# problems: reciprocal, addition, subtraction, multiplication, division

• Aug 13th 2010, 02:55 AM
vavavroom
problems: reciprocal, addition, subtraction, multiplication, division
hi masters, math newbie here. just a background all i know right now is some arithmetic and algebra topics. i'm self-studying mathematics just recently. i have two questions.

1. what is the mathematical process of getting the reciprocal?

i'm longing for a concrete mathematical solution other than just simply flipping off the numerator and the denominator to get the reciprocal.

1/3 -> (a missing mathematical process here) -> 3/1

i thought it could be derived using my solution but something is missing. here's my solutions:

1/3
= 1^1 / 3^-1
= 1^1 * 3^1 / 1^-1
= 3^1 / 1^-1
= 3/1

(though i actually flip 3, am i self-contradicting?) the solution above is correct i guess.

3/1
= 3^1 / 1^-1
= 1^1 / 1^-1 * 3^-1
= 1^1 / 3^-2
= 1/0.111111111

the solution above does not permit me to arrive to the reciprocal of 3/1 which is 1/3 by using the same process as the first solution stated above.

masters, i hope you can give me some light to this problem. i have been solving this for about one night and i haven't searched yet the internet. but i will search for it later after i post. thank you masters in advance.

2. by accident, i devised another way of adding, subtracting, multiplying, and dividing by using the inverse property of these operations as illustrated below. is there any other way known in mathematics today that let us operate in these operations by using the givens with little to no modifications?

when i was reading involutions in algebra i came across a thought that any number has it's own exponent of 1. so if i'm not mistaken 5 = 5 / 1 = 5^1 / 1^-1 and 1 / 5 = 1^1 / 5^-1. just correct me if i'm wrong though. so with that thought i come up with this method.

multiplication:
normal: a * b * , ... n = product
mine : a^1 / b^-1 / , ... n^-1 = product

normal: -235.98 * 72.123 * -93.8765 = 1,597,739.122
mine : -235.98^1 / 72.123^-1 / -93.8765^-1 = 1,597,739.122

normal: 2^3 * 4^5 * 6^7 = 2,293,235,712
mine: (2^3)^1 / (4^5)^-1 / (6^7)^-1 = 2,293,235,712

the mine solution:
1. raise the first given by 1, just for clarity
2. raise the other given by -1
3. operate in division to find the product

division:
normal: a / b / , ... n = quotient
mine: a^1 * b^-1 * , ... n^-1 = quotient

normal: -235.98 / 72.123 / -93.8765 = 0.034853349
mine: -235.98^1 * 72.123^-1 * -93.8765^-1 = 0.034853349

normal: 2^3 / 4^5 / 6^7 = 2.790816472x10e-8
mine: (2^3)^1 * (4^5)^-1 * (6^7)^-1 = 2.790816472x10e-8

the mine solution:
1. raise the first given by 1, just for clarity
2. raise the other given by -1
3. operate in multiplication to find the quotient

i thought i could apply exponent modifications in addition and subtraction but to no success so i change the method.

normal: a + b + (-c) + , ... n = sum
mine: a - (-b) - c - , ... n = sum

normal: -235.98 + 72.123 + -93.8765 = -257.7335
mine: -235.98 - -72.123 - 93.8765 = -257.7335

the mine solution:
1. retain the sign of the first given
2. change the positive sign to negative sign and vice-versa for the other given
3. operate in subtraction to find the sum

subtraction:
normal: a - b - (-c) - , ... n = difference
mine: a + (-b) + c + , ... n = difference

normal: -235.98 - 72.123 - -93.8765 = -214.2265
mine: -235.98 + -72.123 + 93.8765 = -214.2265

the mine solution:
1. retain the sign of the first given
2. change the positive sign to negative sign and vice-versa for the other given
3. operate in addition to find the difference

i know for the fact that the mine solution operates internally the normal solution which is the basis of the method. so far the mine solution works in all real numbers with consistency, and i apply it as part of my proofing system.

i haven't tested it on imaginary numbers because i don't know yet how and i'm very confused with imaginary numbers. the number itself is an idea and an idea is an imaginary. within an imaginative number we have three types of sub-imaginative numbers in the name of imaginary number, negative number, and zero. imagine?

so masters if you know how to use the imaginary numbers on these operations please enrich me.

forgive me if my questions might not make sense. i know i'm no math genius in any possible way, so i trust in you. once again, thank you masters for your time.
• Aug 13th 2010, 05:47 AM
Sudharaka
Quote:

Originally Posted by vavavroom
hi masters, math newbie here. just a background all i know right now is some arithmetic and algebra topics. i'm self-studying mathematics just recently. i have two questions.

1. what is the mathematical process of getting the reciprocal?

i'm longing for a concrete mathematical solution other than just simply flipping off the numerator and the denominator to get the reciprocal.

1/3 -> (a missing mathematical process here) -> 3/1

i thought it could be derived using my solution but something is missing. here's my solutions:

1/3
= 1^1 / 3^-1
= 1^1 * 3^1 / 1^-1
= 3^1 / 1^-1
= 3/1

(though i actually flip 3, am i self-contradicting?) the solution above is correct i guess.

3/1
= 3^1 / 1^-1
= 1^1 / 1^-1 * 3^-1
= 1^1 / 3^-2
= 1/0.111111111

the solution above does not permit me to arrive to the reciprocal of 3/1 which is 1/3 by using the same process as the first solution stated above.

masters, i hope you can give me some light to this problem. i have been solving this for about one night and i haven't searched yet the internet. but i will search for it later after i post. thank you masters in advance.

2. by accident, i devised another way of adding, subtracting, multiplying, and dividing by using the inverse property of these operations as illustrated below. is there any other way known in mathematics today that let us operate in these operations by using the givens with little to no modifications?

when i was reading involutions in algebra i came across a thought that any number has it's own exponent of 1. so if i'm not mistaken 5 = 5 / 1 = 5^1 / 1^-1 and 1 / 5 = 1^1 / 5^-1. just correct me if i'm wrong though. so with that thought i come up with this method.

multiplication:
normal: a * b * , ... n = product
mine : a^1 / b^-1 / , ... n^-1 = product

normal: -235.98 * 72.123 * -93.8765 = 1,597,739.122
mine : -235.98^1 / 72.123^-1 / -93.8765^-1 = 1,597,739.122

normal: 2^3 * 4^5 * 6^7 = 2,293,235,712
mine: (2^3)^1 / (4^5)^-1 / (6^7)^-1 = 2,293,235,712

the mine solution:
1. raise the first given by 1, just for clarity
2. raise the other given by -1
3. operate in division to find the product

division:
normal: a / b / , ... n = quotient
mine: a^1 * b^-1 * , ... n^-1 = quotient

normal: -235.98 / 72.123 / -93.8765 = 0.034853349
mine: -235.98^1 * 72.123^-1 * -93.8765^-1 = 0.034853349

normal: 2^3 / 4^5 / 6^7 = 2.790816472x10e-8
mine: (2^3)^1 * (4^5)^-1 * (6^7)^-1 = 2.790816472x10e-8

the mine solution:
1. raise the first given by 1, just for clarity
2. raise the other given by -1
3. operate in multiplication to find the quotient

i thought i could apply exponent modifications in addition and subtraction but to no success so i change the method.

normal: a + b + (-c) + , ... n = sum
mine: a - (-b) - c - , ... n = sum

normal: -235.98 + 72.123 + -93.8765 = -257.7335
mine: -235.98 - -72.123 - 93.8765 = -257.7335

the mine solution:
1. retain the sign of the first given
2. change the positive sign to negative sign and vice-versa for the other given
3. operate in subtraction to find the sum

subtraction:
normal: a - b - (-c) - , ... n = difference
mine: a + (-b) + c + , ... n = difference

normal: -235.98 - 72.123 - -93.8765 = -214.2265
mine: -235.98 + -72.123 + 93.8765 = -214.2265

the mine solution:
1. retain the sign of the first given
2. change the positive sign to negative sign and vice-versa for the other given
3. operate in addition to find the difference

i know for the fact that the mine solution operates internally the normal solution which is the basis of the method. so far the mine solution works in all real numbers with consistency, and i apply it as part of my proofing system.

i haven't tested it on imaginary numbers because i don't know yet how and i'm very confused with imaginary numbers. the number itself is an idea and an idea is an imaginary. within an imaginative number we have three types of sub-imaginative numbers in the name of imaginary number, negative number, and zero. imagine?

so masters if you know how to use the imaginary numbers on these operations please enrich me.

forgive me if my questions might not make sense. i know i'm no math genius in any possible way, so i trust in you. once again, thank you masters for your time.

Dear vavavroom,

Let me shed some light into this problem.....

First of all the reciprocal of a number(say x) is the number which multiplied gives the multiplicative identity 1. If the reciprocal of x is y.............

By definition, $x\times{y}=1\Rightarrow{y=\frac{1}{x}}$ hence we call $\frac{1}{x}$ the reciprocal of x. Hope you understood!
• Oct 24th 2012, 10:51 PM
Salahuddin559
Re: problems: reciprocal, addition, subtraction, multiplication, division
Mathematical process is to assume numerator as 1000000000... (as many as you need for precision), and divide it by the denominator. After division, put the decimal place in exact location (corresponding to the number of zeroes).

Salahuddin
Maths online