# 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)

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Updated on 2019 Oct 9.

DISCLAIMER: This is not professional advice. The ideas and opinions presented here are my own, not necessarily those of my employer.