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I know this intuitively obvious, and I might be overthinking this, but is this really something you can say without arguing further? If you take for instance eigenvector eigenvalue equation [$Av=A\lambda \Rightarrow v=\lambda$] is definitely false.

I've thought about using the fact that integers have unique factorization to say that $x$ factors uniquely as does $y$ and $z$, and because $xy = xz$ the factorization of $y$ and $z$ must equal one another $\Rightarrow$ $y=z$. However this argument also relies on [ $xy=xz \Rightarrow y=z$] in the last implication.

Then I thought of showing it inductively by using $1 \cdot b= 1 \cdot a \Leftrightarrow a = b$. Asssuming [$(n-1) \cdot b= (n-1) \cdot a \Leftrightarrow a = b$]. Then observing that $n\cdot a = n\cdot b \Leftrightarrow (n-1) \cdot b+b = (n-1) \cdot a+a \Leftrightarrow ((n-1) \cdot b - (n-1) \cdot a) +b=a$ $ \Leftrightarrow a=b$ and this seems to hold.

My question is if there are any other ways to arrive at the same result?

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    $\begingroup$ Your claim is not true on the integers $\Bbb Z$. For example $0\cdot 2=0\cdot 3$ does not entail $2=3$. What you need is that $x$ is neither $0$ (nor a divisor of $0$). $\endgroup$ – Hagen von Eitzen Oct 20 '15 at 11:40
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$xy=xz\Leftrightarrow xy-xz=0\Leftrightarrow x(y-z)=0\Leftrightarrow x=0\lor y-z=0\Leftrightarrow x=0\lor y=z$

Edit: Thus $x\neq0\Rightarrow(xy=xz\Leftrightarrow y=z)$

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  • $\begingroup$ This is perfect. Just what I was looking for. Big thanks! $\endgroup$ – Penman Oct 20 '15 at 11:47
  • $\begingroup$ Just as a sidenote/extension: This proof makes use of the fact that if $ab = 0$, either $a$ or $b$ must be $0$. That property is valid on any integral domain, so for example it would be valid too in the ring of polynomials over a field or in the Gaussian integers, but not e.g. in $\mathbb{Z}/4\mathbb{Z}$, since $2\cdot 2 = 4 = 0$. $\endgroup$ – Fryie Oct 20 '15 at 13:20

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