Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 127400 by mohammad17 last updated on 29/Dec/20

prove that :∣∣z_1 ∣−∣z_2 ∣∣≤∣z_1 +z_2 ∣

provethat:∣∣z1∣−∣z2∣∣⩽∣z1+z2∣

Commented by mohammad17 last updated on 29/Dec/20

?

?

Commented by liberty last updated on 29/Dec/20

observasi the inequality ∣ z_1 +z_2 ∣ ≤ ∣ z_1 ∣ + ∣z_2 ∣ is known as triangle inequality .Now from the identity z_1 = z_1 +z_2 +(−z_2 ) gives ∣z_1 ∣ = ∣z_1 +z_2 +(−z_2 )∣ ≤ ∣z_1 +z_2 ∣+∣−z_2 ∣ since ∣z_2 ∣=∣−z_2 ∣ then ∣z_1 ∣−∣z_2 ∣ ≤ ∣z_1 +z_2 ∣...(i) but because z_1 +z_2 =z_2 +z_1 , (i) can be written in the alternative form ∣z_1 +z_2 ∣=∣z_2 +z_1 ∣ ≥ ∣z_2 ∣−∣z_1 ∣ =−(∣z_1 ∣−∣z_2 ∣) or ∣z_2 ∣−∣z_1 ∣ = ∣∣z_1 ∣−∣z_2 ∣∣ ...(ii) replacing z_2 by −z_2 we get ∣z_1 −z_2 ∣ ≥ ∣ ∣z_1 ∣−∣−z_2 ∣ ∣ or ∣z_1 −z_2 ∣ ≥ ∣ ∣z_1 ∣−∣z_2 ∣ ∣

observasitheinequality∣z1+z2∣⩽∣z1∣+∣z2∣isknownastriangleinequality.Nowfromtheidentityz1=z1+z2+(−z2)gives∣z1∣=∣z1+z2+(−z2)∣⩽∣z1+z2∣+∣−z2∣since∣z2∣=∣−z2∣then∣z1∣−∣z2∣⩽∣z1+z2∣...(i)butbecausez1+z2=z2+z1,(i)canbewritteninthealternativeform∣z1+z2∣=∣z2+z1∣⩾∣z2∣−∣z1∣=−(∣z1∣−∣z2∣)or∣z2∣−∣z1∣=∣∣z1∣−∣z2∣∣...(ii)replacingz2by−z2weget∣z1−z2∣⩾∣∣z1∣−∣−z2∣∣or∣z1−z2∣⩾∣∣z1∣−∣z2∣∣

Terms of Service

Privacy Policy

Contact: info@tinkutara.com