{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 4 solutions

# HW 4 solutions - COT 3100 Discrete Mathematics HW#4...

This preview shows pages 1–3. Sign up to view the full content.

COT 3100 Discrete Mathematics HW #4 Solutions Section 1.7 3. Proof: If x y, then max(x,y)=x, min(x,y)=y. Thus max(x,y)+min(x,y)=x+y. If x<y, then max(x,y)=y, min (x,y)=x. Thus max(x,y)+min(x,y)=x+y. In either case, max(x,y)+min(x,y)=x+y holds. 4.Proof: If a b and a c, b c , min(a,min(b,c))= min(a,b)=a. min(min(a,b),c)=min(a,c)=a. If a b and a c, b >c , min(a,min(b,c))= min(a,c)=a. min(min(a,b),c)=min(a,c)=a. If b a and b c, a c , min(a,min(b,c))= min(a,b)=b. min(min(a,b),c)=min(b,c)=b. If b a and b c, a >c , min(a,min(b,c))= min(a,b)=b. min(min(a,b),c)=min(b,c)=b. If c a and c b, a b , min(a,min(b,c))= min(a,c)=c. min(min(a,b),c)=min(a,c)=c. If c a and c b, a >b , min(a,min(b,c))= min(a,c)=c. min(min(a,b),c)=min(b,c)=c. 5 . Proof: Since | || | | | , we have | | | | | || | | | . Meanwhile | | | | , | | | | | | must hold. 6. Proof: Since 1+2=3, 3 satisfied the requirement, which indictes that the statment is true. This is a constructive proof. 11. Proof: If is irrational, then we can set and .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Suppose is rational, then we set x= and /4. We have which is irrational.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}