99% of people can't prove this for real a,b,epsilon

99% of people can't prove this for real a,b,epsilon.

If a < b - epsilon for all epsilon > 0, then you have a contradiction. Which implies a < 0, I guess.

Proof:
Consider the case when epsilon is greater than b. Then a

But what if epsilon ISN'T greater than b?

Then it's not for all epsilon

So you haven't fulfilled the antecedent and you haven't proved anything

uhh wat?

Take a=1, b=2, [math]\epsilon = \frac{1}{2}[/math].

It's supposed to be true for all epsilon, that isn't even true for the epsilon you provided

What antecedent? It's a fucking tibetan leapfrog blog comment

>If ___, then ___.
>what antecedent?

'You are a retard if you can't solve these math problems"

Calc 2? Why?

Well it's not true for all epsilon, it's trivial to find a counter-example if [math]\epsilon[/math] < b.

MAYBE I could two years ago lel
>guess I'm a retard
Dont worry I'll git gud baby

But what if b

This looks like it is your homework, but school doesn't start till Michaelmas.

It looks like a Calc 2 review quiz you'd get on day 1 of Intro ODEs. Cheating on this is a mistake.

What if it's not?

Then putting epsilon = 1 implies a < 0

But what if epsilon = 100?

:thinking:

Proving [math] ¬q \implies ¬p [/math] seems kind of trivial.

But what if q is necessarily false?

just use substitution faggot

Case 1: b

This is the actual answer. Assume something that is not true, then anything in the implication is always true.

E.g. the statement "If seven is even then 7 < 0" is true.

What then? The contrapositive is logically equivalent to the original statement. It doesn't matter what p and q are.

There are no reals [math]a, b[\math] such that the antecedent holds, since [math]b - \epsilon[/math] can be made less than any real. Also, since no such [math]a[/math] exists, the consequent is always false. Thus, the statement is vacuously true.

Actually a can be less than or equal to 0

Checkmate atheists

No you literal brainlet, for all [math]\varepsilon > 0 [/math], the statement [math] a < b - \varepsilon [/math] is valid. So also in the case [math]\varepsilon = b [/math], in which case [math] a > 0 [/math]. You don't have to prove anything "for all [math]\varepsilon[/math]" here, since it's in the assumption.

The fuck are these replies? [math]\forall \varepsilon >0: a

correct ones.
are both correct.

>lb/ft^3
so I guess you like burgers

They're correct replies. If you rewrite the equation as [math] \epsilon0[/math], maybe you see the contradiction more easily.

[math]\epsilon0 [/math]

The hypothesis implies that a is smaller than any real number which can't happen. (a,b,ε are all supposed to be real)