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## Which is Greater: a^b or b^a?

Published on Sunday, September 17, 2017 in , , , ,

Here's an interesting mental math challenge. Given two different positive real numbers, which we'll call a and b, which is greater, ab or ba? If you're able to calculate both exponents mentally, such as 23 vs 32, then that's probably the simplest way to go. What happens if one or both exponential expressions are too hard to mentally calculate?

To solve this, we'll need to find a general rule. Some advanced calculation will be required to find it. However, once we do the work to find the rule, you'll see that almost no math will be needed to solve these types of problems!

SEARCH FOR THE RULE: Consider that 23 < 32, but 34 > 43. Just looking at the placement of smaller and bigger numbers, there doesn't seem to hard and fast rule which applies. The best place to start is just by assuming what we wish to discover:

$a^{b}>b^{a}$

From that point, let's see if we can separate a and b somehow. The quickest way is to raise both sides to the power of 1/ab:

$\\(a^{b})^{\frac{1}{ab}}>(b^{a})^{\frac{1}{ab}}\\ \\a^{\frac{b}{ab}}>b^{\frac{a}{ab}}\\ \\a^{\frac{1}{a}}>b^{\frac{1}{b}}$

In English, then, when the ath root of a is greater than the bth root of b, then ab > ba. Don't worry, though. You won't have to do any roots in your head. Instead, just look at this Desmos graph where y=xth root of x. If you click in the box marked 1 at the upper left, Desmos highlights two points, the point with the minimum y value, which is (0, 0), and the point with the maximum value, (2.718, 1.445).

Wait a minute! The value of 2.718 sounds familiar. Could the maximum value of xth root of x really be e? Sure enough, Wolfram|Alpha verifies that the maximum is e!

Look at both sides of the graph, then. From 0 to e, the graph increases. From e on upwards, the value of the xth root of x will steadily decrease. In fact, the value will just keep getting closer and closer to 1.

This means we've found the start of our rule. When you have two numbers, both of which are equal to or greater than e, the smaller number x will always yield a greater xth root of x than the higher number. This tells us that the ath root of a is greater than the bth root of b when a is the smaller number. Working backwards to our original assumption, this means that:

When a and b are both equal to or greater than e, ab is always greater than ba when a is the smaller number and b is the larger number.
There's our rule!

WORKING WITH INTEGERS: What about when one or both of a and b are less than e? In the case of positive integers, this means we only have to consider the cases involving 1 and 2.

The case of 1 is easy. Assume that a equals 1 and b has a value of 2 or more. 1b will always be 1, and b1 will always be b, which is 2 or more. That covers every case involving 1.

What about 2? Let's work through each case individually, always assigning a to a value of 2, and b a value other than 2. If b is 1, we just covered that case. b can't be 2, as it would be the same as a. What about when b is 3? We already covered the fact that 23 < 32, so we remember that special case. What happens when b is 4? This is a very unique case, as 24 is exactly equal to 42! In fact, this is a well-known special case, as it's the only time when ab equals ba when a and b are different integers. When b is 5 ore greater, as you can see in this graph, we can fall back on the rule we set above.

A CLASSIC CHALLENGE: Which is greater, πe or eπ? There's a problem you probably never thought you'd solve in your head!

Let's go see if our primary rule applies. Are a and b both equal to or greater than e? e, of course, is exactly equal to e (2.71828 and so on). π, as any Grey Matters readers should already know, is roughly 3.1415, and therefore also greater than e.

This means we can apply our primary rule! e is smaller than π, which means that eπ must be greater than πe. A quick verification with Wolfram|Alpha shows that this is correct.

About 4 years ago, Presh Talwalkar wrote up several approaches to this classic problem. Interestingly, this approach wasn't included. The 4th method does work with the xth root of x as we did, but it uses a deeper approach involving calculus.

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## Estimating Square Roots: Improved Accuracy

Published on Sunday, September 10, 2017 in , , ,

Back about 5 years ago, I taught a simple method for estimating the square roots of non-perfect squares. I recently learned an improvement that makes this far more accurate, and even a little more impressive.

The new approach works well with the original approach I taught. I'll start with a review of the original method I taught, and then I'll delve into the details of the improved method.

Quick Refresher: As mentioned in the original post, you should be comfortable with mentally squaring 2-digit numbers, and being able to find the square roots of perfect squares. You'll also need to know the squares of the numbers from 1 to 31 off the top of your head, in order to handle the numbers from 1 to 1,000.

As an example number, we'll say you're given 130, and asked to find the square root. You start by partitioning the given number into the sum of the largest perfect square equal to or less than the number, plus any remaining amount. The largest perfect square equal to or less than 130 is 121, so we partition it into 121 + 9:

$\sqrt{130} = \sqrt{121 + 9}$

If you consider the sum now under the square root at this point to consist of numbers x and y (x = 121 and y = 9 in our example), then the original square root approximation is as follows:

$\sqrt{x +y} \approx \sqrt{x}+\frac{y}{2\sqrt{x}}$

This may seem harder than it is. Remember that x is a perfect square, so sqrt(x) is an integer, so it will be easy to work with. Applying this formula to our example, and recalling that sqrt(121) = 11, we get this approximation:

$\\ \sqrt{121 +9} \approx \sqrt{121}+\frac{9}{2\sqrt{121}} \\ \hphantom{spaces} \ \ \ \ \approx 11\frac{9}{22}$

If you square that estimation, you get roughly 130.167. That's not bad, but how can we do better?

Improved Version: This past Monday, UK-based math tutor Colin Beveridge wrote a post called The Mathematical Ninja takes a square root, which is the improved version I've been mentioning.

The whole part (11 in our above example) will not change, so we'll only focus on changing the fractional part. If you consider, as in Colin's post, the fractional part to be ab, then the improved estimation is as follows:

$\frac{ab}{b^{2}+a}$

So, for the new numerator (top number of the fraction) in our example, you'd multiply 9 × 22 to get 198. For the new denominator (bottom number of the fractions), you'd have to square the original bottom number, 22 squared = 484, and then add the original top number, 9, for a total of 493. Our new estimation for the square root of 13 becomes:

$\sqrt{130} \approx 11\frac{198}{493}$

If you square that estimate, you get roughly 129.997, which is much closer to 130 than 130.167!

If this approach interests you, do take the time to read Colin's original post, as it has plenty of nice touches. Most importantly, he suggests using the closest perfect square above or below the given number, which will improve the accuracy of your estimate as compared to always using the perfect square less than or equal to the given number.

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## Yet Again Still More Quick Snippets

Published on Sunday, August 27, 2017 in , , , , , , , , ,

I apologize for the irregular posting over the past few months. I've had to deal with some personal issues (don't worry, everything is fine!). The good news is that, with this entry, everything should start returning to normal.

Having said that, let's dive into August's snippets!

James Grime and Katie Steckles made a video about a seemingly simple game:

First, if it's on Grey Matters, you know all is not always what it seems. Long-time fans of Grey Matters may remember this when I described it under the name Wythoff's Nim. It winds up having some very interesting math behind it. James went on to make a solo video explaining the mathematics behind it in more detail:

• We can't ignore Katie Steckles' game video after all that! Katie teaches 2 games (or does she?). The first one involves numbered fishes, and the second one involves cards with stars and moons on them:

It's a little bit surprising that these are actually the same game! Back in June's snippets, there was a multiplication version of this. Like this and Scam School's game of 15, they all go back to Tic-Tac-Toe. If you want to see some other interesting variations of this same idea, read Martin Gardner's Jam, Hot, and Other Games column.

• There's usually more than one way to use your knowledge. In my tutorial about mental division, I teach a simple method for mentally dividing the numbers 1 through 6 by 7. Presenting it as an exacting feat of mental division is one thing. How else could you present it? Take a look at how Scam School presents the same feat:

If you watch the full explanation, you'll notice another difference between the way I teach it and the way Brian teaches it. He puts emphasis on the last digit, which works well for performing the feat this way. In my version, I teach how to work out the first few digits, as you'll need those first when giving the answer verbally. This is a good lesson in the benefits of changing your point of view!

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## Solar Eclipse Mental Feat!

Published on Friday, July 28, 2017 in , , , ,

On August 21, 2017, there will be a total solar eclipse visible in the United States, which hasn't happened since 2012, and won't happen again until 2024!

It's the perfect time to present an impressive knowledge of the moon, such as being able to estimate the moon phase for any date in 2017! Does learning this feat sound difficult? Surprisingly, it's much easier than you may think.

For 2017 only, the phase of the moon formula is simple: (Month key number + date) mod 30. The result is the age of the moon in days from 0 to 29. I'll explain each part of this formula below.

Month key number: January's key number is 3, February's key number is 4, and all other months' keys are their traditional numbers; March is 3, April is 4, May is 5, and so on up to December, which is 12.

Date: This is simply the number represented by the particular date in the month. For the 1st, add 1. For the 2nd, add 2. For the 3rd, add 3, and so on.

mod 30: If you get a total of 30 or more, simply subtract 30. Otherwise, just leave the number as is.

As an example, let's try the published date of this post, July 28th. July is the 7th month and the date is the 28th, so we work out 7 + 28 = 35. Since that number is greater than 30, we subtract 30 to get 35 - 30 = 5. This result tells us that the moon's age will be 5 days on July 28th, 2017.

But what does it mean to say that a moon is some number of days old? Here's a simple explanation:

0 days = New moon (the moon is as dark as it's going to get)

0 to 7.5 days = Waxing crescent (Less than half the moon is lit, and it's getting brighter each night)

7.5 days = 1st quarter moon (Half the moon is lit, and getting brighter each night)

7.5 to 15 days = Waxing gibbous (More than half the moon is lit, and getting brighter each night)

15 days = Full moon (The moon is as bright as it's going to get, and will start getting darker each night)

15 to 22.5 days = Waning gibbous (More than half the moon is lit, and it's getting darker each night)

22.5 days = 3rd quarter moon (Half the moon is lit, and it's getting darker each night)

22.5 to 29 days = Waning crescent (Less than half the moon is lit, and it's getting darker each night)

So, our 5-day-old moon would be a waxing crescent (Less than half the moon is lit, and it's getting brighter each night). Sure enough, Wolfram|Alpha confirms this estimate! Moon Giant confirms this estimate, as well.

There are a couple of finer points to note. First, this simple method happens to work only in 2017. The method won't be this simple again until 2036 and then 2055. If you want to learn to calculate the moon phase for any date in the 1900s, you can learn the full feat over in the Grey Matters Mental Gym.

Second, remember that this is an estimate. The actual error margin is plus or minus one day. So, getting an estimate of 5 days means that the moon is somewhere from 4 to 6 days old. When the overlap includes a first quarter moon, a full moon, or a 3rd quarter moon, I usually describe this as, "While it might not be technically accurate, most people would look up and describe it as a 1st quarter moon" (or full moon, or 3rd quarter moon).

This method is a simplification of John Conway's original moon phase estimation formula from Winning Ways for Your Mathematical Plays, vol. 4. Practice it and have some fun amazing your friends and family during the coming solar eclipse!

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## Review: Perfectly Possible

Published on Sunday, July 09, 2017 in , , , , , , ,

Many regular Grey Matters readers will be familiar with Michael Daniels' Mind Magician site, where he teaches numerous math and memory feats, such as calculating cube roots in your head instantly. He's recently written a new ebook on the 4-by-4 magic square, titled Perfectly Possible. I found it to be well worth the time and money invested, and wanted to share my thoughts with Grey Matters readers.

This is going to be a difficult review, as I can't give too much away, but I also want to share with you the quality of this method. I'll start with the qualities promoted by Michael Daniels himself:

• Completely impromptu. No set-up, gimmicks, or cribs.
• New, improved method - minimal memory and the simplest of calculations.
• Suitable for close-up or stage performances.
• Produces elegant magic squares.
• Can be immediately repeated for different totals.
• Includes a browser application that helps you to learn and practice (Internet connection not required).
Let's clarify a few points here. Yes, it is completely impromptu. This is a calculation method, but the calculations are minimal, quick, and will quickly become second nature during practice. Speaking of practice, the included browser application is very handy. It's similar to the magic square practice app posted at mindmagician.org, but streamlined for the new routine.

What does "elegant magic squares" mean? One problem with many magic square approaches is that the number can appear unbalanced, such as when 12 of the numbers are less than 15, and the other 4 are over 30. This can give your audience clues about the method. With the Perfectly Possible method, you don't have to worry about that. You're guaranteed a balanced magic square. Elegant also means that you're guaranteed at least 36 different ways in which some combination of 4 squares gives the magic total. Under the right circumstances, this method can yield as many as 52 different combinations!

As with any magic square, the ability to repeat the square immediately with different totals is, of course, essential. Even more impressive, though, is that if 2 people give you the same total, you can still generate a different magic square! Naturally, the same total requires the numbers used to be in the same general range, but this method will allow you to put different numbers in each of the squares with very little difficulty.

That quality is really what makes Perfectly Possible stand out. Unlike the rigid approaches behind most magic squares, the ability to take multiple approaches gives the performer more freedom while disguising the method very effectively. When a change is as constrained as the magic square, finding an approach like this that offers you remarkable degrees of freedom like this is incredible!

If you're interested in creating magic squares, I can't recommend Michael Daniels' Perfectly Possible ebook enough. It's available for \$6 on its own, or \$8 in combination with Mostly Perfect, its predecessor. If you're seriously consider this as a performance piece, I would also recommend the Unknown Mentalist's Why A Magic Square Should Not Be A Magic Square ebook. It teaches many very effective original presentations that disguise the principle, and will help preserve the mystery by showing you how to prevent audiences from simply searching for "magic square" on the internet during or after your performance.

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## Squaring Numbers from 100-199

Published on Sunday, July 02, 2017 in , , ,

Over in the Mental Gym, I have a tutorial on squaring numbers, starting with simpler techniques for multiples of 10 and 5, and working up to squaring numbers as large as 125.

Naturally, I always like to see how much farther I can go, especially when I can still keep things relatively simple. With the technique I'll be teaching in this post, if you're comfortable with squaring numbers up to 125, you're ready to move on to squaring numbers up to 199 (Well, actually 200, since that's not difficult to square).

Rightmost Two Digits

We're going to generate the answer from right to left in this technique, working with no more than 2 digits at a time. To get the rightmost 2 digits of the answer, simply square the rightmost 2 digits of the given number. For example, if you're given the number 112 to square in your head, you'd square the rightmost 2-digits of that number, 12, to get 144. You write down the rightmost 2 digits, 44 in this example, and remember the remaining digits to the left, such as the 1 in this example (underscores are used to hold places for numbers not yet written):

$\\written:&space;\textunderscore&space;\textunderscore&space;\textunderscore&space;\textunderscore&space;44&space;\\&space;remembered:&space;1$

Middle Two Digits

For the middle 2 digits, simply double the rightmost 2 digits of the given number, then add the number you remembered, if any. Write down the 2 rightmost digits of this answer to the left of the digits previously written, and remember any digits to the left of that. Returning to our 112 example, we look at the rightmost 2 digits, 12, and double that to get 24. Adding the number we remembered from the previous step, 1, we get a total of 25. The rightmost 2 digits of this answer are 25, and there's no digits to the left of those to remember:

$\\written:&space;\textunderscore&space;\textunderscore&space;2544&space;\\&space;remembered:&space;\{nothing\}$

Leftmost Digit(s)

For the leftmost digits, take any amount you have remembered at this point, and simply add 1 to it. Write down that total to the left of all the digits you've previously written, and you're done! In our 112 example, we didn't remember anything for this stage, so we just write down 1, resulting in:

$\\written:&space;12544$

You can check for yourself that 1122=12,544.

Tips

Single-digit numbers: When working on either the rightmost 2 digits or the middle 2 digits, you may wind up working with a single-digit answer. These steps always require working with 2 digits, so for single-digit answers, just place a 0 to the left of it to make it a 2-digit number.

For example, when squaring 103, you start by squaring 3 to get 9, with nothing to remember:

$\\written:&space;\textunderscore&space;\textunderscore&space;\textunderscore&space;\textunderscore&space;09&space;\\&space;remembered:&space;\{nothing\}$

Next, you'd double 3 to get 6, again with nothing to remember:

$\\written:&space;\textunderscore&space;\textunderscore&space;0609&space;\\&space;remembered:&space;\{nothing\}$

Since you don't remember anything from the previous steps, just add a 1 to the left of this answer:

$\\written:&space;10609$

This tells us that 1032=10,609.

Remembering multiple digits: Just so you have an example of working with larger numbers, let's try squaring 178. Start by squaring 78, which is 6,084. Write down the 84, and remember the 60:

$\\written:&space;\textunderscore&space;\textunderscore&space;\textunderscore&space;\textunderscore&space;84&space;\\&space;remembered:&space;60$

Next, double 78 to get 156, and add the 60 you remembered from the previous step, giving a total of 216. Write down the 16, and remember the 2:

$\\written:&space;\textunderscore&space;\textunderscore&space;1684&space;\\&space;remembered:&space;2$

Finally, add 1 to the number you're remembering and write that down to the left of the previous digits. In this case, we add 1 and 2 to get 3:

$\\written:&space;31684$

The result of 1782 is 31,684.

Interest for 2 time periods: This technique is especially handy for quickly calculating how many times the principal will grow at interest rates of 99% or less for 2 time periods. The only additional step is to put a decimal point between the ten-thousands and thousands place. For example, since 1032=10,609, that means that principal invested at 3% per year for 2 years will grow to 1.0609 times its original size. Although you're not likely to ever see it happen, principal invested at 78% per year for 2 years would grow to 3.1684 times its original size, because 1782 is 31,684.

Practice this, have a little fun with it, and you'll have an impressive new skill!

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## Magic, Math and Memory Videos!

Published on Sunday, June 25, 2017 in , , , , , ,

I recently ran across a number of videos I figured would be interesting to regular Grey Matters readers, so I thought I would share them.

We'll start things off with a little math magic, courtesy of Tom London and his appearance on America's Got Talent earlier this week:

Yes, I could explain the method, but I don't want to ruin the fun and the mystery. Just enjoy the magic of the prediction for what it is, since that's how it's meant to be enjoyed.

If you want mathematical explanations, however, I highly recommend checking out PBS Digital Studios' Infinite Series. These are videos on assorted advanced mathematical topics, yet they're taught in a very accessible way. Back in March, I discussed a puzzle which required the understanding of Markov chains to solve. Compare that to their video Can a Chess Piece Explain Markov Chains?, which also happens to employ my favorite chess piece, the knight:

If you enjoy Grey Matters, you may also the work of 4-time USA memory champion Nelson Dellis, who focuses on both mental and physical fitness. He has a series of memory technique videos, as well as interviews with masters of mental skills. Both of these are available on his YouTube channel, as well. As a taste of his skilll, watch his video, Memorizing 28 names in less than 60 seconds!:

Curious how he's able to do that? He explains in the next video in the series, HOW TO // "Memorizing 28 names in less than 60 seconds!".

At this point, I'll wrap things up so you can get started on a potentially mind-expanding journey.

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## Yet Still More Quick Snippets

Published on Sunday, June 11, 2017 in , , ,

It's now been a few months since Grey Matters was back, so now it's time to bring Quick Snippets back!

This time around, we have plenty of mathy goodness, so it's best to just jump right in!

• Besides the Clay Institute's famous selection of Millennium Problems, which will make you a millionaire if you prove or disprove any one of them, there's a lesser known set, known as John Conway's \$1,000 problems. Not long ago, the 5th conjecture, which claims that working through a certain procedure (described in the link an video below) will always end in a prime, was disproven by physicist James Davis. The Numberphile video below details the problem and James Davis' counterexample:

For more about the million-dollar Millennium Problems, watch the BBC's Horizon documentary, "A Mathematical Mystery Tour of Unsolved Mathematical Problems."

• Speaking of fun discoveries in recreational mathematics, check out Allan William Johnson's "Magic Square of Squares", discovered back in 1990, and just recently posted over at Futility Closet.

• James Grimes introduces us to a different sort of "Square of Squares", in his latest Numberphile video, "Squared Squares". The challenge here is to make a perfect square shape from a set of smaller square shapes:

• Presh Talwalkar, of Mind Your Decisions, posted an interesting puzzle recently. It's titled, "The Race To 32,768. Game Theory Puzzle". Read the article up to the point where you're challenged to work it out yourself, or watch the video below up to the 2:12 mark, and try and figure it out for yourself. If you get stuck, try going over my Scam School Teaches the Game of 15 post for inspiration.

• Late last month, mathematical video maker 3Blue1Brown posted a must-watch video on the visualization of all possible Pythagorean triples. Even if you remember everything from you math classes about Pythagorean triples, this video is both eye-popping and an eye opener:

• We'll wrap this set of Quick Snippets up with help from Mathologer. His videos are always interesting, but his latest one is one of those unusual approaches to math that makes you appreciate its beauty. This video is titled, "Gauss's magic shoelace area formula and its calculus companion", and it teaches an simple but unusual method for working out the area of any polygon that doesn't intersect itself. The host even goes on to show how this approach can be adapted in calculus to work out the area contained by curves!