What does a proof look like? • Assumptions • Conclusion • Proof Example 1 • Assumption ○ V={(x_1,x_2,x_3 )|x_1,x_2,x_3∈Rand x_1+x_3=0} ○ ∀ x,y∈V, x+y is defined by ○ z=x+y if z=(x_1+y_1,x_2+y_2,x_3+y_3 ) ○ tx is defined by tx=(tx_1,tx_2,tx_3 ) for every x∈V,t∈R • Conclusion ○ V is a vector space • Proof: Axiom 1 (∀x,y∈V:x+y∈V) ○ let z=(z_1,z_2,z_3 )=x+y=(x_1+y_1,x_2+y_2,x_3+y_3 ) ○ z_1+z_3=x_1+y_1+x_3+y_3=(x_1+x_3 )+(z_1+z_3 )=0 ○ ⇒z∈V Example 2 • Assumption ○ V={(x_1,x_2,x_3 )|x_1,x_2,x_3∈Rand x_1+x_3=1} ○ ∀ x,y∈V, x+y is defined by ○ z=x+y if z=(x_1+y_1,x_2+y_2,x_3+y_3 ) ○ tx is defined by tx=(tx_1,tx_2,tx_3 ) for every x∈V,t∈R • Conclusion ○ V is not a vector Space • Proof: ∃x,y∈V:x+y∉V Axiom 5 • To show Axiom 5 does not hold, • we have to prove for every O∈V, • there is an x∈V with O+x≠x Example 3 • Assumption ○ V={all functions f:[0,1]→R} • Conclusion ○ V is a vector space • Proof: Axiom 3（∀f,g∈V:f+g=g+f） ○ Let h=f+g and k=g+f ○ Both h and g has a domain of [0,1] ○ h(x)=f(x)+g(x)=g(x)+f(x)=k(x)
