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Question Number 746 by 123456 last updated on 06/Mar/15

l e t s ⊞ : {(R^{+})}^{2} \to R^{+}

d e f i n e d b y x ⊞ y = \sqrt{⌊ x ⌋ ⌈ x ⌉} + y

1 . x ⊞ y \overset{?}{=} y ⊞ x

2 . x ⊞ (y ⊞ z) \overset{?}{=} (x ⊞ y) ⊞ z

3 . \exists e, \forall x \in R^{+}, x ⊞ e = x ?

4 . \exists e, \forall x \in R^{+}, e ⊞ x = x ?

Commented by prakash jain last updated on 06/Mar/15

1 . y = 5, x = 4.5

x ⊞ y = \sqrt{4 \times 5} + 5 = 5 + \sqrt{20}

y ⊞ x = 10

x ⊞ y \neq y ⊞ x

2 . x = 4.5, y = 5.5, z = 6

x ⊞ (y ⊞ z) \neq (x ⊞ y) ⊞ z

Answered by prakash jain last updated on 06/Mar/15

\sqrt{⌊ x ⌋ ⌈ x ⌉}

⌊ x ⌋ = x - a w h e r e a = {x} = f r a c t i o n a l p a r t

⌈ x ⌉ = x + 1 - a

(x - a) (x + 1 - a) = x^{2} - a x + x - a - a x + a^{2}

= x^{2} - 2 a x + a^{2} + x - a

y = a^{2} - 2 a x + x - a = (a^{2} - a) + x (1 - 2 a) < 0 for a > 0.5

So \forall x ∣ {x} > . 5 ∄ e \in R^{+}, x ⊞ e = x

Statment 3 is false .

4 . e = 0

0 ⊞ x = x