power laws: how to show that expression (a - b)^2 is given by (a + (-b))^2 ?

[DDDman]

How can I show that:
(a) The expression (a - b)^2 is given by (a + (-b))^2
(b) The expression (a - b)^3 is given by (a + (-b))^3
and vice versa.
Can it be done by power of laws because the simple thing is to multiply the minus by plus?
Is there another way to show it?

 

[HallsofIvy]

The powers are not really relevant. If x= y then trivially x^2= y^2 and the point is just to prove that x- y= x+ (-y). What are your definitions of "a- b" and "-b?

 

[JeffM]

How can I show that:
(a) The expression (a - b)^2 is given by (a + (-b))^2
(b) The expression (a - b)^3 is given by (a + (-b))^3
and vice versa.
Can it be done by power of laws because the simple thing is to multiply the minus by plus?
Is there another way to show it?

Induction

\(\displaystyle (a - b)^1 = a - b = a + (-\ b) = \{a + (-\ b)\}^1.\)

\(\displaystyle \therefore \exists \text { non-empty set } \mathbb K \text { such that }\)

\(\displaystyle x \in \mathbb K \implies x \in \mathbb Z^+ \text { and } (a - b)^x = \{a + (-\ b)\}^x.\)

\(\displaystyle \text {Let } k \text { be an arbitrary member of } \mathbb K.\)

\(\displaystyle (a - b)^{(k+1)} = (a - b)^k * (a - b)^1 \text { by laws of exponents.}\)

\(\displaystyle (a - b)^{(k+1)} = (a - b)^k * \{a + (-\ b)\}^1 \text { by first line of proof.}\)

\(\displaystyle (a - b)^{(k+1)} = \{a + (-\ b)\}^k * \{a + (-\ b)\}^1 \ \because \ k \in \mathbb K.\)

\(\displaystyle \therefore (a - b)^{(k+1)} = \{a + (-\ b)\}^{(k+1)} \text { by laws of exponents.}\)

\(\displaystyle \therefore (k + 1) \in \mathbb K.\)

\(\displaystyle \therefore n \in \mathbb Z^+ \implies (a - b)^n = \{a + (-\ b)\}^n. \text {Q.E.D.}\)