krit.club logo

Number and Algebra - Deductive proof

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Deductive Proof: A process of demonstrating that a statement is true by using established axioms, definitions, and previously proven theorems in a logical sequence.

•

Direct Proof: A method where you start with a known statement (the hypothesis) and use algebraic manipulation to reach the conclusion. For example, to prove the product of two even numbers is even, you define them as 2n2n and 2m2m and show their product is of the form 2k2k.

•

Proof by Contradiction: A method where you assume the negation of the statement you are trying to prove is true. If this assumption leads to a logical impossibility (a contradiction), the original statement must be true.

•

Proof by Exhaustion: A method where a statement is proven by testing every possible case within the given domain. This is only practical for finite and small sets.

•

Counter-example: A single case that proves a general statement is false. To disprove a statement like 'All prime numbers are odd', one only needs to point to the number 22.

•

Standard Definitions: In proofs, integers are often defined as: Even (2k2k), Odd (2k+12k + 1), and Rational (pq\frac{p}{q} where p,q∈Z,q≠0p, q \in \mathbb{Z}, q \neq 0).

📐Formulae

n=2k,k∈Z  ⟹  n is evenn = 2k, k \in \mathbb{Z} \implies n \text{ is even}

n=2k+1,k∈Z  ⟹  n is oddn = 2k + 1, k \in \mathbb{Z} \implies n \text{ is odd}

x∈Q  ⟺  x=ab, where a,b∈Z,b≠0x \in \mathbb{Q} \iff x = \frac{a}{b}, \text{ where } a, b \in \mathbb{Z}, b \neq 0

Consecutive Integers: n,n+1,n+2,…\text{Consecutive Integers: } n, n+1, n+2, \dots

💡Examples

Problem 1:

Prove that the square of any odd integer is also an odd integer.

Solution:

Let nn be an odd integer. By definition, n=2k+1n = 2k + 1 for some integer k∈Zk \in \mathbb{Z}. Squaring both sides: n2=(2k+1)2n^2 = (2k + 1)^2 n2=4k2+4k+1n^2 = 4k^2 + 4k + 1 n2=2(2k2+2k)+1n^2 = 2(2k^2 + 2k) + 1 Let m=2k2+2km = 2k^2 + 2k. Since kk is an integer, mm is also an integer. n2=2m+1n^2 = 2m + 1. Therefore, n2n^2 is odd.

Explanation:

This is a direct proof. We used the algebraic definition of an odd number and expanded the square to show the result still fits the definition of an odd number (2m+12m + 1).

Problem 2:

Prove by contradiction that if n2n^2 is even, then nn is even.

Solution:

Assume the opposite: Suppose nn is odd. If nn is odd, then n=2k+1n = 2k + 1 for some k∈Zk \in \mathbb{Z}. n2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1. This implies n2n^2 is odd. However, this contradicts the given premise that n2n^2 is even. Since the assumption that nn is odd leads to a contradiction, nn must be even.

Explanation:

In proof by contradiction, we show that the negation of the conclusion leads to a statement that conflicts with our given information.

Problem 3:

Disprove the statement: 'For all n∈Z+n \in \mathbb{Z}^+, n2+n+11n^2 + n + 11 is a prime number.'

Solution:

To disprove this, we look for a counter-example. Let n=11n = 11. n2+n+11=112+11+11n^2 + n + 11 = 11^2 + 11 + 11 121+11+11=143121 + 11 + 11 = 143 Since 143=11×13143 = 11 \times 13, it is not a prime number. Therefore, the statement is false.

Explanation:

A counter-example is the most efficient way to disprove a universal statement. Here, setting nn to be a multiple of the constant term often reveals a composite number.