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Prithu biswas
Prithu biswas
11:37 AM
@user21820 For abs I am trying to prove:
∀a∈ℝ ∃!b∈ℝ ( ( a ≥ 0 ⇒ b = a ) ∧ ( a ≤ 0 ⇒ b = −a ) ).
And for that I am first trying to prove:
∀a∈ℝ ∃b∈ℝ ( ( a ≥ 0 ⇒ b = a ) ∧ ( a ≤ 0 ⇒ b = −a ) ).
I thought it is something very trivial. But now when I am trying to prove it , it seems almost impossible. Maybe it is not trivial or I am missing something very simple.
 
Prithu biswas
Prithu biswas
12:06 PM
So , I am trying to show: 
Given a∈ℝ
    ...
    ( a ≥ 0 ⇒ b = a ) ∧ ( a ≤ 0 ⇒ b = −a )
I am just trying to find which "b" will make the statement:
( a ≥ 0 ⇒ b = a ) ∧ ( a ≤ 0 ⇒ b = −a )
Valid.
 
Prithu biswas
Prithu biswas
12:51 PM
Given a∈ℝ
    If a ≥ 0
        ...
        b = a
    If a ≤ 0
        ...
        b = -a
 
 
2 hours later…
user21820
user21820
2:55 PM
@Prithubiswas You can prove "∃b∈ℝ ( ( a ≥ 0 ⇒ b = a ) ∧ ( a ≤ 0 ⇒ b = −a ) )" in each case.
But it would be easier if you split cases using trichotomy rather than using "a ≥ 0 ∨ a ≤ 0".
 

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