Bashyboy
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Claim: x \vec{0} = \vec{0}
Let x be some arbitrary scalar in \mathbb{R}.
x \vec{0} = x ( \vec{0} + \vec{0}) \iff
x \vec{0} = x \vec{0} + x \vec{0} \iff
x \vec{0} - x \vec{0} = x \vec{0} + x \vec{0} - x \vec{0} \iff
\vec{0} = x \vec{0}
Because we had supposed x was arbitrary, meaning that we have not assumed anything about x beyond the fact that it is merely in \mathbb{R}, then this statement must be true of all real numbers.
Let x be some arbitrary scalar in \mathbb{R}.
x \vec{0} = x ( \vec{0} + \vec{0}) \iff
x \vec{0} = x \vec{0} + x \vec{0} \iff
x \vec{0} - x \vec{0} = x \vec{0} + x \vec{0} - x \vec{0} \iff
\vec{0} = x \vec{0}
Because we had supposed x was arbitrary, meaning that we have not assumed anything about x beyond the fact that it is merely in \mathbb{R}, then this statement must be true of all real numbers.
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