Problem:
Solution
If p∈⟨f−qg,g⟩, then p=a(f−qg)+bg=af+(b−aq)g∈⟨f,g⟩
If p∈⟨f,g⟩, then p=af+bg=a(f−qg)+(b+aq)g∈⟨f−qg,g⟩
Therefore ⟨f−qg,g⟩=⟨f,g⟩.
Solution
If p∈⟨f−qg,g⟩, then p=a(f−qg)+bg=af+(b−aq)g∈⟨f,g⟩
If p∈⟨f,g⟩, then p=af+bg=a(f−qg)+(b+aq)g∈⟨f−qg,g⟩
Therefore ⟨f−qg,g⟩=⟨f,g⟩.
No comments:
Post a Comment