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I am a self learner so excuse me if I am asking a seemingly easy question , But I ve been stuck at this point for couple of days , I think I understand mathematical induction and what the author is trying to build is a natural numbers set with a defining axioms - 5 axioms - then building subsequent theories over them . But I cannot understand at all the part encircled in yellow , where did a1 and b1 come from ? What do they stand for ? Again sorry if I am asking a rather obvious or easy question , I am really passionate about learning math with a full thorough understanding , thanks in advance

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Let me elaborate and say what I already understand so I can explain what I don’t understand . Induction is using an assumed set which adhere to the assumed property then prove that that set is the general set in question . The author alrdy postulated that for x+y the result MUST be unique : becuz generaly x+y is a number producing another number Let it be c and because :

1-ANY OTHER NUMBER BUT 1 is a successor : that’s why he’s using x’ and y’ .

2-from theory 2 I know that x!= x’ then x+y is either -> x+1 = x’ or x+y producing another successor (x+y)’ all numbers but 1 are successors axiom 3.

So now it’s induction time and the author starts with y then goes to x , the second part of proof I understand … What I don’t understand is in order to prove set M containing all possible Ys , he stated that y= 1 has a1 ,b1 with an inherent property that they are equal thus it belongs to the desired set M with postulated property that I don’t understand where did he get it from saying each y has a.y and b.y which are equal , but for other Ys he stated that they inherently DONT have such property but because he postulated that they belong to set M so he proves that their a and b are equal thus y’ being their successors Must also belong to the set and again becuz all numbers but 1 are y’ so they must belong to natural set .

why using two different methods to prove 1 and other Ys belong to the set .. ? Is that how analysis works , I can postulate anything I want ..?

Mauro ALLEGRANZA
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skipping tutorial
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  • See here for details. – Mauro ALLEGRANZA Aug 29 '23 at 07:04
  • SO after pushing through till i reached multiplication , i stumbled upon the same problem , proving uniqueness , then i realized it , the assumption that Ys have Ay and By .. these are the available postulated combinations , it doesnt matter if its more than two , the author meant by these two being equal thus the x+y or x.y combination resulting into another number is unique as the two available combinations are indeed equal to begin with .. i think i got it now – skipping tutorial Aug 30 '23 at 17:24

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