Math Homework Help

Find this two digit number

In a two digit number, the ten’s digit is three times the units digit.

Labels: Algebra, Problems and Solutions

A football problem

Labels: Algebra, Math Homework Help, Problems and Solutions

Fundamental Theorem of Arithmetic

Fundamental Theorem of Arithmetic :

Every composite number can be expressed ( factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a product of primes. Actually it says more. It says that given any composite number it can be factorised as a product of prime numbers in a‘unique’ way, except for the order in which the primes occur. That is, given any composite number there is one and only one way to write it as a product of primes,as long as we are not particular about the order in which the primes occur.

So, for example, we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other possible order in which these primes are written. This fact is also stated in the following form:

The prime factorisation of a natural number is unique, except for the order of its factors.

In general, given a composite number x, we factorise it as x = p1p2 ... pn, where p1, p2,..., pn are primes and written in ascending order

If we combine the same primes, we will get powers of primes.

Once we have decided that the order will be ascending, then the way the number is factorised, is unique.

The Fundamental Theorem of Arithmetic has many applications, both within mathematics and in other fields.

Labels: Algebra, Arithmetic, Theorems

Euclid’s Division Lemma

Euclid's Division Lemma
Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0≤r
This result was perhaps known for a long time, but was first recorded in Book VII of Euclid’s Elements. Euclid’s division algorithm is based on this lemma.
Euclid’s division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers. Recall that the HCF of two positive integers a and b is the largest positive integer d that divides both a and b.
Let us see how the algorithm works, through an example first. Suppose we need to find the HCF of the integers 455 and 42.

We start with the larger integer, that is, 455.
Then we use Euclid’s lemma to get 455 = 42 × 10 + 35.
Now consider the divisor 42 and the remainder 35, and apply the division lemma to get 42 = 35 × 1 + 7.
Now consider the divisor 35 and the remainder 7, and apply the division lemma to get 35 = 7 × 5 + 0.

Notice that the remainder has become zero, and we cannot proceed any further. We claim that the HCF of 455 and 42 is the divisor at this stage, i.e., 7. You can easily verify this by listing all the factors of 455 and 42.Why does this method work? It works because of the following result. So, let us state Euclid’s division algorithm clearly.To obtain the HCF of two positive integers, say c and d, with c> d, follow the steps below:

Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤rStep 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

This algorithm works because HCF (c, d) = HCF (d, r) where the symbol HCF (c, d) denotes the HCF of c and d, etc.Euclid’s division algorithm is not only useful for calculating the HCF of very large numbers, but also because it is one of the earliest examples of an algorithm that a computer had been programmed to carry out.

Remarks :
1. Euclid’s division lemma and algorithm are so closely interlinked that people often call former as the division algorithm also.
2. Although Euclid’s Division Algorithm is stated for only positive integers, it can be extended for all integers except zero, i.e., b ≠ 0.

Labels: Algebra, Arithmetic

Some Properties Of Ratios

1) Let a:b>c:d and c:d be two ratios. Then,

i) a:b> c:d, if ad>bc,

ii) a:b

iii) a:b = c:d, if ad=bc

2) A ratio a:b is called a ratio of

i) greater inequality ifa>b,

ii) less inequality if a < b

iii) equality if a=b

3) If the same positive quantity is added to both the terms of a ratio of greater inequality, then the ratio is decreased.

4) If the same positive quantity is added to both the terms of a ratio of less inequality, then the ratio is decreased.

5) If the same positive quantity is subtracted to both the terms of a ratio of greater inequality, then the ratio is increased.

Roots of a Polynomial

A root or zero of a function is a number that, when plugged in for the variable, makes the function equal to zero. Thus, the roots of a polynomial P(x) are values of x such that P(x) = 0.

The Rational Zeros (Roots) Theorem

We can use the Rational Zeros Theorem to find all the rational zeros of a polynomial.

1. 
   Arrange the polynomial in descending order


2. Write down all the factors of the constant term. These are all the possible values of p.


3. Write down all the factors of the leading coefficient (The coefficient of the first term of a polynomialwhen writing in descending order.)
   These are all the possible values of q.


4. Write down all the possible values of p/q . Remember that since factors can be negative p/q, and - (p/q)must both be included. Simplify each value and cross out any duplicates.


5. Use synthetic division or remainder theorem to determine the values of p/q for which P(p/q) = 0. These are all the rational roots of P(x).
   

Example:

Find all the possible rational roots of

Labels: Algebra

Remainder Theorem and Factor Theorem

Remainder Theorem
Let p(x) be any polynomial of degree n>0 ,and a any real number. If p(x) is divided by ( x-a), then the remainder is p(a).

The Remainder Theorem can be proved as follows.

Proof
Let us suppose when p(x) is divided by ( x-a), the quotient is q(x) and remainder is r(x).

So we have,
p(x)=(x-a) q(x)+r(x), where r(x)=0 or degree of r(x)< degree of x-a.
Since degree of ( x-a) is1, either r(x)=0 or degree of r(x)=0
So r(x) must be a constant,say r.

Thus for all values of x,
p(x)=(x-a) q(x)+r ........(1), where r is a constant.

In particular, when x=a,
p(a)=0.q(x)+r
=0+r
=r
Hence the theorem.

Factor Theorem
Let p(x) be a polynomial of degree n>0. If p(a)=0 for a real number a, then (x-a) is a factor of
p(x). Conversely, if (x-a) is factor of p(x), then p(a)=0.

Proof

First part:
Let p(a)=0
Then by remander theorem, r=0
So equation (1) becomes
p(x)=(x-a) q(x)
==>(x-a) is a factor of p(x).

Second Part:
By remainder theorem,

p(x)=(x-a) q(x)+r
ie. p(x)=(x-a) q(x)+p(a)
Since (x-a) is a factor, p(a) must be zero.
This proves the theorem.

Example
Find the remainder when p(x)= x^2 +3x+1 is divided by x+1.
Determine whether( x-2) is a factor of p(x) or not.

Solution
x+1= x-(-1)
[Always check whether the divisor is in the form of (x-a)or not. Otherwise rewrite that in the form of(x-a)]

So, here a=-1
There fore by remainder theorem, the required remainder is
p(a)= p(-1)
=(-1)^2+3(-1)+1
=1-3+1
=-1

We know, by factor theorem,

if (x-2) is a factor of p(x), then p(2) must be zero.

Here p(2)=2^2+3(2)+1
=4+6+1
=11 which is not equal to zero.
So (x-2) is not a factor of p(x).

Labels: Algebra

Value of a Function

Let f:A-->B be a function. Let y0 be an element in B. then y0 is called a value of f provided there is some element, x0 in A, such that y0 = f(x0); that is, y0 is a value of
the function f if it corresponds, with respect to the rule of f, to some x0 in the set A = Dom(f).

Example
Find the value of the following function when x=-2

Find the Domain and Range of this function.

Labels: Algebra, Functions

Range and Domain of Functions

Yesterday we have learned what is a function.
Today let's discuss about range and domain of a function.



Answers


Try to do more problems from your text.

Labels: Algebra, Calculus, Definitions, Functions

Functions

Today we can discuss a topic from functions.

Definitions:

Function
Let A anb B be two non empty sets. A function "f" from a set A to a set B is a rule so that to each element x in A there corresponds exactly one element y in B, under f ,then we say that f is a functin from A to B and write
f:A -> B

y is called the image of x under f and is denoted by f(x). x and y are respectively called the independent variable and the dependent variable. We also say that y is a function of x and write
y=f(x)

Examples:

1. 
   In the family of circles, the area A of the circle is a function of radius r of the circle.
   Here radius r is the independent variable and area A is the dependent variable
2. 
   The speed of a chemical reaction increases 2 times with the addition of every 5 milligrams of a catalyst. Here the amount of catalyst is the independent variable and speed of the chemical reaction is the dependent variable.
   
   

Labels: Algebra, Calculus, Definitions

10 Trees Problem

There are 10 trees spaced out equally along my street with one at each end. If I run at a steady speed from one end of the street, I can reach the 5 th tree in just 5 seconds.

How long would it take me to run down the whole street at the same pase?
( No, it is not 10 seconds)





Did you get the answer?

When I reach the 5 th tree I cover 4 trees.
So,time taken to cover 4 trees is 5 seconds.
So, time taken to cover 1 tree is 5/4 seconds.
When I reach the 10 th tree I cover 9 trees.
So, time taken to cover 9 trees( and this is same as the time taken to run down the whole street) is 9 multiplies 5/4
ie 45/4
or
11.25 seconds

Labels: Algebra, Puzzles

Operations with Fractions

Addition and Subtraction

When adding or subtracting fractions, if the denominators are same, simply add or subtract the numerators and write the denominator given.

If the denominators are different, convert each fraction so that its denominator is equal to the Lowest Common Multiple of the denominators of the given fractions. This means multiplying both the numerator and denominator by the same term. For example, say the LCM is 24 and one fraction in the problem is 5/6, then to convert that fraction to denominator of 24 you must multiply numerator and denominator by4, so the fraction becomes 20/24. You follow this same procedure for each term, add or subtract as indicated by the sign of each term and divide the total by the LCM

For example:


Multiplication of Fractions is straightforward. You just multiply the numerators and multiply the denominators and then reduce the fraction to its lowest term, if possible

When we divide Fractions, we actually multiply the numerator by the reciprocal of the denominator. A reciprocal is a fraction turned upside down. For example, 3/4 divided by 5/6 = 3/4*6/5 = 18/20 = 9/10

Labels: Algebra, Arithmetic

Help with fractions

It appears that many students, even some students of grade 9 and grade 10 are struggling with fractions. So I have prepared this lesson.Not in a thorough but simplified manner.

A term which is of the form a/b is called a fraction. The top Numbers a’ is called the numerator and the bottom number ‘b’ is called the denominator.

Proper fraction: If the numerator is less than the denominator, it is called a proper fraction
Example:1/2,4/7

Improper Fraction: If the numerator is greater than the denominator, it is called an improper fraction.

Example:5/2,9/7

Mixed Fraction: - A term consisting of an integer (5) followed by a proper fraction (2/3), written as 5 2/3 is called a mixed fraction.

see you tomorrow

Labels: Algebra, Arithmetic