Presentation on theme: "Why do rational numbers have either repeating or terminating decimal expansions?"— Presentation transcript:
Why do rational numbers have either repeating or terminating decimal expansions?
When you have finished this presentation you should understand: 1.Why rational numbers MUST have either repeating or terminating decimal expansions 2.How to convert a decimal into a fraction, including o repeating decimals o a fractional approximation for an irrational number 3. The math behind why these techniques work
Rational – A number is rational if and only if it can be expressed as the quotient,,of two integers, with the denominator p not equal to zero. This fraction will produce a decimal with either a repeating or terminating decimal expansion. – 1, ½, 0.49, 0.33 Irrational – any real number that cannot be written as a simple fraction, including infinite and non-repeating decimals. – π,, 0.123459382958601923...
Terminating Repeating What makes a rational number have either a terminating or repeating decimal expansion?
When dividing (as shown to the right) with p
"name": "When dividing (as shown to the right) with p
Take. If we assumed 10 goes into 21 one time we would have a remainder of 11 because 21=(10)(1)+11 But 11 is greater than 10 so it must go in more times. 21=(10)(2)+1 so that we have a remainder of 1 which is less than 10 So anytime you divide a number by 10 you can have remainders 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 but once you get to 10 then that number can be divided again so there will be a remainder of 0, or if you have a remainder of 1 you can divide the number again to have a remainder of 1, etc.
How to convert a terminating decimal to a fraction. We can take the fractional equivalence of the decimal so that the ‘tenths place’ gets put over 10, the ‘hundredths place’ gets put over 100, etc. Then we can find a common denominator for all the fractions and add the numerators over the common denominator to get the fraction. Decimal to Fraction Convertor
How to convert a repeating decimal to a fraction. Set your repeating decimal equal to x Start with your repeating decimal and multiply both sides by 10 factor length. In the example to the right 12 repeats itself so we have a factor length of 2 since 12 has 2 units in it and we multiply by 10 2 =100. Now we can subtract the two equations to eliminate the repeating portion of the decimal. Solve for x and simplify the fraction!
The very last step of this process returns a fraction sometimes this is a very large fraction and we need to simplify it. The common factors aren’t obvious so we can use the Euclidian Algorithm to find the Greatest common Factor – Given two numbers a and b we can find the greatest common factor – This algorithm is essentially rewriting each number as a sum of a product and a remainder. – The algorithm stops when the remainder eventually reaches zero. Euclidian Algorithm Game
First consider any two integers a, b such that neither are 0. There exists unique integers q 1 and r 1 such that a=bq 1 + r 1 where 0≤ r 1 < b, r is the remainder needed so that when b is multiplied q times a=b q 1 + r 1. If r 1 =0 then q 1 is the greatest common divisor. If not we can repeat the algorithm so b= r 1 q 2 + r 2 now 0≤ r 2 < r 1. If the remainder r 2 is still not 0 then we can continue to repeat the process until we get 0 for a remainder r 1 = r 2 q 3 + r 3 0≤ r 3 < r 2. r 2 = r 3 q 4 + r 4 0≤ r 4 < r 3. r 3 = r 4 q 5 + r 5 0≤ r 5 < r 4. …. r k-2 = r k-1 q k + r k 0≤ r k < r k-1 r k-1 = r k q k+1 Notice that 0 ≤ r k
"name": "First consider any two integers a, b such that neither are 0.",
"description": "There exists unique integers q 1 and r 1 such that a=bq 1 + r 1 where 0≤ r 1 < b, r is the remainder needed so that when b is multiplied q times a=b q 1 + r 1. If r 1 =0 then q 1 is the greatest common divisor. If not we can repeat the algorithm so b= r 1 q 2 + r 2 now 0≤ r 2 < r 1. If the remainder r 2 is still not 0 then we can continue to repeat the process until we get 0 for a remainder r 1 = r 2 q 3 + r 3 0≤ r 3 < r 2. r 2 = r 3 q 4 + r 4 0≤ r 4 < r 3. r 3 = r 4 q 5 + r 5 0≤ r 5 < r 4. …. r k-2 = r k-1 q k + r k 0≤ r k < r k-1 r k-1 = r k q k+1 Notice that 0 ≤ r k
It may seem obvious that the answer is 3 but what if it wasn’t so obvious, we can use the Euclidean Algorithm: Since the remainder is 0 the greatest common devisor of 12 and 99 is 3
Now that we know the procedure and have seen an example why does this work? Notice that the very last step is r k-1 = r k q k+1 so r k goes into r k-1, q k+1 times. Also r k divides r k-2 because r k-2 = r k-1 q k + r k and r k divides r k-1 and itself. We can continue in this manner seeing that r divides all steps including a and b.
Decimal approximation of rational or irrational numbers. Start with a decimal, find the inverse. You now have a fraction with a decimal on the bottom, subtract the whole number from this decimal and take the inverse of this new decimal which will be less than 1. Repeat this process until you need to take the inverse of a relatively large number. We can stop here because this is essentially 0… WHY?
Why is almost equal to 0? LIMITS The limit of a function f(x) describes what happens to the function as x gets microscopically close to a value. Limits
Lim x ∞ As x gets large we can see that the graph of is getting smaller and smaller. When x=10, =0.1, when x=100, = 0.01 and when x=1000000,. = 0.000001. As x gets larger, moving towards infinity we can see that is getting closer to 0. It will never reach 0 but it is so close that it doesn’t make a difference and we can call it 0.
Let ƒ be a function defined on an open interval containing c (exce pt possibly at c) and let L be a real number. Then lim x c f(x) =L means that for each real ε > 0 there exists a real δ > 0 such that for every x with 0 < |x − c| < δ, we have |ƒ(x) − L| < ε.
We pick point c, we want to find out what f(x) is as we approach that value c. To do this we can make a δ- neighborhood around c so that as we are in this neighborhood all values of f(x) will be in a corresponding ε-neighborhood around f(x). We can make these neighborhoods smaller and smaller to get a more exact value of f(x) when x is approaching a point c. The limit exists when all values of x after a certain point in the domain fall in the ε-neighborhood of f(x).
When we continuously took the inverse of the decimal we eventually ended up with Using what we have learned about limits we can see that this is essentially 0 therefore we can write the fraction by finding common denominators and we find that 0.141527 in fraction form is approximately 76/537