Recent content by mb55113

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    Proving abs(x-y) < ε for all ε>0 in Real Analysis

    Ok...I tried the contrapositive...I think that i got it so suppose that abs(x-y)=m>0. Therefore we make ε=m/2 which makes abs(x-y)=m>(m/2)=ε and therefore we have found an ε>o and which makes abs(x-y)>= ε...is that right?
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    Proving the Triangle Inequality in Real Analysis: abs(abs(x)-abs(y))<=abs(x-y)

    Homework Statement Prove: abs(abs(x)-abs(y))<=abs(x-y) Homework Equations Triangle Inequality: abs(a+b)<=abs(a)+abs(b) The Attempt at a Solution This is what i have so far: Let a=x-y and b=y. Then abs(x-y+y) <= abs(x-y)+abs(y) which becomes abs(x)-abs(y)<=abs(x-y). From...
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    Proving abs(x-y) < ε for all ε>0 in Real Analysis

    Prove that abs(x-y) < ε for all ε>0, then x=y. I really do not know how to start this... I have tried to do the contra positive which would be If x does not equal y, then there exist a ε>0 such that abs(x-y) >= ε. Can someone help me and lead me to the right direction.