# The Feynman Lectures on Physics - Algebra - Summary

Click to see the full chapter. This is my summary, quoting and paraphrasing the content.

## Addition and multiplication

`ubrace(1 + 1 + ... + 1)_(b " times") = b`

`ubrace(a + a + ... + a)_(b " times") = b * a`

`ubrace(a * a * ... * a)_(b " times") = a^b`

Now as a consequence of these definitions it can be shown that all of the following relationships are true:

`a + b = b + a`

`a + (b + c) = (a + b) + c`

`ab = ba`

`a(b + c) = ab + ac`

`a(bc) = (ab)c`

`(ab)^c = a^cb^c`

`a^ba^c = a^(b+c)`

`(a^b)^c = a^(bc)`

`a + 0 = a`

`a * 1 = a`

`a^1 = a`

## The inverse operations

`a + b = c => b = c - a`

`ab = c => b = c/a`

`b^a = c => b = root(a)(c)`

`a^b = c => b = log_ac`

## Abstraction and generalization

`b = 3 - 5 = 0 - 2 => text(negative numbers)`

`(3 - 5) + 5 = 3`

`b = a^(3 - 5) => ba^5 = a^(3 - 5)a^5 => ba^5 = a^3 => `

` => b = a^3/a^5 = 1/a^2 => text(rational numbers)`

`(3/5) * 5 = 3`

`b = a^(3/5) => b^5 = a^((3/5)*5) => `

` => b = root(5)(a^3) => text(real numbers)`

All rules still work.

## Approximating irrational numbers

`log_b(ac) = log_ba + log_bc`

`log_ba = (log_ca) / log_cb`

Using base 10. Approximate `10^(1/2) = sqrt(10)` using `a' = 1/2(a + 10/a)`. Then keep taking square roots to find `10^(1/4), 10^(1/8), 10^(1/16), ..., 10^(1/1024)`. Approximate smaller powers by `10^n ~~ 1 + 2.3025n text( as ) n -> 0`.

The resulting values can be used to calculate `10^x text( and ) log_(10)x`, and with extra multiplication `a^x text( and ) log_ax`.

As `n -> 0`:

`10^n ~~ 1 + 2.3025n => 10^(n/2.3025) ~~ 1 + n`

`10^(1/2.3025) ~~ e`

`log_e(1 + n) ~~ n`

`e^n ~~ 1 + n`

## Complex numbers

`x^2 = -1 => x = +-i => text(complex numbers)`

Now every algebraic equation can be solved!

`(p+iq)+(r+is) = (p+r)+i(q+s)`

`(r+is)(p+iq) = (rp−sq)+i(rq+sp)`

`10^(r+is) = 10^r10^(is)`

`10^(is)=x+iy`

`10^(-is)=x-iy`

`10^(is)10^(-is) = 10^0 = 1 = (x + iy)(x - iy) = `

` = x^2 + y^2`

Start from `10^(i/1024)=1 + 2.3025i/1024` and keep squaring the number to find `10^i`. The resulting values can also be used to calculate `log_(10) i`.

## Imaginary exponents

Calculate `10^(ip/8) = x + iy` for successive values of `p` and plot the results.

`10^(is) = e^(2.3025is)`

`e^(itheta) = cos theta + i sin theta`

`x + iy = re^(itheta)`