What do the following mean..

1) ∀x ∈ ℝ. (x > 0 ⇒ ∃y ∈ ℝ. 0 < y ≤ x ∧ f(y) = 0)

2) ∀x, y ∈ ℝ. (x ≠ y ⇒ f(x) ≠ 0 ∨ f(y) ≠ 0)

3)Write with symbols: "Vectorsainbare linearly dependant"

AND why is this wrong (for 3)):

∃λ ∈ ℝ.a= λb

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- Oct 21st 2009, 09:29 AMmetlxStatements
What do the following mean..

1) ∀x ∈ ℝ. (x > 0 ⇒ ∃y ∈ ℝ. 0 < y ≤ x ∧ f(y) = 0)

2) ∀x, y ∈ ℝ. (x ≠ y ⇒ f(x) ≠ 0 ∨ f(y) ≠ 0)

3)Write with symbols: "Vectors*a*in*b*are linearly dependant"

AND why is this wrong (for 3)):

∃λ ∈ ℝ.*a*= λ*b*

- Oct 23rd 2009, 08:51 AMmetlx
can someone help me with this one?

i tried:

1) the function has a upper limit but then i get lost.. don't know what f(0) = 0 means.. is it a minimum (inf)?

2) no clue.. injective function?

3) ∃λ ∈ ℝ.*a*= λ*b ; a ≠ 0,b ≠ 0, λ ≠ 0*maybe? - Oct 23rd 2009, 09:48 AMclic-clac
Hi

1) says that no matter how close you are from $\displaystyle 0$ in $\displaystyle \mathbb{R}^+,$ there is always a lower positive real number where the function nullifies. Example: $\displaystyle x\rightarrow\sin(\frac{1}{x})$

2) This is not injectivity: for instance, consider the constant function $\displaystyle c_1:\mathbb{R}\rightarrow\mathbb{R}$ equal to $\displaystyle 1.$ Then for all $\displaystyle x,y\in\mathbb{R},\ c_1(x)=c_1(y)=1,$ so "$\displaystyle c_1(x)\neq 0\vee c_1(y)\neq 0$" is true. But of course $\displaystyle c_1$ is not injective.

Assume there are two different reals where $\displaystyle f$ value is $\displaystyle 0.$ Is the formula true? Conclude.

3)

Quote:

∃λ ∈ ℝ.*a*= λ*b*

You have no condition about $\displaystyle a$ and $\displaystyle b,$ so they can be the zero vector. To say $\displaystyle a,b$ are linearly dependent, use two scalars in your formula, or consider two cases: one with $\displaystyle a\neq 0$ and the other with $\displaystyle a=0.$