Question

Prove by contraposition that if a + b \(\ge\) 15, then a\(\ge\) 8 or b\(\ge\) 8

Answer 1

8+8=16

Answer 2

...

Answer 3

that's not proving...

Answer 4

a>=8 .......i
b>=8.......ii
Adding i and ii
a+b>=16

Answer 5

Since a+b>=15 is always true for a+b>=16.
a>=8 and b>=8

Answer 6

Hmmmmm, you'd have to show that \[a\lt 8 \wedge b \lt 8 \Rightarrow a+b\lt 15\]

Answer 7

If a < 8 and b < 8 then a + b < 15

Answer 8

Weird because it doesn't seem true, yet....

Answer 9

\(a + b \ge 15\)

=>

\(a \ge 8 \) or \(b \ge 8\)

is true only if \((a, b) \in \mathbb N \)

Answer 10

@ganeshie8 What's the counter example, if \(a, b \in \mathbb{R}\)?