Discrete Mathematics
1.6.30
Suppose A: Allen is a bad boy, H: Hillary is a good girl, D: David is happy
(A∨H)∧(¬A∨D)→(H∨D) by Resolution.
1.7.30
(i)If m^2=n^2, then m^2-n^2=(m+n)(m-n)=0. When the product of two
number is 0,at least one of them is 0,so m+n=0 or m-n=0,which means
m=n or m=-n
(ii)if m=n, then m^2=n^2. On the other hand, if m=-n, then m^2=(-1.7.12
A rational number can be expressed as a/b, where a and b are two
integers and b≠0, while an irrational number can’t.
We assume the product of a nonzero rational number and an irrational
number is rational. We assume the nonzero rational number is a/b,
1.7.30
(i)If m^2=n^2, then m^2-n^2=(m+n)(m-n)=0. When the product of two
number is 0,at least one of them is 0,so m+n=0 or m-n=0,which means
m=n or m=-n
(ii)if m=n, then m^2=n^2. On the other hand, if m=-n, then m^2=(-where
a and b are two integers and ab≠0, and the irrational number is R, and
the product is c/d, where c and d are two integers and cd≠0.
So (a/b)*R=c/d, which means R=bc/(ad). But a, b, c, d are integers, so bc
and ad are integers, too. This implies that R is a rational number, and the
contradiction occurs. Therefore, we know that the product of a nonzero
rational number and an irrational number is irrational by contradiction.
n)^2=n^2.
Therefore, m^2=n^2 and (m=n or m=-n) are logically equivalent.
1.7.37
There is an error in step (2). When squaring both sides of an equation,
one should check the restrict of x. Because root(x+3) is always positive or
0 , so 3-x should be positive or 0. Therefore, 6 is an invalid solution.