[NumTh Expository] The Principle Behind Casting Out Nines

Introduction

Think of a number ... say 685932.  Divide by  9  and take the remainder.      685932 ¸ 9      = 76214 r 6  Add up the digits,  divide by  9  and take the remainder.      6+8+5+9+3+2 = 33 ¸ 9 = 3 r 6  What do you notice?  Try this with any other positive whole number.

Discussion

     Did you see that a number and its sum of digits always have the same remainder when divided by  9?  This is the principle behind the method of “casting out nines”, used in the past for checking arithmetical calculations.  Why does this work?  Where is its magic?

     The decimal number system that we use today is based on the number  10, which is just  1  larger than  9.  Observe that  9, 99, 999, 9 999, 99 999, ... etc are all divisible by  9.  Hence, the powers of 10, namely 10⁰ = 1,  10¹ = 10,  10² = 100,  10³ = 1 000,  10⁴ = 10 000,  10⁵ = 100 000,  etc  all leave a remainder of  1  when divided by  9.   Thus in our example,

     685932 = 6´10⁵ + 8´10⁴ + 5´10³ + 9´10² + 3´10¹ + 2´1

                  = 6´(99999+1) + 8´(9999+1) + 5´(999+1) + 9´(99+1) + 3´(9+1) + 2´1

                  = 6´99999+8´99995´999+9´993´9 + 6´1+8´1+5´1+9´1+3´1+2´1

                  = 9 ´ something + 6+8+5+9+3+2

As you can see, all the “´1” allow us to separate out the digits, and then the stuff with 9, 99, 999 etc can be lumped together as 9 ´ some whole number, but we do not need to care too much about this multiple of 9 as it would not make any difference to the remainder.  It is now obvious that  685932 and 6+8+5+9+3+2=33 will have the same number when divided by 9.

     Let us generalise the argument.   If two numbers  x  and  y  have the same remaider when divided by 9,  we say that  x  and  y  are congruent modulo 9, and we write     x  º  y  (mod 9).  Congruence is an equivalence relation and “º” behaves in many ways similar to “=”.

Theorem

For an arbitrary number  n   with digits  [dk...d3d2d1d0]                                         n º dk + ... + d3 + d2 + d1 + d0   (mod 9)

                 n = d_k ´10⁴ + ... + d₃´10³ + d₂´10² + d₁´10 + d₀.    

Since  10^k º 1 (mod 9)  for all integers  k > 0,  we have

                 n º d_k´1 + ... + d₃´1 + d₂´1 + d₁´1 + d₀

                 n º    d_k  + ... +   d₃   +    d₂  +   d₁    + d₀     (mod 9).   © (Q.E.D.)

As an example of application of this principle, please refer read thisarticle.

Suitable Levels

* Primary School Mathematics Olympiad

* syllabuses that involve congruences and Number Theory

* anybody who is interested

[S1_20150510AESFRN] Figuring out Different Significant Figures

Question

Vishal estimated a number as 2000 when he rounded the number to 1, 2 or 3 significant figures. Ethan was shocked by Vishal's answer and he asked Vishal what is the smallest and largest possible number such that the estimated number is 2000.  Can you help Vishal answer Ethan?

Introduction

     This is taken from ShingLee’s Secondary 1 Mathematics (7th edition) textbook.  It is a good question to test if students really understand the meaning of significant figures!  If you have not done so, please read my article Significant Figures Made Easy.

     To round certain number to  n  significant figures:-

Step 1:  Identify the first non-zero digit from the left. Step 2:  From this digit, start highlighting  n  significant figures. Step 3:  Look at the digit just after the last highlighted digit.  Round up if              this is 5 and above, otherwise do nothing. Always remember to include the place-holder ‘0’s.

Solution

     This question requires one to think backwards.  You are given the number  2000.  If it is correct to  1  significant figures, then what are the possible numbers that round to it?  If it is correct to  2  significant figures, then what possible numbers round to it?  If it is correct to  3  significant figures, what are the possible numbers round to that?  The question is not specific about what sort of “number” they meant, so I shall assume that they refer to real numbers.

Technical Remark

     In case you are wondering, for example,  why in the answer for  2  significant figures i.e. the interval
1 950 < x < 2 050, the  x cannot be made equal 2 050, this is because it would be rounded to  2 100.  A number like  2 049.9999999  (seven ‘9’s after the decimal point)  will be rounded down to  2 000.  Even if there were one trillion ‘9’s after the decimal point, as long as the number of decimals terminates (i.e. is finite), it will still be rounded down to  2 000.

     Once a number touches  2 050  exactly, it will be rounded up to  2 100  for  2  significant figures.  So technically, there is no largest number, because you can keep on adding more and more  9’s  behind as long as it is a finite number.  But you cannot add an infinite number of  9’s  because that would touch  2 050. 
2 050  is not the largest number, as it is not among the possible numbers, but  1 950  is the lowest number, because it is among all the possible numbers.   Because there is actually no largest number, I used the word “largest” in quotation marks.  [ Optional: The interested reader may want to check up infimum  and supremum ]  These are some very fine details that the textbook authors and editors forgot.  Even if the editor has a PhD, the PhD could be in “mathematics education” but not “mathematics”, and may have forgotten the rigours of the real numbers system.  The answers given at the back of the book are not only wrong, one pair of answers was missing.  This caused some confusion among parents and students.  I suspect they employed a non-professional part-timer to do up the answers in order to save costs.

Suitable Levels

* Lower Secondary Mathematics (usually secondary 1 ~ equivalent to about grade 7)

* other syllabuses that involve estimation and approximation

[S1_Expository] Significant Figures Made Easy

Introduction

     When you read a textbook on the topic of significant figures, you often get a bunch of complicated rules which seem pretty arbitrary and do not make sense.  Frankly, I too found these rules confusing.  They do not explain why those rules exist.  But there is sure-fire short-cut way to round a number to a given number of significant figures, based on understanding two key concepts: (1) the meaning of “significance” and (2) the role of ‘0’.  I present these below.

What does “significant” figure mean?

     The word “significant” simply means important.  Suppose you sold a sofa to somebody for

$7 287.  If the buyer wrote you a cheque for  $1 287,  there is an error in the leftmost first digit and you would get  $6 000  less.  That is bad.  If, instead, the buyer wrote you a cheque for  $7 281,  there is an error in the fourth digit and you would get  $6  less.  Bad, but not so bad.  Compare the 7 on the left against the 7 on the right.  Which digit is more significant?  As you can see, the digits on the left are more significant, and decrease in significance as we move towards the right.

The Roles of ‘0’ (Zero)

     Our modern Arabic-based place-value numerical system evolved over many centuries.  It is definitely far superior than Roman numerals.  But what is the symbol ‘0’ for?

     One usage of ‘0’ is to denote nothing, or the absence of something.  In Buddhism, which originated in India, there is this concept of sunyata (from Sanskrit), which means emptiness.  On a stone inscription in a Hindu temple at Gwalior in India, there is a small circle which stands for the digit zero.  Perhaps that was where ‘0’ originated from.  Is life an illusion?  Why then do we use something (a symbol ‘0’) to represent nothing?  Hmmmm .... Anyway, zero as a number is quite convenient to use in everyday life.  If you check up a website of a store for their inventory of jPhone VII mobile phones, and you see ‘0’, you know that they do not have that model of mobile phones in stock.  Without the number zero, they would have to make two separate lists: one for the models in stock and another for those out-of-stock.  By the way, all modern computing, storage and communication depends on 1’s and 0’s in various forms.

     Another usage of ‘0’ is that of a place-holder.  The ancient Chinese had a decimal place-value system for calculations in sand, but they used blank spaces on a grid instead of ‘0’s.  It is difficult to distinguish between say  507 (five hundred and seven)  and  5007 (five thousand and seven), especially if the grid was erased.  As you can see, ‘0’ is used to “hold the forte” in our decimal system.  Without the ‘0’,  507  and  5007  would both collapse to become  57, which means something entirely different.  Similarly, for the decimal  0.0009876,  the leading zeros function as place-holders.  The first significant figure is  9,  the second significant figure is  8,  and so on.  In general, placeholder ‘0’s are needed on the left of the decimal point, as well as to the right of the decimal point. 

Rounding to significant figures

     With the above understanding, here is the easy and sure-fire way to round to a certain number of significant figures:-

Step 1:  Skipping zeros on the left if necessary, identify the first non-zero digit.

Step 2:  From this digit, start highlighting the desired number of significant figures.

Step 3:  Look at the digit just after the last highlighted digit.  If this is 4 or below, do nothing.  If
              it is 5 and above, add one unit to the last highlighted digit.  Carry to the left if
              necessary.

I have highlighted the required significant figures (or digits) in yellow.  After we have identified the first non-zero digit, we just highlight the required number of significant digits, without caring about whether or not anyone of them is zero.  Then we just look at the digit after that to see if we need to round up.

Summary

     To round certain number to  n  significant figures:-

Step 1:  Identify the first non-zero digit from the left. Step 2:  From this digit, start highlighting  n  significant figures. Step 3:  Look at the digit just after the last highlighted digit.  Round up if              this is 5 and above, otherwise do nothing. Always remember to include the place-holder ‘0’s.

Suitable Levels

* Lower Secondary Mathematics (usually secondary 1 ~ equivalent to about grade 7)

* other syllabuses that involve estimation and approximation

[#Critique] Is #pi #infinite?

Article / Resource
The infinite life of pi - Reynaldo Lopes

My Comments
This is a well-produced video in which the team took great care in producing an interesting narrative.  The circles were drawn imperfect and shaky, so as to catch the attention of viewers and to highlight them.  A couple of points they missed out:-

1) p being irrational, does not only have an infinite number of digits in its decimal expansion, the digits are also non-repeating.  A rational number (fraction) like 1/7 also has an infinite number of digits, but they repeat: 1/7 = 0.142857 142857 142857 ...

2) near the end of the video, the animators depicted universe < p.  I find the artistic licence disturbing.  I am sure there is some way to artistically depict the known universe having less number of atoms than number of digits of p.

To answer the question in the title of this article "Is p infinite?": No, but her number of decimal digits is.