Elementary School Maths

Details

Last Updated: Sunday, 20 October 2024 23:06

Published: Tuesday, 31 January 2023 00:39

Hits: 460

Elementary School Maths :

This is Maths starting from grade1 and continuing grade8. Grade 6 to Grade 8 is known as middle school and not elementary school, but until 8th grade, maths being taught is still elementary in nature, so we'll club it all in Elementary maths. In fact all the maths until 8th grade can be grasped by a kid by his 5th grade very easily, and is a 6-12 month effort in elementary school. I would highly advise to complete all 8th grade maths by 5th grade, so that the kid can start learning real high school maths by start of his 6th grade. If your kid is practicing maths sample papers for grade 5, I would highly suggest practicing 8th grade maths papers, as the content is almost the same from 5th grade to 8th grade, it's just that the difficulty level increases a little bit.

This is the sequence of study material from grade 1 to grade 8. Basic Algebra, Basic Trignometry and Basic Plotting are the main areas covered here.

Basic Algebra:

From mathsisfun website, we can start algebra from this pre algebra section: https://www.mathsisfun.com/algebra/index-pre-algebra.html

Then we can move on to algebra 1 section: https://www.mathsisfun.com/algebra/index.html

Numbers:

Natural numbers: numbers starting from 1 to infinity. We don't include 0 since earliest humans didn't know 0 when they started counting. They also didn't know about decimals, fractions, etc. So you can think of natural num as those occuring naturally, i.e complete num with no decimals or fractions. ex: 1, 2, 3, ....

Whole numbers: numbers starting from 0 to infinity. They are whole, no decimals or fractions. ex: 0, 1, 2, 3, ....

Integers: Same as whole numbers except they can be negative. ex: -2, -1, 0, 1, 2, ..

$\text{Rational numbers: N}$start color #7854ab, start text, R, a, t, i, o, n, a, l, space, n, u, m, b, e, r, s, end text, end color #7854ab: : N umbers that can be expressed as a fraction of two integers.

ex: 4/5, 1.45, 7/9, 7/9, 0.777777, √36, etc.

When a number has digits that repeat for ever, we put a bar on the top of repeating numbers to indicate it's repeating. So, 0.77777... repeating for ever is rep as 0.7 and a line on top of 7.

NOTE: For number 0.7777, it looks like it can't be represented by a fraction, as it goes on forever. However, it can be proved to be a fraction as follows:

x=0.7777, 10x=7.77777 => 10x-x=7 => 9x=7 => x=7/9

In fact, any number of form where set of numbers is repeated infinitely can be rep as fraction.

ex: x=2.3475757575... => 100x=234.757575... => 100x-x=234.7575 - 2.347575 => 99x=234.75-2.34=232.41 => x=232.41/99 => x=23241/9900 => This fraction can be reduced further if it has common factors.

Brain Teaser: One very weird and common sense defying number is 0.99999. If we take x=0.9999, then 10x=9.9999 => 10x-x=9 => 9x=9=> x=1. So, the fraction to give 0.999 is 1. But dividing 1/1, we never get 0.999. So, how did 0.999 become 1? We know 0.999 is not equal to 1. In limiting case, it becomes 1. But it's formally proved to be 1, which looks incorrect as dividing 1 by 1 will never give you .9999. Does that mean that .9999 is irrational (see below for irrational numbers), since we couldn't find 2 integers p and q such that p/q=0.99999. Go figure it out !! There is a link here (look at the bottom of the page in the link): https://cs.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.html

Irr$\text{ational numbers: N}$start color #7854ab, start text, R, a, t, i, o, n, a, l, space, n, u, m, b, e, r, s, end text, end color #7854ab: : N umbers that can't be expressed as a fraction of two integers. Thousands of years back, it was thought by the smartest mathematicians that all numbers are rational, as they could not see how a number can't be represented by a fraction. Their thinking was that if any decimal number existed, it can be put on a number line and by choosing higher numerator and denominator for the fraction, they can get closer and closer to it. However they were proved wrong later. These new set of numbers were so unorthodox, they they called them irrational. Number pi was such an example, and was proved rigorously in last 200 years or so to be irrational. Similarly many square roots as √2, √3, etc are irrational. Irrational numbers are totally different class of numbers. All whole numbers and integers fall under the umbrella of Rational numbers.

One quality of irrational numbers is that the decimal numbers don,t ever repeat in any pattern (for ex, if the decimal numbers repeated, we could employ the process for rational numbers to find such a fraction)

ex: Π (pi), √2, √3, √6, etc

There is very simple proof of why √2 can't be rep as a fraction. The proof assumes that √2 = p/q, and then by contradiction concludes that it can't be fraction. Proving pi is irrational is much more difficult.

Basic Arithmetic: 4 kinds of operations allowed on numbers => addition, subtraction, multiplication and division.

3 ways to represent numbers => Integers (both +ve and -ve), Decimal, Fraction (we won't talk about irrational numbers for now). The table below shows all possible combination of 4 operations on these numbers. These operations form the basic of Maths education, so make sure your kid is comfortable with all the operations below. It's not necessary to learn the multiplication table, but it does help to know multiplication table from 1 to 10.

Once all above arithmetic is understood well by the kiddo, there's not much left in Maths algebra.

add/sub rules: -(+ve) = -ve, +(-ve)=-ve, +(+ve)=+ve, -(-ve)=+ve

mult/div rules: +ve mult/div +ve = +ve,  -ve mult/div +ve = -ve, +ve mult/div -ve = -ve, -ve mult/div -ve = +ve, I haven't found a good explanation of why -ve multiplied or divided by -ve becomes +ve ??

distributive property: This is one of the most important properties and difficult one to grasp for kids. A*(B+C) = A*B + A*C

Successive Distribution: An extension of distributive property is applying it successively. Ex: (A+B)*(C+D) => Here we treat (A+B) as one variable "P". then this thing becomes P*(C+D). We can apply distributive property to get P*C+P*D. Now we substitute P back We get (A+B)*C+(A+B)*D. Now we can again apply distributive property to get A*C+B*C+A*D+B*D.

So, (A+B)(C+D) = A*C+B*C+A*D+B*D => This can be extended to any number of variables and any number of nested parenthesis.

Order of operations: Given a complex arithmetic as 2*5+7+ (4+5)/9 + 2, student should be able to know the BODMAS (aka PEDMAS) rule and Left to right rule. Basically first look for precedence of operators, and if operators have same precedence then use left to right rule (i.e when * and / appear in an equation, then left to right rule used to do computation, since * and / are at same precedence). This is just a convention, so that different people don't interpret the same equation differently and come up with different answers. The prudent thing to do with equations is to enclose them in parenthesis, so that it's always clear, which ones you want to do first.

ex: a-(b+c).d+e = a-bd-cd+e

operation on fractions: As explained in the table above, add/sub, mult/div on fractions are same as on decimals. However, few points to keep in mind:

A. (a+b)/c => is same as (a+b)*1/c => same as a/c + b/c (distributive property)

B. c/(a+b) => This cannot be reduced any further. Distributive property applies to numerator, not to denominator. i.e c/(a+b) is not the same as c/a + c/b

• c/a + c/b = c*(1/a+1/b) = c(a+b)/ab = c/(a+b) [(a+b)^2/ab)] => My multiplying both numerator and denominator by (a+b). We see the extra term in sq bracket which is never 1, so c/(a+b) ≠ c/a + c/b

C. (a+b)/(c+d) => is same as a/(c+d) + b/(c+d) => Distributive property applies to numerator here. Denominator remains intact.

Percentage, ratio can now be introduced. They are just different way of writing fractions or decimals. No new concept here. Make sure the student understands that x% is just x * 1/100 (i.e % is just other way of writing out of 100 parts).

Prime numbers: Prime numbers (PN) are numbers which are only divsible by 2 numbers, 1 and itself. 1 is not a PM as it has only divisor: 1 (while PN as per defn needs 2 divisors). 2 is the only PN which is even. PN are very important concept to develop, and a kid should be able to figure out prime numbers for all numbers less than 100. Make sure they know all the prime numbers correctly between 1 to 100. We write all numbers b/w 1 to 100, and then start striking off numbers which are multiples of 2, 3, 5 and 7. This will yield PN as shown below (Maybe ask kids in 5th grade to write a simple program in python to test if a number is prime or not).

All PN < 100: (there are 25 such numbers)

• 2, 3, 5, 7
• 11, 13, 17, 19
• 23, 29
• 31, 37
• 41, 43, 47
• 53, 59
• 61, 67
• 71, 73, 79
• 83, 89
• 97

Composite number: Composite number is defined as number which has more than 2 factors. Basically, it's numbers which ar not prime (with the exception of 1, which has only 1 factor). So, 1 is neither prime nor composite. All other numbers which are not prime are composite.

Co prime numbers: Co prime numbers are numbers that have only one common factor which is 1, i.e they are prime wrt each other. i.r 8 and 15 and coprime since factors of 8 are 1,2,4,8 while factors of 15 are 1,3,5,15. So, by themselves 8 and 15 are not prime, but thay are coprime as a pair as only common factor b/w them is 1. 

Factorization: Factorization refers to expressing a integer as a multiple of 2 or more integers. Important thing to note is that we say integers, which means +ve and -ve numbers can be factors of a integer. Also, 1 and the number itself are always factors of a given number, since 1*number=number.

ex: 36=2*18, here 2 and 18 are factors of 36. To find all factors of 36, we find out all integers which completely divide the number (without leaving a remainder). So, in this case, 1,2,4,6,12,18 and 36 are all factors of 36. On top of this, all -ve counterpart of these numbers are also factors of 36, since 36=(-2)*(-18). So, -1,-2,-4,-6,-12,-18,-36 are also factors of 36, though we don't usually write -ve numbers as factors (not useful). But strictly speaking, they are factors.

Prime factorization: prime factorization can also be grasped well by a kid in 5th grade. We saw how to find factors of a integer. Of al the factors of a integers, the factors which are prime are called prime factors. It should be stressed that any number can be factored into prime numbers, and there is one and only one unique way of factorizing a number into prime numbers. Most of the smaller numbers can be factorized into prime numbers 2, 3 and 5.

ex; 36 = 2*2*3*3 = 2^2 * 3^2 (integers 2 and 3 are prime factors here)

We can find all factors of a given number from it's prime factors, since all other factors are formed from these prime numbers. Let' say a number is factored into prime numbers as follows:

N = p1^e1 * p2^e2 * ... * pn^en where N is a given +ve integer whose prime factors are p1,p2,..,pn, and their corresponding powers are e1,e2,..,en.

All possible factors with p1 prime number are 1, p1, p1^2, p1^3, ...., p1^e1

Similarly all possible factors with p2 prime number are 1, p2, p2^2, p2^3, ...., p2^e2

All the way to pn are 1, pn, pn^2, pn^3, ...., pn^en

We can line up all of these prime factors into a column, with 1st col being 1, p1, p1^2, ..., p1^e1,  2nd col being 1, p2, p^2, ..., p2^e2, and so on. Let's try to figure out all possible factors of N (not just prime factors):

• From first row of all col except the last one,we get (en+1) factors: 1st factor of N is 1st row of all col = 1*1*1...*1 (N times) = 1, 2nd factor = 1*1*1...*1*pn = pn, 3rd factor = 1*1*1...*1*pn^2 = pn^2, ... en th factor= 1*1*1...*1*pn^2 = pn^en = pn^en
• Now if we take second last col, we get (e_n-1 + 1) factors for each factor we got above.
• If we continue this way, the very first col has (e1+1) factors.
• So, total number of all factors possible is (e1+1)*(e2+1)*....*(en+1)

As an ex: 162=2^1*3^4 => This has total (1+1)*(4+1)=10 factors which are 1*3, 1*3^2, ..., 1*3^4 for 1st row (total 5 factors possible), then 2*3, 2*3^2, ..., 2*3^4 for 2nd row (total 5 factors possible). There are no other factors possible, as every factor is accounted for. None of the factors are repeated here, as they are all made from prime numbers with different exponent to each of the prime numbers. So, total number of unique factors possible is 10.

This is all the reason, why prime factors so important in factorizing any number. First they allow us to compute all possible factors, and secondly they allow us to have a unique representation of any number as multiple of other numbers.

HCF, LCM: HCF can be taught to help the kid learn how to reduce fraction to it's lowest ratio, while LCM can be taught to help him learn how to add or subtract fractions when denominators are different. There are techniques based on prime factorization that help them determine HCF and LCM. Look up on Khan Academy. Of course what I've seen is that kids don't really use LCM HCF for these fractions. They use it only when they are directly asked HCF or LCM of 2 numbers. My son still reduces fractions the long way (by dividing it by a small number as 2,3 etc and then repeatedly dividing it), instead of using HCF !!

Fraction reduction: Fraction reduction can be done by repeatedly dividing numerator and denominator by same number, until they are prime wrt each other. This is an an important area that kids will need to be comfortable with, since they will need to identify equivalent fractions. One simpler way to reduce fractions is to do prime factorization of numerator and denominator, and then cancel the common terms.

ex: 52/72 = 2*2*13/2*2*2*3*3 = 13/18

Other way to reduce fractions by using HCF.

ex: 52/72 => Find HCF of (52,72). Then divide both 52 and 72 by their HCF.

Mean, Median and Mode:

Mean, Median and Mode are simple concepts.

Mean: It is the avg value of a given sample. It's calculated by summing the value of all the samples divided by the number of samples.

ex: If kids in a class have scores of 70, 90, 10 and 30, then Mean = (70+90+10+30)/4 = 200/4 = 50.

Quantities which are formed division of other quantities can't be averaged by just dividing it by number of samples.

Ex: density = weight/volume.  Let's say we have sample A with density of 2g/cm^3 and sample B with density of 4g/cm^3. If we mix them in ratio 1:1 by weight, what is the mean density of the mixture. It seems like mean density should be the avg of 2 densities, so mean should be 3g/cm^3.

Mean density = Total_weight/Total_volume => That is how any mean is defined.

Let's we take x g of sample A, then sample B is also x g. So, total weight=2*x g. Total volume = x/2 cm^3 + x/4 cm^3 = x(1/2+1/4)=3*x/4, So mean density = 2*x/((3/4)*x) = 8/3 = 2.66g/cm^3 and 3g/cm^3 as expected. However, if they were mixed in 1:1 by volume, then

Median: Median is the mid point of a sample where half the samples values are below that number, and the other half are above that number.

ex: For above ex, our median has to be a number which is greater than 10, 30 but less than 70 and 90. We may choose such number to be any number as 40, 50, 60, etc and they will all be median. Generally we choose the avg of 2 middle numbers as median, so here median=(30+70)/2= 50.

Mode: Mode is the easiest. It's the number in the sample that is repeated the most times.

ex: In above ex, since each number is repeated only once, each number is a mode. However, we has sample, 70,90,10,30,70, then 70 will be the mode since it's repeated 2 times.

Powers:

Powers are just an extension of multiplication, atleast for powers of integers (i.e where exponent is an integer, and base can be any decimal). Any number in form x^y = x*x* ... *x (i.e x repeated y times). (x^y)^z is same as x^(y*z) as x^y is repeated z times. x^(y^z) is different than (x^y)^z as in x^(y^z), we first calculate a=(y^z) and then do x^a.

Below table shows various x^y possibilities where x, y can be any real number. Here x is called the base, and y is called the exponent. Any fraction can be treated as integers with a division or as decimals. So, there is no separate table for fraction base or fraction exponents.

When y is a decimal (i.e exponent is decimal), it's hard to understand what it means. We'll understand it in the table below.

 NOTE: whenever exponent is a real number, it's not easy to solve such powers, and we resort to trial and error. So, the 3rd col of above table where exponent is real is not really expected to be solved by students w/o a calculator.

Operations: All basic operations of add, sub, mult and div can be done on powers too.

Add: ex: 2^4 + 2^5 = 2^4 *(1 + 2^1) = 3*(2^4) => explain distributive property in solving these

Sub: ex: 3^5 - 2^4 => These can't be solved by factoring out common terms as 2 and 3 are prime wrt each other.

Mult: 2 kinds: 1 with same base, and other with same exponent

Same base: ex: 2^5 * 2^6 = 2^(5+6) = 2^11 => exponents add up during multiplication, of same base

Same exponent: ex: 2^5 * 3^5 = (2*3)^5 = 6^5 => bases an be multiplied, if same exponent. Converse also true: 6^5 = (2*3)^5 = 2^5 * 3^5

Div: ex: 3^6/3^5 = 3^(6-5) = 3^1 => exponents subtract during division, of same base. this is the reason why x^0 is always 1, as X^n / X^n = 1 => X^(n-n) = 1 => X^0 = 1

Also -ve exponents can be understood the same way. 3^4 / 3^6 = 3^(4-6) = 3^(-2). However 3*3*3*3/(3*3*3*3*3*3) = 1/(3*3) = 1/(3^2) => 3^(-2) = 1/(3^2)

Exponent: Here, we raise an exponent to further exponent. In this case, the exponents multiply.

ex: (2^3)^5 = (2^3) multiplied 5 times = 2^(3*5)=2^15.

So, (x^y)^z = x^(y*z)

Brain Teaser: What is 0^0. This is a question which has found various different answers from Mathematicians. We know 0^y where y≠0 is 0. We also know that x^0 where x≠0 is 1.

In limiting case y->0 for the eqn 0^y, we get 0^y->0, For other eqn x^0, in the limiting case x->0, we get x^0->1. So, we get 2 different answers which makes our task difficult. What if we take limit x->0 and y->0 simultaneouly for x^y. then we see that it approaches to 1. this video shows it: https://www.youtube.com/watch?v=r0_mi8ngNnM&t=701s

However, it looks very weird, since 0 raised to something like 0 gets us to "1" and NOT "0". Defies common sense, doesn't it. The problem is if we take limit some other way, then we get different answer of either 0 or 1, depending on whose limit we take, x or y. This link tries to explain on what the value is: https://cs.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.html

In nutshell, looks like for all practical purposes 0^0 =1, but if you want to be accurate, then it's undefined or indeterminate.Just like 0/0 is undefined or indeterminate. We may say that x/y, where we start taking limit of x->0 and y->0, and prove that x/y=1, but that's incorrect. We can argue the same way that 0^0 is not 1, but indeterminate. The answer is nobody knows !!

Expressions:

So far, we talked in terms of constant numbers, but we can also have all arithmetic operations on variables. i.e var x multiplied by var y = x.y. We can have multiplication/division of variables (x.y/z) or addition/subtraction (x+y-z). These equations written in terms of variables are called algebraic expressions. They have constants and coefficients too. A variable can take any value, it is not fixed but a constant is a fixed value.

ex: xy + 2ab + 4 => This is an algebraic expr. x,y,a and b are var here. 4 is a constant. 2 is also a constant and is called the coefficient of the term a.b.

There are many therems and other cool maths properties based on poly and their degree. More complex ones are part of high school maths (shown in "high school maths" section). Some basic eqn with poly are shown next.

Basic equations: If we have 2 variables x,y, we can write reduced form equations for powers of 2.

1. (x+y)^2 = x^2 + 2*x*y + y^2 => trinomial with deg=2 (degrees are explained in high school maths section)
2. (x-y)^2 = x^2 - 2*x*y + y^2 => trinomial with deg=2
3. (x+y)*(x-y) = x^2 - y^2 => binomial with deg=2
4. (x+y)^3 = x^3 + 3*x^2*y + 3*x*y^2 + y^3 => Quadnomial with deg=3
5. (x-y)^3 = x^3 - 3*x^2*y + 3*x*y^2 - y^3 => Quadnomial with deg=3

Square root: square root of any number is such a number which when squared gives that number, i.e square root of 9 is 3, since 3*3=9. square root is denoted by √ .This is same as where exponent is 1/2. i.e √9 = (9)^(0.5) = (9)^(1/2)

square roots are encountered very often in solving many kinds of equations, not so much for cubic root or higher powers of root. So, we try to keep a table for square root of numbers handy. We should be able to figure out square root of any number under 100. This is a good exercise for student. First we start with square root of numbers which have integer square root.

ex: √1 = 1, √4 = 2, √9 = 3, √16 = 4, √25 = 5 and so on

Then square root of prime numbers. We will keep a table handy. The way to find square root of a prime number is difficult, but we can use indirect way to verify square root of prime numbers, by squaring the answer and checking if it comes close to the prime number itself.

ex:

√2 = 1.414 (if we square 1.414, we should get 2, i.e 1.414 * 1.414 gets us very close to 2, so the answer is correct)

√3 = 1.732

√5 = 2.23

√7 = 2.64

and so on

Now, we can calculate square root of any non prime number by factoring it into prime numbers or numbers which have a integer square root.

ex:

√6 = √(2*3) = √2 * √3 = 1.4 * 1.7 =

√8 = √(2*2*2) = √2 * √2 * √2 = 1.4 * 1.4 * 1.4 => However there is an easier way by noting that 8 can be factored into a number which has an integer square root. So, 8 = √(4*2) = √4 * √2 = 2*1.4=2.8

√10 = √(5*2) = √5 * √2 = 2.2 * 1.4

and so on for larger numbers.

Best way to find square root of any number is to write it as prime factors, and then group them in pairs. Ones without any pair, remain as square root, while others being in pair come out of square root.

Cubic root and other higher nth root:

These are usually not expected to be solved by hand, and a calculator can be used. However for simple integer ones, they should be solved by student.

ex: cubic root of 8 = ³√(8) = 8^(1/3) = a number which when multiplied 3 times gives 8. such a number is 2. so, cubic root of 8 is 2.

ex: 4th root of 16 = ⁴√(81) = 81^(1/4) = a number which when multiplied 4 times gives 81. such a number is 3. so, 4th root of 81 is 3.

Equations:

solving equations is one of the skills that should be taught in elementary school.

ex: Find a number such that when it's multiplied by 3 and added with 7, it becomes 19. The student may first try to solve it by trial and error, and come with an answer 4. Then introduce concept of variable "x", and show him how this equation 3*x + 7 = 19 can be solved to give an answer of 4.

Equations of 1 variable: Here there is only 1 variable "x" that we are solving for.

Linear equations: Here x doesn't have higher powers to it, and is relatively easy to solve. These are expected to be solved by 7th or 8th grade student. You add/subtract or mult/div by same thing on both sides, until you get "x" by itself on one side. It's called linear, because if you draw it on a plot, it turns out to be linear or a straight line. Linear equations have only 1 solution for x. However, you can also have cases where there is no solution or there are infinite solutions.

ex: 4*(x+2) +7 = 3*x - 10 => gives 1 solution for x, which is x=-25. No other x will satisfy this equality

ex: 4*x + 2 = 4*x + 3 => Here 4*x cancels on both sides, and for this equality to be true, 2 has to be equal to 3, i.e 2=3. Since that's not possible, there is no solution

ex: 4*x + 6 = 2*(2*x+3) => Here 4*x cancels on both sides, and we get 6 on both sides, i.e 6 =6. since this is always true, this equation is satisfied for any x, so it has infinite solutions.

Inequality:

As we have equality in equation, we can have inequality (i.e < or >) in equations too. These can be solved the same way as above, just that it has infinite solutions, as compared to equality which usually has just one solution.

ex: 6*x+7<2*x+11 => 4*x<4 => x<1 => This implies that any real number less than 1 is the solution to this inequality. We can plug in x=0 to check if our answer is correct. We that we get 7<11, which is correct. Next we put x=2 which is not a solution., so our inequality should be incorrect. We get 19<15 which is incorrect as expected. So, our solution looks correct.

NOTE: one weird case with inequalities happens when you multiply or divide the 2 sides by a -ve number. i.e if x<3, and we multiply both sides by -1, then it becomes -x<-3. However, this incorrect, as x<3 => x can be 2,1,0,etc. So, for x=2, -x=-2, but -2 is not less than -3, so -x<-3 is incorrect. This happened because we multiplied both sides by a -ve number. Whenever, we multiply or divide both sides by a -ve number, we need to change the direction of inequality sign. So, if x<3, then if we multiply both sides by -1, we need to do -x>-3 (i.e < became >).

CAUTION: We don't change the direction of inequality sign when doing + or -. We do it only when doing * or / by a -ve number. If we find this confusing, then we should do + or - to achieve. Let's see this by an example:

ex1: -2*x+5<7 => -2*x<2 => x>-1 (by reversing the direction from < to > when dividing by -2)

ex2: -2*x+5<7 => -2*x<2 => -2*x+2*x<2+2*x => 0<2+2*x => -2<2*x => -1<x => x>-1 (here we got the same answer but we used + and - instead of divide by -ve number, so we didn't have to do any reversal of inequality sign. sometimes this is easier to understand, and I recommend solving inequalities this way)

Equations of 2 variable: Here there are 2 variables "x" and "y" that we are solving for. Here too, we can have higher powers of x and y.

Linear equations: Here x and y have powers of 1, i.e they are both linear.

ex: find 2 numbers whose sum is 5. this can have infinite solutions, as one of the answers in 2 and 3, while other solution is 4 and 1, and many more decimal and integer solutions. However, if we add 1 more constraint to the numbers that their difference has to be 1, then there is only 1 solution which is 2 and 3. So, for solving an equation in 2 variables, we need at least 2 equations to solve it uniquely.

There are 2 ways to solve these kind of equations:

1. Find y in terms of x, by using 1st eqn and then substitute for y in 2nd eqn.

ex: x+2y=7, 2x+5y=15 => Here from 1st eqn, y=1/2*(7-x). Now substitute y in 2nd eqn. i.e 2x+5*1/2(7-x)=15 => solve for x in this eqn, and then find y by using either 1st or 2nd eqn

2. Here we try to cancel x or y by multiplying 1st or 2nd eqn such that one of the variables has the same coefficient, and then add or sub the 2 eqn to cancel the variable out.

ex: x+y=7; x-y=3 => Add both RHS and LHS, which cancels out y, giving x=5, then solve for y=2.

More Equations: We'll learn solving more complicated eqn in high school maths section

Basic Geometry:

Good material on geometry is on this link: https://www.mathsisfun.com/geometry/index.html

There are different kinds of 2D and 3D figures. We'll look at 2D figures here => Line, triangle, quadrilateral, circle, pentagon, etc.

Quadrilaterals => Any 2D figure enclosed with 4 sides which are straight lines is a quadrilateral. Ex are: square, rectangle, etc.

Link showing different kind of Quadrilaterals => quadrilaterals

There are very few fundamental concepts in Geometry. All known theorems are derived from these few fundamental theorems. We'll learn these fundamental ones, and derive everything else from these fundamental theorems or concepts.

1. concept of angles: straight line has 180 degrees, perpendicular lines have 90 degrees, and around a line, the total angle is 360 degrees. This can be thought of as 1st concept or theorem.

2. Concept of Lines: These are straight lines.

• Intersecting lines: They have opposite angles the same. They are known as opposite angles. The angles next to each other add up to 180 degrees and are known as complimentary angles. This can be thought of as 2nd concept or theorem.
  
• Parallel lines: They have transversal line crossing them. In this case, corresponding angles are equal. This can be thought of as 3rd concept or theorem.
  

3. Triangle:

Theorem for sum of angles: Sum of 3 angles of triangles is 180 degrees: proof by drawing parallel line. This can be thought of as 4th concept or theorem.

Find remaining angles of a triangle, given 2 interior angles or outside angles. This is a good exercise.

Rght angle triangle: Right angle triangles are an important category of triangles, as they have many special properties.

pythagoras theorem (c^2 = a^2 + b^2) and it's proof (where a, b and c are sides of a right triangle). 

Proof: Drop a line at right angle from vertex to hypotenuse. It may be proved that 3 triangles formed are similar to each other (if one angle is X, other angle is (90-x). This is true for all 3 triangles). This similarity property may be used to calculate ratios of sides, and find height of triangle. Rearranging it yields Pythagoras Thm. Also calculating area of triangle using all 3 sides as base, and multiplying it by height yields eqn for Pythagoras Thm.  More proofs here: https://en.wikipedia.org/wiki/Pythagorean_theorem

- Sine, cosine and tangent are advanced topics for high school. We'll learn these later in "high school maths" section.

Before we start with any problems on triangles, we should learn how to draw triangles using ruler, compass and protractor. There are many kinds of triangles possible:

Based on above table, only the cells in green are the ones where we have just the right info to draw a unique triangle. In all other cases, we have either insufficient info, or extra info.

Impossible triangles: Triangles which are not possible to draw is something that students should know. Third side of a triangle should always be less than the sum of other 2 sides of triangle, and it should also be more than the difference of the other 2 sides of triangle.

ex: Given triangle ABC, |AB-BC| < AC < AB+BC => Here we take the +ve diff of the sides, hence modulus

4. Perimeter: Perimeter of square, rectangle, triangle, parallelogram, etc is just the sum of the sides. Perimeter of circle is called circumference and is shown to be 2*Π*R, where R is the radius. The proof is based on calculus.

5. Area: Area of square, rectangle, triangle, parallelogram (explain how area of triangle is 1/2 of area of rectangle, and how area of parallelogram, is same as that of a rectangle with same height). Area of Circle is Π*R^2, where R is the radius. The proof is based on calculus.

6. Volume: Volume of cube, rectangular pyramid, (volume cylinder, cone, sphere are high school topic. Proof of these require calculus).

Basic Plotting:

some introductory material on plotting is on this link: https://www.mathsisfun.com/data/index.html

Equation of Straight Line:

Introduce X axis and Y axis, and how to find where a coordinate lies.

Then introduce students to an equation of line, which is of form y = m*x+c. Here m is the slope and c is the y intercept. Show him how y=m*x is the same line as y=m*x+c, but just shifted by c units up or down. y=m*x line always passes thru origin, as x=0, yields y=0. In many textbooks, this eqn is rep as y=m*x+b (i.e y intercept is written as "b" instead of "c").

NOTE: slope "m" is taken as positive for going up a hill left to right (i.e push up a hill or push means +ve). Line going down the hill from left to right is treated as -ve slope. When 2 points are used to find a slope, then you don't have to worry about -ve or +ve slope, as using (y2-y1)/(x2-x1) will automatically give you the correct sign of the slope.

Eqn of staright line can be written in different forms as shown below. One form is more convenient than the other depending on what is given (i.e m, c, coordinates (x1,y1), etc).

1. Slope intercept form: This is the form we use when m and c are given. This is what we saw above: y=m*x+c. This is the most widely used form, and easiest to visualize.
2. Point Slope form: This is used when we have slope m given, as well as one of the coordinates given. Then (y-y1)=m(x-x1). If m is not given, but instead, other coordinate (x2,y2) given, then we can use the 2 coordinates to find m=(y2-y1)/(x2-x1). This is also referred to as 2 point form.
3. Standard form: This is of the form Ax+By=C. This eqn is actually derived from other form called "intercept form", where the x-intercept and y-intercept are given as (a,0) and (0,b). Then we can use 2 point form to write: y-0=(x-a)*((b-0)/(0-a) => y=(x-a)(-b/a) => x/a+y/b=1. We write this in form b*x+y*a=ab or Ax+By=C which looks simpler than writing in x/a+y/b=1.

Most important is to show that the linear equations of 1 variable that we introduced in "equations" section above, can also be solved by plotting the straight line, and looking for the coordinate where y=0. That X coordinate is the solution of that linear eqn.

The next section on "high school maths" has plots for straight line, quadratic functions and exponential functions.

Basic Transformations: Translation, Rotation and Reflection:

One very popular topic in Maths and IQ/Gifted Talented tests or any general Brain test is Translation, Rotation and Reflection. A figure is rotated, reflected and /or translated, and kids are asked on how would the final figure look like. This can get very complicated depending on axis across which the required operation is done. Fortunately, for elementary and high school maths, very simple operations are done, which can easily by figured out by using formula below

Translation:

Translation is the easiest. We just move the figure horizontally left or right by said number of units. We just need to add or subtract that number of units to x coordinate. Sometimes the figure is moved vertically up or down. in this case, we add or subtract that number of units to y coordinate. if we translate in some other direction (i.e not horizontally or vertically, but in a slanted direction, then we have to first move into x direction horizontally and then y direction vertically.

IMP: For any function f(x), we can obtain a plot for f(x+b), where x is replaced by "x+b" in the function f(x), the new plot of f(x+b) will be shifted by "b" towards the left compared to the original function f(x). This is easy to see, as whatever f(x1) was for a given x1, now we get the same value of f(x+b) for x=x1-b, so that x+b becomes x1-b+b=x1, so the whole function is shifted.

Similarly given f(x), the plot of f(ax) will be expanded/compressed version of f(x) where x axis is expanded/compressed, where whatever value of f(x1) was, will now be at x=x1/a. NOTE, the y values of function don't change/ Where they occur on x axis changes.

So, for any modified function f(ax+b), we can just use the above 2 observations to convert it to f(a*(x+b/a)). Now we use the shift rule to shift functio left by b/a, and then scale the x-axis by a, to get the final plot of f(ax+b).

Reflection:

Reflection is next easy one after translation. Easiest way to solve reflection problem is to look at vertices of figure to be reflected. Start with vertex 1, drop a line perpendicular to the reflecting surface starting from that vertex, and then extend that line by the same length inside the reflecting surface (on the other side). That way you get mirror reflection of that one vertex. Now repeat the process with other vertices. Finally connect all these reflected vertices to get the reflected figure. It doesn't matter whether reflection is across x axis, y axis or any slanted line. Same procedure gets applied.

Most common reflection: Reflection around y=x line. Here points (x,y) get reflected to become points (y,x)

Another teaser question: Reflection around line: y=mx+c. This is more general case of y=x case above (i.e m=1, c=0). This has to be solved to figure out what the new coords will be after reflecting.

Rotation:

Rotation is the hardest of all transformations. It requires a lot of visualization, and easily gets confusing. You can rotate any object by any degree around any point. Rotation by 90 degrees (clockwise and anticlockwise), and 180 degrees (clockwise and anticlockwise rotations for 180 degrees are the same) are very common, and those are the ones we'll discuss below. It's very important to know which point are we rotating the figure around. The rotated shape will appear in different places depending on which point is it rotated around.

1. Rotation by 90 degrees around origin: Let's consider a point (x1,y1) in 1st quadrant (i.e where both x and y are +ve). Let's make a right angle triangle for that point, with base=x1 and height=y1. If we rotate this triangle 90 degrees clockwise, then the triangle rotates to a new position, and it's base now has length y1 and height has length x1. So, the new coordinate after 90 degrees rotation is (y1, -x1). Note the sign changed since y coordiante of new triangle is now -ve. Similarly when we rotate 90 degrees anticlockwise, the result is the same except that x coordinate of new triangle becomes -ve, so new coordinates are (-y1, x1).

We repeat the same process with other vertices, and then connect them to get the rotated figure.

2. Rotation by 180 degrees around origin: This is very simple. You consider a point (x1,y1) in 1st quadrant (i.e where both x and y are +ve). If we rotate this point 180 degrees clockwise or anticlockwise, the the new point becomes (-x1, -y1).

3. Rotation around any point (X0, Y0): So far we looked at rotation around origin. Rotation around any other point looks complicated, but it's actually very simple. You consider a rigid bar from point (X0, Y0) which is attached to the figure to be rotated. Now we rotate this bar by 90 degrees or 180 degrees, and with that the figure also rotates. This becomes the new coordinates of the figure.

To find out new coordinates for any point (x1,y1), we repeat the same exercise as before. However, now we shift our origin to (X0, Y0). Then we get new coordinates for (x1,y1) as (x1-X0, y1-Y0). Since we know how to do rotation around origin, we can now rotate all the vertices of the figure and get the new vertices.

NOTE: We treated (x1, y1) or (X0,Y0) as positive, but above formula are true irrespective of whether (x,y) are +ve or -ve.  So (-3,2) with 90 degree clockwise rotation will become (2, -(-3))=(2,3) while for 180 degree rotation, it will become (-(-3),-2)=(3,-2)

That's the end of Elementary Maths. That wasn't too hard for you, was it