# Maths 101

### The Weirdness in Maths

I find it quite bizarre that there are a lot of mysteries and confusions and “weirdness” in the many wonders of mathematics. One example would be the fact that 0.999… = 1. This means that the number 0.9, followed by an infinite number of 9s, are all equal to 1. But not in the sense that this number is rounded up to 1. It is just equal to 1. But how could that be? We have all learned from middle school that the comparison of decimals involves comparing the integral part first before comparing the decimal part. And in doing so, the greater whole number, the greater its value; and then comparing the decimal part digit after digit after the decimal point. But in this case it seems that 0.999… is of a smaller value since its whole number value is 0 and 1 has an integral part of 1; this meant that 1 would have been greater number. But it seems that there exists a proof that proves that they are indeed both equal:

Let x = 0.999… Therefore, 10x = 9.999… (Multiplying by 10 means that you have to move the decimal point one place to the right). Also if you subtract these two numbers, we have (10x-x) = (9.999…-0.999…). Simplifying it, we have 9x = 9. (The decimal portion of both are both identical; hence they cancel out). Dividing both sides by 9, x = 1. And by transitive property, if x = 0.999… and x = 1, then it must be that 0.999… = 1.

Q.E.D.

There doesn’t seem to be an error in the proof; all it took was a little Algebraic Magic. However it still seems erroneous at first glance. How come a number with an integral part of 0 be equal to 1? 1.000… = 1 I may say seems more valid, but not 0.999… = 1. And that’s where the problem with infinity arises. Remember that there are an infinite number of 9s after the decimal point. This means that there is no last digit to this number. Think about it; what number would you add to 0.999… so that you can reach the value of 1? Well, that would be 0.000… where the last digit has to be 1; or is it 0.000…1 and that there’s an infinite number of 0s between the decimal point and 1? But how could this be the case since we have already talked about that there is “no last digit”. If your number contains a finite number of 0s instead of an infinite number of 0s, then adding this to 0.999… will result into a number greater than 1. So instead, we expect to add 0.000… instead (infinite number of 0s after the decimal point) to 0.999… to produce 1. But 0.000… is also equal to 0. And adding any number to 0 is always the number itself, which is 0.999… Then again, we are using the transitive property which equates that this number 0.999… is indeed equal to 1.

This is all linked to the peculiar, yet magnificent power of infinity. Here’s another one: how many types of infinity are there? Perhaps I could be more specific by saying that there are a lot of infinities out there that exist, much much greater than other infinities. Here’s an example: how many positive integers are there? The set of all positive integers contains 1, 2, 3, and so forth. And now how many positive real numbers are there? Well, this is a bit much harder since the set of positive integers is just a subset of the set of all positive real numbers. Ok so if we want to list the set of all positive real numbers, we have all these positive integers. What else then? Well, there’s 0.1, 0.2, 0.3,… 0.9. Oh but wait, I think we missed something: 0.01. It, too, is a positive real number. So 0.02, 0.03, …, 0.09 also count. But what if I tell you that you missed another positive real number, which is 0.001? With that said, our listing methods seem to be resulting into a failure. Hence, even if there is an infinite number of elements in this set, it is of a greater cardinality (the number of elements in a given set) than the cardinality of the set of positive integers. Hence, we say that the set of positive real numbers is uncountably infinite and the set of all positive integers is countably infinite. And being uncountably infinite over countably infinite provides proof that there are indeed infinities much greater than others.

There are yet a lot of “weirdness” in maths, and many of them are said to be linked with infinity. But just looking at it, it just seems so beautiful! Ever wonder why the slope of a vertical line on a Cartesian plane is undefined? Ever wonder why the sum of a series tends to converge (meaning to approach a certain definitive number), while others tends to diverge (the opposite of converge)? And have you ever wondered (elementary question) why one is not able to give a definitive number when dividing by 0? All these questions lie to some of the foundations of infinities and I find these very captivating. But yet again, there are still other “mysteries” out there in Mathematics that I also find attractive. All these are to be discussed and explored further in my future blogs.

– Royalle

Date of Post: 12.27.16

### BEDMAS 101

It has always been a pet peeve for me when there are people around me who still do not know the BEDMAS rule. But then I always think this over the moment I hear that someone beside me in the library would argue with all their might to their friend that that 2 + 8 x 3 is 30 and not 26. But hey, I would then understand that Math is a tricky subject and that anything Math-related especially these types of problems can be quite confusing. Nevertheless, rather than being mad at that guy over at the library, it would be better if I teach them the ways of the BEDMAS.

Now **BEDMAS**, which stands for ** B**rackets,

**xponents,**

__E__**ivision/**

__D__**ultiplication and**

__M__**ddition/**

__A__**ubtraction, represents the order of operations. It is a very handy mnemonic to easily remember the precedence among operations. There are other names for BEDMAS and one in particular would be PEMDAS, which is**

__S__**arentheses,**

__P__**xponents,**

__E__**ultiplication/**

__M__**ivision and**

__D__**ddition/**

__A__**ubtraction. Both work the same way, as long as the order of operations is still followed.**

__S__So how does this work? The rule states that any operation found within brackets or parentheses must be worked out first. It is also important to note that the innermost grouping symbol has to be operated first before the outermost grouping. For instance, in calculating 3 x [2 – (4 + 5)], we solve first for 4 + 5 to get 9. Replace the (4 + 5) in the original expression as 9 to make the expression much simpler by reducing the number of grouping symbols. So we then have 3 x [2 – 9]. Solving the operation inside the brackets which is 2 – 9 = -7. Replace [2 – 9] with -7, which leaves us with 3 x -7 and therefore can easily be solved to get -21.

The next precedence would be Exponents. After getting rid of the grouping brackets, then it is time to attack exponents. For instance, calculate 40 / (6 – 2^2)^3. Note here that “/” means division and “^” means an exponent raised to the power of. So here we start solving for the bracket which is 6 – 2^2. Here we see an exponent, which we should solve first. 2^2 is 4 and it should replace 2^2 here. Therefore 6 – 4 = 2. This was initially part of the bracket in the original expression and we should replace this with the number we got which is 2. Our simplified expression would thn be 40 / 2^3. Now recall that Exponents must go before any Multiplication, Division, Addition or Subtraction. Therefore 2^3, which is also equal to 2 x 2 x 2, is 8. Replace 2^3 with 8 to get 40 / 8, which is 5.

Now always keep mind to solve for bracketed operations first, beginning from the innermost operation to the outermost. Then any exponents found must be solved immediately before solving for any operation with Division, Multiplication, Addition or Subtraction. Division and Multiplication have interchangeable precedences, meaning that the first operation that appears after reading the expression from left to right has to be performed first. A simple example, for instance, would be 6 / 3 x 2. Here we first operate 6 / 3 to get 2 before multiplying it to 2 to get 4. If we have 3 x 9 / 27, therefore we first solve 3 x 9 before dividing the result by 27, which is equal to 1. Addition and subtraction, however have lower precedences than Division and multiplication; thus, division and multiplication would have to be performed first before operating addition and subtraction, while of course keeping the interchangeable precedence premise among division and multiplication. Moreover, like division and multiplication, addition and subtraction also have interchangeable precedences, performing first the operation that appears when the expression is read from left to right.

*Here’s a recap of the rules of BEDMAS:*

- Begin performing bracketed operations in an expression, starting from the innermost going to the outermost brackets (or parentheses or braces or any other grouping symbol for that matter).
- Simplify exponents whenever possible.
- Solve for division and multiplication operations. Precedences are interchangeable.
- Finish up with addition and subtraction. Precedences are also interchangeable.
- Repeat back to step 1 if there are any other further brackets and continue performing the cycle.

Now go and test yourself if you can calculate the final result of the following expressions using BEDMAS rule, and then check your answers:

- (2^7 – 1 x 5) + 3^4
- 1 + 1 + 1 + 1 + 1 x 0
- 180 / [2 + (8 x 1/2)]^2

Answers:

- 214
- 4
- 5

– Royalle

Date of Post: 08.20.16