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Chapter 3: Problem 54

The difference of the squares of any two consecutive integers is odd.

Short Answer

Expert verified
The difference of the squares of any two consecutive integers is odd. We denoted the first integer as 'n' and the consecutive integer as '(n+1)'. We computed the squares of both integers and calculated the difference as \((n+1)^2 - n^2\). After simplifying the expression, we got '2n + 1', which is an odd integer. Hence, the statement is proven to be true.

Step by step solution

01

Define the consecutive integers.

Let's denote the first integer as 'n', then the consecutive integer will be '(n+1)'. These are two consecutive integers, where 'n' is an arbitrary integer.
02

Compute the squares of the integers.

We need to compute the squares of both integers. The square of the first integer 'n' will be \(n^2\), and the square of the consecutive integer '(n+1)' will be \((n+1)^2\).
03

Compute the difference of the squares.

Now, we need to calculate the difference between the squares of these consecutive integers. The difference will be \((n+1)^2 - n^2\).
04

Simplify the expression.

Let's simplify the expression we got in step 3. We have: \[ \begin{aligned} (n+1)^2 - n^2 &= (n^2+2n+1)-n^2 \\ &= (n^2 - n^2) + 2n + 1 \\ &= 2n + 1 \end{aligned} \]
05

Analyze the result.

Observe that the result, '2n+1', is the product of an even integer '2n' and an odd integer '1', which results in an even integer. By adding 1 to an even integer, we always obtain an odd integer. Therefore, the difference of the squares of any two consecutive integers is odd. So, we have successfully proved that the difference of the squares of any two consecutive integers is odd.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Consecutive Integers
In mathematics, consecutive integers are numbers that follow each other in order, without any gaps. Simply put, they are numbers that come one after another. When we define consecutive integers, we often start with a variable, say, "n".
For example, if "n" represents your first integer, then "n+1" would naturally follow as the next consecutive integer.
  • If "n" is 3, then the consecutive integer is 4.
  • If "n" is 10, then the consecutive integer is 11.
Consecutive integers are useful in various mathematical concepts because of their straightforward and predictable nature. This makes them particularly handy when proving theorems or exploring patterns, such as calculating the difference of their squares.
Difference of Squares
The difference of squares is a significant concept in mathematics. It describes the result of subtracting the square of one number from the square of another number. A specific interesting case is the difference of squares of consecutive integers.
When dealing with two consecutive integers, "n" and "n+1", the formula for their difference of squares becomes:
\[(n+1)^2 - n^2\]
Upon expanding this, it simplifies to\[2n + 1\]
This shows a neat property: the outcome changes at every integer step, which is crucial in understanding patterns generated by numbers.
  • The term "2n" always produces an even result, as multiplying any integer by 2 yields an even number.
  • Adding 1 makes the entire expression odd, ensuring that every difference of squares of consecutive integers is odd.
Mathematical Proof
A mathematical proof is a logical process used to demonstrate the truth of a statement beyond any doubt. When proving the property about the difference of squares of consecutive integers, we use algebraic manipulation to reach our conclusion.
The goal here is to show the expression \[(n+1)^2 - n^2\]
always results in an odd number, no matter what integer "n" is chosen.
Here's the proof in brief steps:
  • Define your integers, "n" and "n+1".
  • Calculate their squares: "n^2" and "(n+1)^2."
  • Subtract the first square from the second, resulting in:\[(n^2+2n+1) - n^2 = 2n + 1\]
  • The term "2n" being even added to 1 is always odd; hence, the statement is true.
Such proof is essential because it stands as a universal statement across all integer values. This approach is a cornerstone of discrete mathematics, allowing us to confirm patterns and truths in a systematic way.

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Most popular questions from this chapter

Observe that $$ \begin{aligned} 7524 &=7 \cdot 1000+5 \cdot 100+2 \cdot 10+4 \\ &=7(999+1)+5(99+1)+2(9+1)+4 \\ &=(7 \cdot 999+7)+(5 \cdot 99+5)+(2 \cdot 9+2)+4 \\ &=(7 \cdot 999+5 \cdot 99+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111 \cdot 9+5 \cdot 11 \cdot 9+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111+5 \cdot 11+2) \cdot 9+(7+5+2+4) \\ &=(\text { an integer divisible by } 9) \end{aligned} $$ \(\begin{aligned} 7524 &=7 \cdot 1000+5 \cdot 100+2 \cdot 10+4 \\\ &=7(999+1)+5(99+1)+2(9+1)+4 \\ &=(7.999+7)+(5 \cdot 99+5)+(2 \cdot 9+2)+4 \\\ &=(7 \cdot 999+5 \cdot 99+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111 \cdot 9+5 \cdot 11 \cdot 9+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111+5 \cdot 11+2) \cdot 9+(7+5+2+4) \\ &=(\text { an integer divisible by } 9) \\ &+(\text { the sum of the digits of } 7524) \end{aligned}\) Since the sum of the digits of 7524 is divisible by 9,7524 can be written as a sum of two integers each of which is divisible by 9 . It follows from exercise 15 that 7524 is divisible by \(9 .\) Generalize the argument given in this example to any nonnegative integer \(n\). In other words, prove that for any nonnegative integer \(n\), if the sum of the digits of \(n\) is divisible by 9 , then \(n\) is divisible by 9 ,

Assume that \(m\) and \(n\) are particular integers. \(\begin{array}{ll}\text { a. Is } 6 m+8 n \text { even? } & \text { b. Is } 10 m n+7 \text { odd? }\end{array}\) c. If \(m>n>0\), is \(m^{2}-n^{2}\) composite?

If \(0=\) Sunday, \(1=\) Monday, \(2=\) Tuesday, \(\ldots, 6=\) Saturday, then January 1 of year \(n\) occurs on the day of the week given by the following formula: \(\left(n+\left\lfloor\frac{n-1}{4}\right\rfloor-\left\lfloor\frac{n-1}{100}\right\rfloor+\left\lfloor\frac{n-1}{400}\right\rfloor\right) \bmod 7\). a. Use this formula to find January 1 of \(\begin{array}{lll}\text { i. } 2050 & \text { ii. } 2100 & \text { iii. the year of your birth. }\end{array}\)

Prove that for all real numbers \(c\), if \(c\) is a root of a polynomial with rational coefficients, then \(c\) is a root of a polynomial with integer coefficients.

The difference of any two odd integers is even.
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