(A\) if necessary, or use congruent triangles (\(SAS\) criterion) by considering an appropriate reflection.
3. Alternatively, by standard geometry theorems, if two sides of a triangle are congruent, the angles opposite those sides are congruent.

Conclusion: The base angles of an isosceles triangle are equal.

Solution (Algebraic Factorization):

1. Expand \((x-y)(x^2 + xy + y^2)\): \[ (x)(x^2 + xy + y^2) - (y)(x^2 + xy + y^2) \] \[ = x^3 + x^2y + xy^2 - (yx^2 + y^2x + y^3) \] \[ = x^3 + x^2y + xy^2 - yx^2 - y^2x - y^3 = x^3 - y^3. \]

Conclusion: The identity is verified by expansion.

Solution (Contradiction Sketch):

1. Assume \(0.999\ldots \neq 1\). Let \(x = 0.999\ldots\).
2. Then \(10x = 9.999\ldots\). Subtracting \(x\) from \(10x\), we get \(9x = 9\), so \(x=1\).
3. This contradicts the assumption that \(x \neq 1\). Hence \(0.999\ldots = 1\).

Solution (Contraposition or Fundamental Theorem of Arithmetic):

1. Contraposition approach: “If neither integer is divisible by \(p\), then their product is not divisible by \(p\).”
2. If neither integer is divisible by \(p\), then \(p\) does not appear in the prime factorization of either integer. Hence it cannot appear in the prime factorization of their product, so the product is not divisible by \(p\).
3. Therefore, if the product is divisible by \(p\), at least one of the factors must be divisible by \(p\).

Solution (Contradiction):

1. Assume \(\sqrt{2}\) is rational. Then \(\sqrt{2} = \frac{a}{b}\) for some integers \(a, b\) with no common factors (and \(b\neq 0\)).
2. Squaring: \(2 = \frac{a^2}{b^2}\implies a^2 = 2b^2\). Thus \(a^2\) is even, implying \(a\) is even.
3. Let \(a=2k\). Then \(a^2 = 4k^2\). So \(4k^2=2b^2\implies b^2 = 2k^2\). Now \(b^2\) is even, so \(b\) is even.
4. Hence both \(a\) and \(b\) are even, contradicting the assumption that \(\frac{a}{b}\) was in lowest terms.
5. Therefore, \(\sqrt{2}\) is irrational.

Solution (Contradiction - Euclid’s Argument):

1. Assume there are finitely many primes: \(p_1, p_2, \ldots, p_n\).
2. Consider \(N = p_1 p_2 \cdots p_n + 1\). \(N\) is greater than any of these primes.
3. None of the primes \(p_i\) can divide \(N\) evenly (because dividing \(N\) by any \(p_i\) leaves a remainder of 1).
4. Hence \(N\) is either prime itself or has a prime factor not in the original list, contradicting the assumption of a complete prime set.
5. Therefore, there must be infinitely many primes.

Solution (Mathematical Induction):

1. Base Case (n=1): LHS = \(1^3 = 1\). RHS = \(\left(\frac{1\cdot(1+1)}{2}\right)^2 = 1\). OK.
2. Inductive Step: Assume true for \(n=k\): \(1^3 + 2^3 + \cdots + k^3 = \left(\frac{k(k+1)}{2}\right)^2.\)
3. Then for \(n = k+1\): \[ 1^3 + 2^3 + \cdots + k^3 + (k+1)^3. \] Using the induction hypothesis, this becomes \[ \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3. \]
4. Factor out \((k+1)^2\): \[ = (k+1)^2 \left[\frac{k^2}{4} + (k+1)\right] = (k+1)^2 \left[\frac{k^2 + 4k + 4}{4}\right] = (k+1)^2 \frac{(k+2)^2}{4} \] This simplifies to \[ \left(\frac{(k+1)(k+2)}{2}\right)^2. \]
5. Hence the formula is true for \(k+1\). By induction, the identity holds for all \(n\).

Solution (Number Theory - Divisor Logic):

1. Given \(\gcd(a, b) = 1\), no prime factors are common in \(a\) and \(b\).
2. Now, \(\gcd(a, bc) = 1 \iff\) no prime divides both \(a\) and \(bc\).
3. But if a prime divides \(a\) and \(bc\), it must divide either \(b\) or \(c\) (by Euclid's Lemma, since \(p\) is prime). Since it can’t divide \(b\) (as \(\gcd(a,b)=1\) means \(a\) and \(b\) share no prime factors), it must divide \(c\).
4. Hence \(\gcd(a, bc) = 1 \iff \gcd(a, c) = 1\). This is a standard property used in proof of unique factorization.

Solution (Trigonometric Identity):

1. Start with LHS: \(\cos^4 x + \sin^4 x\).
2. Recognize this as \( (\cos^2 x)^2 + (\sin^2 x)^2 \). We can complete the square: \(A^2+B^2 = (A+B)^2 - 2AB\).
3. Let \(A=\cos^2 x\) and \(B=\sin^2 x\). Then \[ \cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\cos^2 x \sin^2 x. \]
4. But \(\cos^2 x + \sin^2 x = 1\). So \[ (\cos^2 x + \sin^2 x)^2 = 1^2 = 1. \]
5. Hence \[ \cos^4 x + \sin^4 x = 1 - 2\sin^2 x \cos^2 x. \]

Conclusion: The identity is proven.

Solution (Sketch of Proof by Induction on \(a\) or Binomial Theorem):

We prove this by induction on \(a \ge 0\). The result extends to negative \(a\).

1. Base Case (\(a=0\)): \(0^p - 0 = 0\), which is divisible by \(p\). Holds.
2. Inductive Step: Assume \(k^p - k\) is divisible by \(p\) for some integer \(k \ge 0\). We want to show \((k+1)^p - (k+1)\) is divisible by \(p\).
   Consider \((k+1)^p - (k+1)\). By the Binomial Theorem: \[ (k+1)^p = k^p + \binom{p}{1}k^{p-1} + \binom{p}{2}k^{p-2} + \cdots + \binom{p}{p-1}k + 1. \] So, \[ (k+1)^p - (k+1) = (k^p + \binom{p}{1}k^{p-1} + \cdots + \binom{p}{p-1}k + 1) - (k+1) \] \[ = (k^p - k) + \left[ \binom{p}{1}k^{p-1} + \binom{p}{2}k^{p-2} + \cdots + \binom{p}{p-1}k \right]. \]
3. For a prime \(p\), all binomial coefficients \(\binom{p}{j}\) for \(1 \le j \le p-1\) are multiples of \(p\). This is because \(p\) divides \(p!\) but does not divide \(j!\) or \((p-j)!\) (since \(j, p-j < p\)).
   Thus, the term in the square brackets is divisible by \(p\).
4. By the inductive hypothesis, \((k^p - k)\) is divisible by \(p\).
   Therefore, \((k+1)^p - (k+1)\) is the sum of two terms divisible by \(p\), so it is also divisible by \(p\).
5. By the principle of mathematical induction, \(a^p - a\) is divisible by \(p\) for all integers \(a \ge 0\). (The argument can be extended for \(a < 0\).)

Solution (Circle Theorems):

1. Let \(ABCD\) be a cyclic quadrilateral inscribed in a circle with center \(O\).
2. Consider opposite angles \(\angle A\) and \(\angle C\).
3. The angle subtended by an arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle.
4. Let arc \(BCD\) subtend angle \(\alpha\) at the center \(O\), and angle \(\angle A\) at the circumference. So \(\alpha = 2 \angle A\).
5. Let arc \(DAB\) subtend angle \(\beta\) at the center \(O\), and angle \(\angle C\) at the circumference. So \(\beta = 2 \angle C\).
6. The sum of angles around the center \(O\) is \(360^\circ\), so \(\alpha + \beta = 360^\circ\).
7. Substituting, \(2\angle A + 2\angle C = 360^\circ\), which implies \(\angle A + \angle C = 180^\circ\).
8. Similarly, \(\angle B + \angle D = 180^\circ\).

Conclusion: Opposite angles of a cyclic quadrilateral are supplementary.

Solution (Geometry / Contradiction):

1. Consider a right-angled triangle \(\triangle ABC\) with the right angle at \(C\). Then \(AB\) is the hypotenuse, and \(AC\) and \(BC\) are the legs.
2. Assume, for the sake of contradiction, that the hypotenuse \(AB\) is NOT the longest side. This means either:
   • Case 1: \(AC \ge AB\), or
   • Case 2: \(BC \ge AB\).
3. Case 1: Assume \(AC \ge AB\).
   By Pythagoras' Theorem, \(AB^2 = AC^2 + BC^2\).
   Since \(BC\) is a length, \(BC > 0\), so \(BC^2 > 0\).
   Therefore, \(AC^2 + BC^2 > AC^2\).
   This implies \(AB^2 > AC^2\). Taking the square root of positive quantities, \(AB > AC\).
   This contradicts our assumption that \(AC \ge AB\).
4. Case 2: Assume \(BC \ge AB\).
   By Pythagoras' Theorem, \(AB^2 = AC^2 + BC^2\).
   Since \(AC\) is a length, \(AC > 0\), so \(AC^2 > 0\).
   Therefore, \(AC^2 + BC^2 > BC^2\).
   This implies \(AB^2 > BC^2\). Taking the square root, \(AB > BC\).
   This contradicts our assumption that \(BC \ge AB\).
5. Since both assumptions lead to a contradiction, our initial premise that the hypotenuse is not the longest side must be false.

Conclusion: Therefore, in a right-angled triangle, the hypotenuse is the longest side.

Solution (Telescoping Series / Induction):

Let \(P(n)\) be the statement \(\sum_{i=1}^{n} \frac{1}{i(i+1)} = 1 - \frac{1}{n+1}\).

1. Base Case (\(n=1\)):
   LHS = \(\frac{1}{1(1+1)} = \frac{1}{1\cdot 2} = \frac{1}{2}\).
   RHS = \(1 - \frac{1}{1+1} = 1 - \frac{1}{2} = \frac{1}{2}\).
   Since LHS = RHS, \(P(1)\) is true.
2. Inductive Step: Assume \(P(k)\) is true for some integer \(k\ge 1\).
   That is, assume \(\sum_{i=1}^{k} \frac{1}{i(i+1)} = 1 - \frac{1}{k+1}\).
   We need to show that \(P(k+1)\) is true, i.e., \(\sum_{i=1}^{k+1} \frac{1}{i(i+1)} = 1 - \frac{1}{(k+1)+1} = 1 - \frac{1}{k+2}\).
   Consider the LHS for \(P(k+1)\): \[ \sum_{i=1}^{k+1} \frac{1}{i(i+1)} = \left(\sum_{i=1}^{k} \frac{1}{i(i+1)}\right) + \frac{1}{(k+1)((k+1)+1)} \] \[ = \left(\sum_{i=1}^{k} \frac{1}{i(i+1)}\right) + \frac{1}{(k+1)(k+2)} \] Using the inductive hypothesis \(P(k)\): \[ = \left(1 - \frac{1}{k+1}\right) + \frac{1}{(k+1)(k+2)} \] Now, we combine the fractions: \[ = 1 - \left(\frac{1}{k+1} - \frac{1}{(k+1)(k+2)}\right) \] \[ = 1 - \left(\frac{k+2}{(k+1)(k+2)} - \frac{1}{(k+1)(k+2)}\right) \] \[ = 1 - \left(\frac{k+2-1}{(k+1)(k+2)}\right) = 1 - \left(\frac{k+1}{(k+1)(k+2)}\right) \] \[ = 1 - \frac{1}{k+2}. \] This is the RHS for \(P(k+1)\).
3. Thus, if \(P(k)\) is true, then \(P(k+1)\) is true. Since \(P(1)\) is true, by the principle of mathematical induction, \(P(n)\) is true for all integers \(n\ge 1\).

(Note: This series can also be proven directly using partial fractions: \(\frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1}\), which leads to a telescoping sum.)

Solution (Trigonometric Identity Manipulations):

Let \(P = \cos(20^\circ)\cos(40^\circ)\cos(60^\circ)\cos(80^\circ)\).

1. We know \(\cos(60^\circ) = \frac{1}{2}\).
   So, \(P = \frac{1}{2} \cos(20^\circ)\cos(40^\circ)\cos(80^\circ)\).
2. Multiply and divide by \(2\sin(20^\circ)\) (assuming \(\sin(20^\circ) \neq 0\), which is true): \[ P = \frac{1}{2} \cdot \frac{2\sin(20^\circ)\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)}{2\sin(20^\circ)} \] Using \(2\sin A \cos A = \sin(2A)\): \[ P = \frac{1}{2} \cdot \frac{\sin(40^\circ)\cos(40^\circ)\cos(80^\circ)}{2\sin(20^\circ)} \]
3. Apply the identity again: Multiply numerator and denominator by 2. \[ P = \frac{1}{2} \cdot \frac{2\sin(40^\circ)\cos(40^\circ)\cos(80^\circ)}{2 \cdot 2\sin(20^\circ)} = \frac{\sin(80^\circ)\cos(80^\circ)}{8\sin(20^\circ)} \]
4. Apply the identity one more time: Multiply numerator and denominator by 2. \[ P = \frac{2\sin(80^\circ)\cos(80^\circ)}{2 \cdot 8\sin(20^\circ)} = \frac{\sin(160^\circ)}{16\sin(20^\circ)} \]
5. We know that \(\sin(180^\circ - x) = \sin(x)\).
   So, \(\sin(160^\circ) = \sin(180^\circ - 20^\circ) = \sin(20^\circ)\). \[ P = \frac{\sin(20^\circ)}{16\sin(20^\circ)} \]
6. Since \(\sin(20^\circ) \neq 0\), we can cancel it: \[ P = \frac{1}{16}. \]

Conclusion: The exact product is \(\frac{1}{16}\).