Prove the following Peano\'s Axioms/Integer Properties: (a) P17 (in image) (b) P

Question

Prove the following Peano's Axioms/Integer Properties:

(a) P17 (in image)

(b) P19 (in image)

You can ONLY use the properties that came BEFORE it in your proof, as well as a few mathematics theorems outside this list of axioms. (Peano axioms only are encouraged.) So you CAN'T use properties listed AFTER something to prove it.

P1: Cancellation of addition: If a c b c then a b. Similarly, if a b a c then b c. P2: 0 is the unique additive identity. 3: 1 is the unique multiplicative identity. P4: For each a in Z, a is unique. P5 a) a. P6: 00. a. 30. P7: 1) a --a. PS: (-a) b a (-b) (ab). DO (-b) ab P10: Multiplication distributes over subtraction: a. (b-c a.b-a .c and (a b).c P11: 1 is in N P12: If a and b are non-zero integers, then a b is non-zero. P 13: Cancellation of multiplication: If a 0, then a b a c implies b c P14: If a b and b c, the a c P15: If a b, then a c b c and a c b c. P16: If a c b c, then a b: if a c b c then a b P17: If c 0 and a b, then ac bc P18: If c 0 and a b, then ac bc P19: If c 0 and ac bc, then a b P20: If c 0 and ac bc, then a b. P21: If 0 a b and 0 c d, then a c b d. P22: 1 is the smallest natural number

Explanation / Answer

Proof:

P17

Given that c>0 and a<b

Take a<b and subtract b on sides, we get a-b<b-b (P4)

a-b<0

Now multiply with c>0 on both sides, we get

(a-b)c<0.c (P10)

ac-bc<0 (P10)

again add bc on both sides, we get

ac-bc+bc<0+bc (P4)

ac<bc

Therefore, if c>0 and a<b, then ac<bc.

P19

Given that c>0 and ac<bc

Take ac<bc and subtract bc on both sides, we get

ac-bc<bc-bc (P4)

ac-bc<0

(a-b)c<0 (P10)

This provides that two cases c<0 and (a-b)<0

But we have c>0, so (a-b)<0

again add b on both sides, we get

a-b+b<0+b (P4)

a<b

Therefore, if c>0 and ac<bc, then a<b.

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