Theorem for FMA Let \(x,y,z\) be nonnegative floating point numbers. Assuming underflow does not occur, then \(xy+z\le o(o(xy)+z)(1+\mathbf{u}^{2})\).
Proof:
Recall from the standard model that \(o(xy)=xy(1+\epsilon),|\epsilon|\le \mathbf{u}\).
Thus, \(xy+z=o(xy)(1+\epsilon)+z,|\epsilon|\le \mathbf{u}\). It follows from the non-negativity of most of the terms that:
Now do the same to \(o(xy)+z\):
The result follows.
Theorem If \(\beta\) is even and \(p\le-e_{\min}\), and if \(k\) is a positive integer such that \(k\mathbf{u}<1\), then \(1-k\mathbf{u}\) is a normal floating point number.
Proof:
Note that \(\mathbf{u}^{-1}\) is a positive integer. We know that \(k\mathbf{u}<1\). Could it be possible that \((k+1)\mathbf{u}>1\)? The answer is no. To see this, assume it is true. Then \(k<\mathbf{u}^{-1}\) and \(\mathbf{u}^{-1}<k+1\). This implies that \(\mathbf{u}^{-1}\) is an integer strictly between the integers \(k\) and \(k+1\). Not possible.
Therefore, the strongest statement we can say is that \((k+1)\mathbf{u}\le1\). From here we get \(\mathbf{u}\le1-k\mathbf{u}<1\). Now \(\beta^{e_{\min}}\le \mathbf{u}\le1-k\mathbf{u}<1\)
Now let’s write it as:
We pick the \(e\) such that \(\beta^{p-1}\le\mu<\beta^{p}\). Using the inequality above, we can show that this \(e<0\). We can write \(\mu=(\mathbf{u}^{-1}-k)\mathbf{u}\beta^{-e+p-1}\). Note that the term in parenthesis is an integer, and so is \(\mathbf{u}\beta^{e-p+1}\) (note that this is why we needed \(\beta\) to be even - to handle round to nearest). Thus \(\mu\) is an integer. The proof is complete.
Theorem Let \(k\) be a positive integer such that \(k\mathbf{u}<1\). Let \(x\) be a floating point such that \(\beta^{e_{min}}\le|x|\le\Omega\). Then:
I hope such an obscure statement comes in handy some day!