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For each $m,n\in \mathbb{N}$ think ${\mathbb{R}}^{m}$ as the space of real column vectors of size $m$ and ${\mathbb{R}}^{m×n}$ as the space of matrices of size $m×n$.
Let $d\in \mathbb{N}$
Let $a:\left\{1,\dots ,{2}^{d}\right\}\to \left\{1\right\}×\left\{0,1{\right\}}^{d}$ be an enumeration (injective and surjective map).
Let $A\in {\mathbb{R}}^{\left(d+1\right)×{2}^{d}}$ be the matrix whose columns are $a\left(1\right),\dots ,a\left({2}^{d}\right)$
Is it true that for each $b\in \left\{1\right\}×\left[0,1{\right]}^{d}$ there exists $x\in \left[0,+\mathrm{\infty }{\right)}^{{2}^{d}}$ such that $Ax=b?$?
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Carmelo Payne
Yes, there is an $x$ with ${x}_{i}\in \left[0,1\right]$.
Essentially, you want to solve
$\sum _{n=1}^{{2}^{d}}{x}_{n}=1,\phantom{\rule{1em}{0ex}}\sum _{⌊\frac{n}{{2}^{k}}⌋\equiv 1\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}2\right)}{x}_{n}={b}_{k}$
and you can see
${x}_{n}=\prod _{k=1}^{d}{f}_{k}\left(n\right),\phantom{\rule{1em}{0ex}}{f}_{k}\left(n\right):=\left\{\begin{array}{ll}{b}_{k}& ⌊\frac{n}{{2}^{k}}⌋\equiv 1\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}2\right)\\ 1-{b}_{k}& ⌊\frac{n}{{2}^{k}}⌋\equiv 0\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}2\right)\end{array}$
is a solution.