Proof

Question 1

Prove that log2(7) is irrational

Answer 1

Assume that log2(7) is rational.
log2(7) = a/b where a,b are any integers with no common factors and b≠0

2^(a/b) = 7
2^(a) = 7^(b)

This cannot be true unless a=b=0 which contradicts our assumption that b≠0.
2^(a) is always even and 7^(b) is always odd hence we have another contradiction….conclusion

Question 2

Prove that √2 is irrational

Answer 2

Assume that √2 is rational.
√2 = a/b where a,b are any integers with no common factors

2 = a²/b² so a² = 2b²

a must be even so we can represent it as a = 2k

4k² = 2b² so b² = 2k so b is also even

a and b are integers with no common factors which is contradiction….conclusion

Question 3

Prove that there are no positive integers a and b with a odd such that a+4b=4√ab

Answer 3

Assume there are positive integers a and b with a odd such that a+4b=4√ab

(a+4b)² = 16ab
a² + 8ab + 16b² = 16ab
a² - 8ab + 16b² = 0
(a-4b)² = 0
a = 4b so a = 2(2b) so a is even which contradicts are initial statement that a is odd….conclusion

Question 4

Prove that there are no integers for a and b for which 21a + 14b = 1

Answer 4

Assume there are integers for a and b which 21a + 14b = 1

if we divide by 7, we get:
3a + 2b = 1/7

an integer plus another integer cannot equal fraction which contradicts are initial assumption that a and b are integers….conclusion

Question 5

Prove that if x is rational and y is irrational then x + y is irrational

Answer 5

Assume that if x is rational and y is irrational then x + y is rational

x = a/b where a,b are any integers with no common factors

a/b + y = c/d
y = c/d - a/b = (bc-ad)/bd

(bc-ad)/bd is rational which contradicts our assumption that y is irrational….conclusion

Question 6

Prove that “if a and b are a irrational and a≠b then (a + b) is always irrational” is false

Answer 6

proof by counter example:
a = √2
b = 1-√2
a + b = √2 + 1 - √2 = 1 which is rational