Some arithmetic

Some Arithmetic PhL 6.25.20 Exercise 1: Consider a2a3 = (a2)*(a3) = (a*a)*(a*a*a) = (a*a*a*a*a) = a5 Now generalize a bit, aman = (a*a*....)*(a*a*....) = (a*a*... *a*a*... ) = am+n // you ADD the two powers m copies n copies m+n copies Note that a could be any complex number. Now this statement that aman = am+n is at first valid for m and n being positive integers. This is not a major step. But then we can analytically continue this to say aαaβ = aα+β where now a, α and β are all complex numbers! That is a larger step! Exercise 2: Consider anbn = (a*a* ....) * (b*b* ....) = (a*b)*(a*b)* ....... = (a*b)n n copies of a n copies of b n copies of (a*b) n copies of a*b reordered We are just reordering terms in a product and multiplication is commutative so order doesn't matter. Here n is a positive integer. I have now proven these two "rules", Here are some extended rules: an * am * ak = an+m+k an * bn * cn = (a*b*c)n ((an)m)k = same with any ordering of the letters = an*m*k Exercise 3: Prove this rule, Well here I go (an)m = (a*a*a....)(a*a*a....) Let an = b. Then (an)m = bm . Hmmm. I don't have proof! I am thinking simple induction proofs! (6.26.20 continue Start with n= 1. The expression is then (a)m = am and there is nothing of interest, Next try n=2. We then have (a2)m which is (a*a)m. Try m = 3. We then have (a2)3 = (a*a)3 = (a*a)*(a*a)*(a*a) = a6 = a2*3 . The expression (a*a)m means "take a*a and do that m times" Is that the same as a2*m ? Yes it is, but I want a cleaner proof! Treat n as fixed and m as variable. Start with m =1. This is then (an)1 = (an) = an . Since this is an identity, it is true, Now try m=2. This is then (an)2. This is (an)*(an). Think of it as [(an)]2 = (an)*(an). Then we should next have [(an)]3 = (an)*(an)*(an) . Like [x]3 = (x)*(x)*(x) where x = an . Then write [(an)]N = (an)*(an)*(an) .... (an) N factors of (an) Can also write this as (an)N = (an)*(an)*(an) .... (an) = an*N with N factors each being (an). Let's now try an induction proof. Want to prove that : 1. If (an)k = an*k , then (an)k+1 = an*(k+1). [ If true for k, then true for k+1.] 2. This is true for k = 1. So assume that (an)k = an*k (assume true for k). Multiply both sides by (an) to get (an)k+1 = (an*k ) * (an) = an*k+n = an*(k+1) We have thus shown that (an)k+1 = an*(k+1). But this is our statement for k+1 ! So we have shown that if our statement is true for k, then it is true also for k+1. So it remains only to show that the statement is true for k = 1. But for k = 1, our statement says (an)1 = an*1 or an = an which is clearly true. I think this is my first post-stroke induction proof! Hurray! Exercise 4: I want to do this again. Here is the induction proof: show this: 1. If an * ak = an+k, then an * a(k+1) = an+(k+1) 2. This is true for k = 1. In words: If this fact is true for k (the fact being an * ak = an+k), then the fact is also true for k+1 ( in which case the fact says an * a(k+1) = an+(k+1). A shorthand notation would then be k k+1. We can't just say that k k+1, we have to prove it! So assume that an * ak = an+k (assume true for k). Given the assumption that an * ak = an+k, I then have to prove that an * a(k+1) = an+(k+1). Here is my proof. Step 1: Write down the statement for k: an * ak = an+k . Step 2: Multiply both sides by a. The truth of an equation is not changed by multiplying both sides by the same number (as long as it is not 0 or ∞ ). We then have, an * ak * a = an+k * a . LHS RHS Step 3: On the LHS we can write ak * a = ak+1. In words, "a to the k times a equals a to the k+1 ". If we now include the factor an this says : an * ak * a = an * ak+1. Step 4: On the RHS we can write an+k * a = an+k+1 = an+(k+1). Step 5: Then writing LHS = RHS we have an * ak+1 = an+(k+1) Step 6: We have now shown that an * ak = an+k an * ak+1 = an+(k+1) . That is, we have shown that if a certain equation is true for k, then it is also true for k+1. The is a key ingredient of an induction proof: we want to show that if something is true for k, then it is also true for k+1. In shorthand, we are showing that k k+1. Exercise 5: I am going to back up now and re-prove this claim: Here is an induction proof. I will first show that the equation is true for m = 1. For m = 1 it says (an)1 = an*1 or (an) = an so this m = 1 case is trivially shown. Now let's try showing that m m+1. We need to show that (an)m = an*m (an)m+1 = an*(m+1) Step 1: Start with (an)m = an*m . This is the statement for m. Step 2: Multiply both sides by an which gives [ leaves equation still true ] (an)m * an = an*m * an LHS RHS Step 3: On the LHS write, (an)m * an = am*n This is really the crux statement! BACK UP. Consider this equation, (an)m = an*m How many powers of a are there on the LHS? We have an multiplied by itself m times. That is an * an * ...... * an = an*m Here are some specific examples, an * an = (an)2 = a2*n an * an * an = (an)3 = a3*n Try this with some small values of n. Start with n = 1, a1 * a1 = (a1)2 = a2 a1 * a1 * a1 = (a1)3 = a3 Now go to n = 2, a2 * a2 = (a2)2 = a4 a2 * a2 * a2 = (a2)3 = a6 // which is a2*3 ' I will now make this broader statement: (an)m = (am)n = an*m = am*n And this seems to follow ((an)m)k = same with any ordering of the letters = an*m*k I created that new fact out of nothing! I think this was a good exponent session! Here is a source of data https://www.rapidtables.com/math/algebra/logarithm/Logarithm_Rules.html Is this useful to me? I will hold judgment . What do I know about logs? logax Wow! I know nothing right now! Here is a starting point, I will write this again: ax = y loga(y) = x Start with ax = y. How do you solve that equation for x? Answer: x = loga(y) How would I prove this claim, Start from scratch with this: ax = M loga(M) = x ay = N loga(N) = y We then have that ax * ay = M*N = ax+y // add exponents Now take the log of both sides loga(M*N) = loga(ax+y) = x+y The loga of ax+y is in fact x+y. The log is the power to which you raise a to get x+y. So we then have loga(M*N) = x+y = loga(M) + loga(N) This is the very famous rule that the log of a product of two numbers is the sum of the logs of the two numbers. Question: What is loga(Mn) ? Answer: n loga(M) Try to follow this development: 1. Define x to be loga Mn. 2. Then ax = Mn. 3. Can I show that this is true: loga M = y ⇒ ay = M ? yes 4. Can I then show this? Now, ax = Mn = (ay)n = any ? This is worse than pulling teeth. Each equation is a battle for me. I don't follow the logic flow. It might be easier for me to start from scratch. Want to show this: loga(Mn) = n loga(M) I know the log of a product is the sum of the logs. So loga(M2) = loga(M*M) = loga(M) + loga(M) = 2 loga(M) loga(M3) = loga(M*M*M) = loga(M) + loga(M) + loga(M) = 3 loga(M) ..... loga(Mn) = loga(M*M*M.....) = loga(M) + loga(M) + ... = n loga(M) OK I am happy now. I think I have not done logs for a long time!