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
√
.
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Suppose
√
is rational, then we set x=
√
and
√
/4. We have
√
which is
irrational.
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 Spring '08
 Staff
 Irrational number

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