Answer to Question #139199 in Discrete Mathematics for Maverick

Question

Use proof by contraposition to show that if x + y ≥ 2, where x and y are real numbers, then x ≥ 1 or y ≥ 1.

Answer 1

$\text{Let }x<1 \text{and }y<1$

$\text{the sum of inequalities is also a true inequality that is}$

$x+y<2$

$\text{If } (x<1 )\land(y<1)\text{ then}\sum{(x+y)}<2$

$\text{The contrapositive version:}$

$\text{if }\sum{(x+y)}\ge2;\text{ then}¬((x<1)\land(y<1))$

$¬(x<1) \text{ is }(x\ge1);\ ¬(y<1) \text{ is }(y\ge1);$

$\text{according to de Morgan’s laws:}$

$¬((x<1)\land(y<1))=¬(x<1)\lor¬(y<1)=(x\ge1)\lor(y\ge1)$

$\text{if }\sum{(x+y)}\eqslantgtr2\text{ then}(x\ge)\text{ or }(y\ge1)$

$\text{proof done}$