c0017p1_0_x

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a < b a + c < b + c a < Lista P1 Ejercicio 0-x. Sean a, b, c . Demuestre que: Se tiene que: b - a > 0 (b - a) + 0 > 0 (b - a) + (c - c) b > 0 ( (b - a) + c) - c > 0 (b +(- a + c) ) - c > 0 (b +( c - a ) ) - c > 0 ( (b + c ) - a ) - c > 0 (b + c ) + (- a - c ) > 0 (b + c ) + (- ( 1a) - (1c ) ) > 0 (b + c ) + ( (-1)a +(- 1) c ) > 0 (b + c ) + (-1)( a + c ) > 0 (b + c ) ( 1 ( a + c ) ) > 0 (b + c ) ( a + c ) > 0 (b + c ) a + c b + c ( a + c ) > 0 a < b a + c < b + c <

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Ejercicio 0 de cálculo

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Page 1: C0017P1_0_x

a < b a + c < b + c

a <

Lista P1 Ejercicio 0-x.Sean a, b, c . Demuestre que: Se tiene que: b - a > 0

(b - a) + 0 > 0 (b - a) + (c - c)

b

> 0

⇒∈

⇒⇒⇒

( (b - a) + c) - c > 0(b +(- a + c) ) - c > 0(b +( c - a ) ) - c > 0( (b + c ) - a ) - c > 0(b + c ) + (- a - c ) > 0(b + c ) + (- (

⇒⇒⇒⇒⇒⇒ 1a) - (1c ) ) > 0

(b + c ) + ( (-1)a +(- 1) c ) > 0(b + c ) + (-1)( a + c ) > 0(b + c ) ( 1 ( a + c ) ) > 0(b + c ) ( a + c ) > 0(b + c )

−−−

⇒⇒⇒⇒⇒

a + c b + c

( a + c ) > 0

a < b a + c < b + c

<

∴⇒