Problem 16 - Precalculus: Mathematical for Calculus

Question

Show that \(n^{2}-n+41\) is odd for all natural numbers \(n\)

Step-by-Step Solution

Answer

The expression \(n^2 - n + 41\) is always odd for any natural number \(n\) because \(n^2 - n\) is even and 41 is odd.

Step 1: Understand what needs to be proven

We need to prove that for every natural number \(n\), the expression \(n^2-n+41\) results in an odd number.

Step 2: Analyze odd and even properties

Recall that an odd number is any integer of the form \(2k+1\) where \(k\) is an integer. An even number is of the form \(2k\). So, our goal is to show that \(n^2 - n + 41\) takes the form \(2k+1\) for any natural number \(n\).

Step 3: Check parity of terms

Consider the expression \(n^2 - n\). If \(n\) is even, then \(n^2\) and \(n\) are both even, making \(n^2-n\) even. If \(n\) is odd, then both \(n^2\) and \(n\) are odd, meaning \(n^2-n\) is also even. Thus, \(n^2-n\) is always even.

Step 4: Express the formula as an odd number

Since \(n^2 - n\) is always even, let's denote it as \(2k\) for some integer \(k\). Thus, \(n^2 - n + 41 = 2k + 41\).

Step 5: Verify the final sum

The number 41 is odd (as it is of the form \(2m+1\)). Therefore, \(2k + 41 = 2k + (2m + 1) = 2(k + m) + 1\), which is of the form \(2x + 1\) for integer \(x = k + m\). This shows that \(n^2-n+41\) is always odd.

Key Concepts

Natural Numbers

Natural numbers are the basic building blocks of mathematics. They are the set of positive integers starting from 1, and include every increasing whole number, like 2, 3, 4, and so on. These numbers are infinite and are often used to count things.

• The smallest natural number is 1.
• There is no largest natural number, as you can always add 1 to any natural number to get another natural number.
• Natural numbers do not include fractions, decimals, or negative numbers.

Natural numbers are crucial in many mathematical concepts, especially in sequences and series. In our exercise, we're dealing with these numbers when trying to prove a property of a quadratic expression. Each value of \(n\) we consider is a natural number.

Parity

Parity refers to whether a number is odd or even. It's an important way to categorize numbers and is determined based on how they divide by 2.

• An even number has a parity of 0 and can be expressed as \(2k\), where \(k\) is an integer.
• An odd number has a parity of 1 and can be expressed as \(2k+1\).

In the given problem, understanding parity helps us determine whether the formula \(n^2 - n + 41\) is odd or even. By looking at the parity of \(n\) and its influence on \(n^2\) and \(n\), we observed that \(n^2-n\) is always even, leading us to conclude the result for each natural number \(n\).

Quadratic Expressions

A quadratic expression is a type of polynomial that involves variables raised to the power of two. The general form is \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable.

• The expression can represent various mathematical structures, such as parabolas when graphed.
• Quadratic expressions are essential for solving problems that involve trajectories or areas.
• A complete quadratic expression can often be simplified or solved for zero by factoring, using the quadratic formula, or completing the square.

In our exercise, the expression \(n^2 - n + 41\) is a specific type of quadratic that interestingly always results in an odd number for every natural number \(n\). This peculiar trait makes it a fascinating example of the interplay between algebra and number theory, showing how simple expressions can have consistent properties.