Does that pattern hold for all natural numbers? Could we claim that n² = n?

Yes, we can, and I’ve written a proof to prove it! The proof uses a valuable concept in mathematics called induction. I remember being introduced to proofs by induction when I was in Junior High. Nowadays, if it is not part of Common Core, it wouldn’t be taught much anymore. Nevertheless, I will use it here to prove that n² = n.

Using a similar proof, we can also prove that n³ = n, n⁴ = n, n⁵ = n, n⁶ = n, and so forth!

Today is the perfect day to review how to use proof by induction so try your hand at proving at least one of those mathematical statements on your own. Use the same steps in my example: prove true for n=1, assume true for n = k, prove true for k + 1, write your conclusion. then have a very Happy April Fools’ Day, Everyone!

Today is also a very good day to review that (x + y)² = x² +2xy + y² and NOT x² + y², a very common error students make. Confession: I remember making that exact error in high school when I definitely should have known better. Using induction to prove something in mathematics is a valid technique, but if you use invalid equations like
(x + y)³ = x³ + y³, you will make invalid conclusions. Thus, today might also be a good day to review the binomial theorem and Pascal’s triangle. (Pascal’s triangle has numbers in its interior, not just 1’s going down the sides, after all.)

My post today was inspired by a post written by Sara Van Der Werf titled Why I’ve Started Teaching the FOIL Method Again. In her post, she not only plays a great April Fools’ joke on her readers, but she explains a tried and true way to multiply binomials and other polynomials.

I read her post exactly one year ago today, and since then, I have been waiting for April Fools’ Day to roll around again so that I could share this post with you. It is my hope that you will enjoy my little prank and learn a little mathematics from it as well.

In elementary school, we learned about improper fractions. Should we call them that? Is it even possible for any kind of number to be IMPROPER? They are simply fractions greater than one. I’ve recently heard the term fraction form used, and ever since I’ve made a point of saying that fractions greater than one are in fraction form.

On Twitter, I’ve found a few people who also don’t like using the word improper to describe any fraction.

This first tweet has a link explaining why it is improper to use the term improper fraction:

Whether it is an improper fraction or mixed number, terminology in maths matters just as much as it does in English, writes Kevin O’Brien https://t.co/B9ZuQThUqf

I always wanted to analyze the "behavior" of any fraction that was called improper! Should this fraction receive some sort of penalty for their deeds? Seriously, knowing their equivalence and when one form may be more appropriate use-wise, is the issue: whether 5/4; 1 1/4 or 1.25

In my 3rd grade class we had a conversation about the term "improper" and how it doesn't fit the fraction. The kids all agreed that fractions can have many different representations and there's nothing "improper" about that. They were super cute.

Agreed. Lots of kids don't think improper fractions are fractions. So call them what they are, fractions greater than one. The term "improper fraction" hides information and adds negative value.

Then again, my students come to me already afraid of improper fractions. Of course my preference would be to *not* have to spend time undoing that damage.

Does the term ‘improper fraction’ lead to misunderstanding?Does it suggest that a /real/ fraction is less than 1?My goal is to use the term ‘rename’ rather than ‘convert’. We aren’t changing anything but the way it looks. #TVDSBmathpic.twitter.com/jlKKx8uN7l

I hope that you will consider not labeling any fraction as improper, as well!

Now I’ll write a little bit about the number 1366:

1366 is a composite number.

Prime factorization: 1366 = 2 × 683

The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1366 has exactly 4 factors.

Factors of 1366: 1, 2, 683, 1366

Factor pairs: 1366 = 1 × 1366 or 2 × 683

1366 has no square factors that allow its square root to be simplified. √1366 ≈ 36.95944

1366 is also the sum of the twenty-six prime number from 5 to 107. Do you know what all those prime numbers are?

By simply changing two clues of that recently published puzzle that I rejected, I was able to create a love-ly puzzle that can be solved entirely by logic. Can you figure out where to put the numbers from 1 to 12 in each of the four outlined areas that divide the puzzle into four equal sections? If you can, my heart might just skip a beat!

If you need some tips on how to get started on this puzzle, check out this video:

Now I’ll tell you a few things about the number 1350:

1350 is a composite number.

Prime factorization: 1350 = 2 × 3 × 3 × 3 × 5 × 5, which can be written 1350 = 2 × 3³ × 5²

The exponents in the prime factorization are 1, 3 and 2. Adding one to each and multiplying we get (1 + 1)(3 + 1)(2 + 1) = 2 × 4 × 3 = 24. Therefore 1350 has exactly 24 factors.

2019 is the hypotenuse of a Pythagorean triple: 1155-1656-2019 so 1155² + 1656² = 2019²

2¹⁰ + 2⁹ + 2⁸ + 2⁷ + 2⁶ + 2⁵ + 2¹ + 2⁰ = 2019

2019 is a palindrome in a couple of bases: It’s 5B5 in BASE 19 (B is 11 base 10) because 5(19²) + 11(19) + 5(1) = 2019, and 3C3 in BASE 24 (C is 12 base 10) because 3(24²) + 12(24) + 3(1) = 2019

Every year has factors that often catch people by surprise. Today I would like to give you my predictions for the factors of 2019: 2019 will have four positive factors: 1, 3, 673, and 2019 However, 2019 will also have four negative factors: -1, -3, -673, and -2019

Which factors, positive or negative, will be your focus in the coming year?

Finally, I’ll share some mathematics-related 2019 and New Year tweets that I’ve seen on twitter. Some of these tweets have links that contain even more facts about the number 2019.

The absolute best thing about 2018 is that not only is 2018 a semiprime number, but it is the 579th semiprime number, and 579 is also a semiprime number. 2019 is a semiprime number, but it’s the 580th, and 580 isn’t a semiprime number, so yaaaaaawn, right?

I feel like twitter needs more posts about beautiful mathematics, so I decided to start a new weekly series featuring the best curves of maths. You'll experience #50FamousCurves, see how to draw them and learn about their various applications. Stay tuned and spread the word! ❤️ pic.twitter.com/z9EiW4aGTt

Print the puzzles or type the solution on this excel file: 10-factors 807-814

One of my education professors taught that you can teach any concept with a picture book.

I recently read the book, Stick and Stone, to a class of 6th graders. Yes, 6th graders. You can get away with reading something way below grade level if you tell them before you start reading that you will use the book to introduce them to something that is definitely NOT below grade level. The first few pages of the book are shared by its publisher here:

As you can see, those first few pages equate stone as a zero and stick as a lonely number one.

The middle part of the book teaches about synergizing, working together to make life good and helping each other through tough times.

By the end of the book Stick and Stone know how to work very well together, “Stick, Stone. Together again. Stick, Stone. A perfect ten.”

The book pretty much ends there, but making a perfect ten is only the beginning of what these two characters can do together. I used this book to teach the class not only about getting along and working together, but also about base 2, or binary as it is also called. Every counting number we know can be represented by using just 1’s and 0’s. I wrote on the board the numbers from 1 to 16 and represented the first few of those numbers in base 2. Then I invited class members to come up with how to write the rest of the numbers in base 2. Some students caught on immediately while the others were able to learn how to do it by watching their classmates and listening to them. Eventually with at least 12 different student’s inputs, we came up with a chart that looked something like this:

Notice that the numbers from 9 to 15 are just 1000 plus the numbers directly across from them in the first column.

Some of the sixth grade students had already heard of binary, so I showed them a little more about base 2: I wrote a bunch of 1’s and 0’s “off the top of my head” onto the board and added the headings to show place values: 1’s place, 2’s place, etc.

Then I told them to sum up the place values that contained a one:

The sixth graders were delighted with the answer.

Stick and Stone are the main two characters, but the book has one other character, Pinecone. At first Pinecone bullied Stone, but after Stick stood up to him, the three of them were eventually able to become friends. You might enjoy finding out more about Pinecone by listening to Sean Anderson read the entire book to his children, one of which seems to really enjoy numbers.

If you used a unique symbol to represent Pinecone, it could look like a 2. Then you also could use the symbols 0, 1, and 2 to represent every counting number in base 3. That’s another concept the picture book Stick and Stone could be used to introduce!

To make a chart for base 3, start with these 3 columns of numbers with 3 numbers in each:

Since this is base 3, where should 10 and 100 go? The bottom of the first column and the bottom of the third column both MUST look like a power of 10. The rest of the chart is easy to fill out. Notice the 1 and 2 look exactly the same in base 10 and base 3. Also since 4 = 3 + 1, 5 = 3 + 2, and 6 = 3 + 3, we can easily fill in the 2nd column. Two more addition facts will finish the third column: 7 = 6 + 1, and 8 = 6 + 2.

Now add what you learned about 4, 5, 6, 7, 8, and 9 to column 1 and put the numbers 10 – 18 in the base 10 second column and numbers 19 – 27 in the base 10 third column. Again the bottom of the first column and the bottom of the third column both MUST look like a power of 10, so we now know where to put 1000.

To fill in the rest of the chart simply add 100 to the base 3 numbers in column 1 to get the the base 3 numbers in column 2. Then add 200 to the base 3 numbers in column 1 to get the remaining base 3 numbers in column 3.

You could do this process again to determine the first 81 counting numbers in base 3 with 81 being represented by 10000.

For base 4, you could do something similar with 4 columns. However, for counting in bases 4, 5, 6, 7, 8, and 9 I would suggest using the very versatile hundred chart. You can give instructions without even mentioning the concept of differing bases. For example, cross out every number on the hundred chart that has 7, 8, or 9 as one or more of its digits. Can you tell even before you get started how many numbers will get crossed out? (100 – 7²) What pattern do the cross-outs make? If you arrange the remaining numbers in order from smallest to largest, then you will have the first 49 numbers represented in base 7. With a minimal amount of cutting and taping you could have a “hundred” chart in base 7. Easy peasy.

This excel file not only has several puzzles, including today’s, but also a hundred chart and even a thousand chart because I know some of you might want to play with 3-digit numbers, too.

Now let me tell you a little bit about the number 810:

810 is a composite number.

Prime factorization: 810 = 2 x 3 x 3 x 3 x 3 x 5, which can be written 810 = 2 × 3⁴ × 5

The exponents in the prime factorization are 1, 4 and 1. Adding one to each and multiplying we get (1 + 1)(4 + 1)(1 + 1) = 2 x 5 x 2 = 20. Therefore 810 has exactly 20 factors.

Factor pairs: 810 = 1 x 810, 2 x 405, 3 x 270, 5 x 162, 6 x 135, 9 x 90, 10 x 81, 15 x 54, 18 x 45 or 27 x 30

Taking the factor pair with the largest square number factor, we get √810 = (√81)(√10) = 9√10 ≈ 28.4604989.

Since 810 has so many factors, it has MANY possible factor trees. If most people made a factor tree for 810, they would probably start with 81 × 10 or 9 x 90. NOT ME! Here are two less-often-used factor trees for 810:

Finally, here is an easy way to express 810 is in a different base:

Make a cake in which you divide 810 by the base number repeatedly, keeping track of the remainders, including zero, as you go.

Keep dividing until the number at the top of the cake is 0.

List the remainders in order from top to bottom and indicate the base you used to do the division.

This method is illustrated for BASE 2 and BASE 3 below:

That’s all pretty good work for a stone, a stick, and a pine cone!

By the way, using that method will also produce the following results:

810 is 30222 BASE 4

810 is 11220 BASE 5

810 is 3430 BASE 6 and so forth.

And just so you’ll know, 810 is the sum of consecutive primes 401 and 409.

807 is palindrome 151 in BASE 26 because 1(26²) + 5(26) + 1(1) = 807.

Anything else? Well, I can figure out a few other things because 807’s has two prime factors, 3 and 269:

We can write ANY number (unless it’s a power of 2) as the sum of consecutive numbers in at least one way. 807 has three different ways to do that:

403 + 404 = 807 because 807 isn’t divisible by 2.

268 + 269 + 270 = 807 because it is divisible by 3.

132 + 133 + 134 + 135 + 136 + 137 = 807 since it is divisible by 3 but not by 6.

I know that one of 807’s factors, 269, is a hypotenuse of a Pythagorean triple, so 807 is also. Thus. . .

(3·69)² + (3·260)² = (3·269)², or in other words, 207² + 780² = 807²

Since 807 has two odd sets of factor pairs, I know that 807 can be written as the difference of two squares two different ways:

136² – 133² = 807

404² – 403² = 807

I don’t usually do this, but today’s puzzle has something in common with 807. Can you tell what it is?

Print the puzzles or type the solution on this excel file: 10-factors 807-814

807 is a composite number.

Prime factorization: 807 = 3 x 269

The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 807 has exactly 4 factors.

Factors of 807: 1, 3, 269, 807

Factor pairs: 807 = 1 x 807 or 3 x 269

807 has no square factors that allow its square root to be simplified. √807 ≈ 28.4077454

Prime factorization: 572 = 2 x 2 x 11 x 13, which can be written 572 = (2^2) x 11 x 13

The exponents in the prime factorization are 2, 1, and 1. Adding one to each and multiplying we get (2 + 1)(1 + 1)(1 + 1) = 3 x 2 x 2 = 12. Therefore 572 has exactly 12 factors.

564 is made from three consecutive numbers so it can be evenly divided by 3. If the middle number is divisible by 3, then a number made from three consecutive numbers will also be divisible by 9. Is 564 divisible by 9? Why or why not?

Prime factorization: 564 = 2 x 2 x 3 x 47, which can be written 564 = (2^2) x 3 x 47

The exponents in the prime factorization are 2, 1, and 1. Adding one to each and multiplying we get (2 + 1)(1 + 1)(1 + 1) = 3 x 2 x 2 = 12. Therefore 564 has exactly 12 factors.

The exponent of prime number 563 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 563 has exactly 2 factors.

Factors of 563: 1, 563

Factor pairs: 563 = 1 x 563

563 has no square factors that allow its square root to be simplified. √563 ≈ 23.727621

How do we know that 563 is a prime number? If 563 were not a prime number, then it would be divisible by at least one prime number less than or equal to √563 ≈ 23.7. Since 563 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, or 23, we know that 563 is a prime number.

The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 562 has exactly 4 factors.

Factors of 562: 1, 2, 281, 562

Factor pairs: 562 = 1 x 562 or 2 x 281

562 has no square factors that allow its square root to be simplified. √562 ≈ 23.7065