You may have heard of the binary numeral system before, but what exactly is it? You may have come across it in some classes, and the binary system is used extensively in computers. You may also have heard that this system is related to our normal numeral system. What is all the commotion about? I will help demystify this system in this post.

A very short answer would be that our normal system is based on ten, while the binary numeral system is based on two. I will explain at length what that means and what it entails.

Our counting system consists of two parts. The first part is a set of digits ranging from 0 to 9. They say this corresponds to the fact that we have ten fingers on our hands, which are also called digits.

The second part is a system for positioning these digits to describe numbers. When we want to write numbers larger than nine, we position the digits further to the left. When we want to write numbers smaller than one, we shift them to the right.

To transition to the binary numeral system, we first have to reduce the total number of available digits from ten to two. We do this by only considering the digits 0 and 1. With these two symbols, we can write 0 to mean zero and 1 to mean one as in our normal system.

Before taking the next step, we should make an observation about our normal system. The ones place is for multiples of one, the tens place is for multiples of ten, the hundreds place is for ten times ten, the thousands place is for ten times ten times ten, and so on. In addition, the tenths place is for multiples of one tenth, the hundredths place is for one tenth of one tenth, and so on.

To transition to the binary numeral system, we should replace all the tens (multiply by ten) and tenths (multiply by one tenth) described above with “twos” and “halves”, respectively.

The ones place should be as before, but the tens place should be replaced with a “twos place”, which corresponds to two. For example, 10 in the binary numeral system represents two because the digit 1 here is in the “twos place”. We can go further. Since the hundreds place corresponds to ten times ten, what replaces the hundreds place should correspond to two times two. So, the hundreds place should be replaced with a “fours place” as four is two times two. For the thousands place, we note that it corresponds to ten times ten times ten. So what replaces that should correspond to two times two times two. That is, the hundreds place should be replaced with an “eights place” as eight is two times two times two.

Recall that the tenths place corresponds to one tenth. So as we want to replace ten with two, the appropriate replacement here would be one half. That is, we should replace the tenths place with a “halves place”.  For instance, 0.1 in the binary numeral system represents one half because the digit 1 here is in the “halves place”. What about the hundredths place? Since the hundredths place corresponds to one tenth or one tenths, we should replace that with one half of one half. So as one half of one half is one quarter, we should replace the hundreds place with a “quarters place”.

Lastly, to calculate what a number in the binary numeral system is, we add the various “places” occupied by 1 together. If 101 were written in the binary numeral system, what number would that equal to? Note that the “fours place” and the “ones place” are occupied by 1. So to calculate 101, we would need to add four and one, which is five. In summary, 101 is five.

If 1.1 were written in the binary numeral system, we can read it like the following. Note that the “ones place” and the “halves place” are occupied by 1. So to calculate 1.1, we would need to add one and one half, which is one and a half. What if 0.11 were written in the binary numeral system? Note that the “halves place” and the “quarters place” are occupied by 1. So to calculate 0.11, we would need to add one half and one quarter, which is three quarters. In summary, 1.1 is one and a half and 0.11 is three quarters.

This could make for good fun. For instance, someone who is sixty years old could write 111,100 as their age. Why is that? Let us consider the places. We discussed the “ones place”, the “twos place”, and the “fours place”. Going further to the left, we can speak of the “eights place”, the “sixteens place”, and the “thirty-twos place”. How come? Because eight is two to the power of three, sixteen is two to the power of four, and thirty-two is two to the power of five.

Looking at 111,100, we notice that the “thirty-twos place”, the “sixteens place”, the “eights place”, and the “fours place” are all occupied by 1. So to calculate 111,100, we add thirty-two, sixteen, eight, and four together to obtain sixty.

So if someone writes 111,100 as their age on their birthday cake, don’t be fooled. They are not actually over a hundred thousand years old; they are much younger!

The binary numeral system presents an alternative that is much like our system but, at the same time, is quite different. Hopefully, this post can further clarify what this interesting numeral system is all about. You may wonder about how we obtained 111,100 above. I may describe how to do this in more detail in another post.

A well-known conjecture in mathematics, known as the Sunflower Conjecture, was introduced and investigated by Erdős and Rado in 1960. Since then, many mathematicians have worked to understand its mysteries. However, the conjecture remains unsolved to this day.

A sunflower brings to mind a particular image: Petals are neatly arranged, fanning out from a circular mass. We can simplify this imagery by drawing a series of loops, each representing one petal.

In this simplistic visualization, a sunflower can be made up of many such loops in such a way that you are reminded of a child’s drawing.

Now, let’s imagine that each of these loops contains some pebbles. If a pebble is contained in more than one of a sunflower’s loops, it will automatically be in its center, assuming that the loops intersect at the center.

In this way, we have effectively described a sunflower as a collection of loops.

However, loops don’t always form a sunflower. Loops can be arranged in such a way that they do not form a sunflower. For example, if there are three loops but only two of them intersect, then those three loops do not form a sunflower.

Something interesting happens here. Imagine a collection of loops and assume that each of them contains, say, seven pebbles. If I have many such loops, will a sunflower always appear? It turns out that if this collection has 130 loops, then no matter how you arrange these loops, you can always find a sunflower with at least three petals. And if you increase the number of these loops further, you will eventually find sunflowers with even larger numbers of petals.

In 1960, Erdős and Rado proved that this always happens.

In this so-called Sunflower Conjecture, they conjectured that the growth rates of these numbers, such as 130 mentioned above, are at most exponential functions on the number of pebbles in each loop. For example, if the Sunflower Conjecture were true for sunflowers with three petals, then we can find some number relating to the aforementioned growth rate, perhaps an extremely large number such as 1,000,000,000, that will enable us to make interesting conclusions.

If I have, say, 1,000,000,000^2000 (means to the power of 2000) loops with 2000 pebbles in each loop, then among all of those loops, I can always find at least three of them that form a sunflower with at least three petals. I can also make the same conclusion when there are 1,000,000,000^7 loops with seven pebbles in each loop. In the above, I stated that 130 such loops are enough, but there are no problems here as 130 is much smaller than 1,000,000,000^7.

What makes this nice is not that 1,000,000,000^7 is much larger than 130, it is that I can find a single number, such as 1,000,000,000, which captures all pebble sizes (from one to infinity) for sunflowers with three petals.

Fundamentally, we are interested in determining the smallest number that captures all the pebble sizes and whether such a number exists.

Deepening this mystery is that no one knows if such numbers even exist.

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Let us take a deeper look into some of the numbers involved.

Recall our earlier example involving 130, with seven pebbles in each loop.

If we increase the number of pebbles in each loop to eight, then it turns out that you will need 258 loops to have the same result. That is, if there are 258 loops with eight pebbles in each loop, you can always find three loops that form a sunflower with three petals.

By how much did the number grow when transitioning from seven pebbles to eight pebbles? We calculate that 258/130 is about 1.98, which is nearly two. What happens if we go further? Further increasing the number of pebbles per group to nine gives 514 as our answer. Comparing this number with the previous one, we obtain 514/258, which is about 1.99.

The sequence 1.98, 1.99 may seem to approach 2, but that turns out to be false. If we were to extend the above sequence of ratios by making further calculations, we would eventually begin to find that many of these ratios are at least 3.16. This is a consequence of a result by Abott, Hansen, and Sauer. This is a far cry from numbers like 1,000,000,000. But no one knows whether these ratios will stop growing.

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Our discussion of sunflowers is indicative of the fact that there are very difficult questions relating to this that can be asked. The Sunflower Conjecture may open the door to many mysteries further ahead.