proof by contradiction

Question 1

Prove by contraction that if n2 is even then n must be even

Answer 1

Assume that if n2 is even n could be odd
If n is odd n =2k+1
Square 2k+1
This means n2 is odd which contrdicticts that n2 is even
Therefore it musty be true

Question 2

Probe by contradiction that a triangle has at most one obtuse angle

Answer 2

Assume that a triangles has more than one obtuse angle
Let the angles of a triangle be x y z
Let x and y be obtuse angles
Now x + y >180
But this us immposib;le since x+y+z = 180 and of x + y > 180 this would mean z is negative which it cannot be be
This contracts the assume therefore much be true

Question 3

Prove by contridiction thwart root 2 is an irrational number

Answer 3

Assume that root 2 is a rational number
Where p and q are integers with no common fa actors other than 1
Now if root 2 = p divided by q the 2q^2 = p^2
As 2q^2 = p^2 must be even and hence p is even as p is even p =2n
If both p and q are even they have a common factor of 2 this contracts the assume therefore true

Question 4

Prove by contraction that of pq is even then at lease one of p and q is even

Answer 4

Assume that if pq is even the neither p nor q is even
This means that p and 1 are odd so p = 2k+1 and q =2X+1
Times them together
This means that pq is odd which contradicts the assumption that pq is even

Question 5

Prove by contradictions that the sum of any two odd square number cannot be a square number

Answer 5

Assume that the sum of two odd square numbers x and y can be a square number z
As x, y and z are all square we can say that
X =m^2 y =n^2 z= p^2
As x and y are odd so are m and n so let m = 2a +1 and n=2b + 1
Now x + y = so squared of both
This means as x+y =z that z + the same which means that z is even
As z is even for some integer c and we have z =2c^2 = 4c^2
This means z is a multiple of 4 but it can be seen from the question that z is not a multiple of 4 so this a contradicts
Therefore..