Answer to Question #139199 in Discrete Mathematics for Maverick

Question #139199
Use proof by contraposition to show that if x + y ≥ 2, where x and y are real numbers, then x ≥ 1 or y ≥ 1.
1
Expert's answer
2020-10-26T18:57:44-0400

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

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

x+y<2x+y<2

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

The contrapositive version:\text{The contrapositive version:}

if (x+y)2; then¬((x<1)(y<1))\text{if }\sum{(x+y)}\ge2;\text{ then}¬((x<1)\land(y<1))

¬(x<1) is (x1); ¬(y<1) is (y1);¬(x<1) \text{ is }(x\ge1);\ ¬(y<1) \text{ is }(y\ge1);

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

¬((x<1)(y<1))=¬(x<1)¬(y<1)=(x1)(y1)¬((x<1)\land(y<1))=¬(x<1)\lor¬(y<1)=(x\ge1)\lor(y\ge1)

if (x+y)2 then(x) or (y1)\text{if }\sum{(x+y)}\eqslantgtr2\text{ then}(x\ge)\text{ or }(y\ge1)

proof done\text{proof done}

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