Please let me know if I set these proofs up correctly. Thanks in advance.
Prove that for every integer x, the integers 3x+1 and 5x+2 are of opposite parity.
Suppose both are even or both are odd.
say both are even
then 3x is odd and 5x is even.
3x odd implies x is odd
and 5x even implies x is even.
Hence both cannot be even.
similiarly both cannot be odd.
therefore 3x+1 and 5x+2 are of opposite parity.
Prove the following statement: Let a, b be elements in a set of integers with a not equal to 0 and b not equal to 0. a|b and b|a if and only if a=b or a= -b
a|b means b=na for some integer n.
b|a means a=mb for some integer m.
so a=mb=m(na) =mna
a=(mn)a implies mn=1
hence m=1 and n=1 or m=-1 and n=-1
so a=b or a=-b
Let a,b,c be elements in a set of integers, where a does not equal 0. Prove that if a |/ bc, then a |/ b and a |/ c.
a |/ ab =a|/b or b|/a
Suppose a|b
Then b=ma for some integer m
which means ab=ma^2 = a|ab as a does not equal to 0
Hence, a|/b
Similarly, b|/a
Hence the proof.
Let n be an element in a set of integers. Prove that 3|(2n^2+1) if and only if 3 |/ n.
3|(2n^2+1)
2n^2+1=3k for some integer k
2n^2=3k-1
3|/(3k-1) i.e. 3|/2n^2
3|/n^2
Hence, 3|/n
1.
a) Let x,y be elements in a set of integers. Prove that if x+y is odd, then x and y are of opposite parity. Hint: Use proof by contrapostive
b) Prove the converse of part a).
c) Combine the statements proven in parts a and b above into one statement.
a)
contrapositive mean we have to prove that if x and y are not of opposite parity then x+y is odd i.e. even.
x and y are not of opposite parity means both are even or both are odd.
when both are even x=2m and y=2n for some integer m and n
then x+y=2m+2n=2(m+n) which is again even being a multiple of 2.
when both are odd x=2j+1 and y=2k+1 for some integers j,k
then x+y=2j+1+2k+1=2(j+k+1) which is even.
hence the proof.
b)we have to prove that if x and y are of opposite parity then x+y is odd
let x=2m and y=2n+1
x+y=2m+2n+1=2(m+n)+1=2k+1 which is odd.
hence the proof.
c) x+y is odd iff x and y are of opposite parity.
Let x,y be elements in a set of integers. Prove that (x+1)y^2 is even if and only if x is odd or y is even.
(x+1)y^2 is even
suppose x is even and y is odd .
then x+1 is odd and y^2 is odd
this implies (x+1)y^2 is odd.
hence x is odd or y is even.
conversely,
suppose x is odd or y is even.
which means x+1 is even or y is even.
i.e x+1 is even or y^2 is even.
since one of them is even their product has to be even.
hence (x+1)y^2 is even.
Prove that for every integer x, the integers 3x+1 and 5x+2 are of opposite parity.
Suppose both are even or both are odd.
say both are even
then 3x is odd and 5x is even.
3x odd implies x is odd
and 5x even implies x is even.
Hence both cannot be even.
similiarly both cannot be odd.
therefore 3x+1 and 5x+2 are of opposite parity.
Prove the following statement: Let a, b be elements in a set of integers with a not equal to 0 and b not equal to 0. a|b and b|a if and only if a=b or a= -b
a|b means b=na for some integer n.
b|a means a=mb for some integer m.
so a=mb=m(na) =mna
a=(mn)a implies mn=1
hence m=1 and n=1 or m=-1 and n=-1
so a=b or a=-b
Let a,b,c be elements in a set of integers, where a does not equal 0. Prove that if a |/ bc, then a |/ b and a |/ c.
a |/ ab =a|/b or b|/a
Suppose a|b
Then b=ma for some integer m
which means ab=ma^2 = a|ab as a does not equal to 0
Hence, a|/b
Similarly, b|/a
Hence the proof.
Let n be an element in a set of integers. Prove that 3|(2n^2+1) if and only if 3 |/ n.
3|(2n^2+1)
2n^2+1=3k for some integer k
2n^2=3k-1
3|/(3k-1) i.e. 3|/2n^2
3|/n^2
Hence, 3|/n
1.
a) Let x,y be elements in a set of integers. Prove that if x+y is odd, then x and y are of opposite parity. Hint: Use proof by contrapostive
b) Prove the converse of part a).
c) Combine the statements proven in parts a and b above into one statement.
a)
contrapositive mean we have to prove that if x and y are not of opposite parity then x+y is odd i.e. even.
x and y are not of opposite parity means both are even or both are odd.
when both are even x=2m and y=2n for some integer m and n
then x+y=2m+2n=2(m+n) which is again even being a multiple of 2.
when both are odd x=2j+1 and y=2k+1 for some integers j,k
then x+y=2j+1+2k+1=2(j+k+1) which is even.
hence the proof.
b)we have to prove that if x and y are of opposite parity then x+y is odd
let x=2m and y=2n+1
x+y=2m+2n+1=2(m+n)+1=2k+1 which is odd.
hence the proof.
c) x+y is odd iff x and y are of opposite parity.
Let x,y be elements in a set of integers. Prove that (x+1)y^2 is even if and only if x is odd or y is even.
(x+1)y^2 is even
suppose x is even and y is odd .
then x+1 is odd and y^2 is odd
this implies (x+1)y^2 is odd.
hence x is odd or y is even.
conversely,
suppose x is odd or y is even.
which means x+1 is even or y is even.
i.e x+1 is even or y^2 is even.
since one of them is even their product has to be even.
hence (x+1)y^2 is even.