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RE: division(x,infinity)=0

The last sentences of 5.2 read:

In the descriptions that follow, Q(x) is the exponent q of the representation of a finite floating-point number x. If x is infinite, Q(x) is +.

which answers your question?

From: stds-754@xxxxxxxx [mailto:stds-754@xxxxxxxx] On Behalf Of Hossam A. H. Fahmy
Sent: 26 February 2011 09:43
To: stds-754
Subject: division(x,infinity)=0

Dear 754 people,

Just wanted to check with you all:
 For decimal FP, the division of a finite x by infinity should be zero but what is the resulting exponent?

In clause 6.1 on page 34 (infinity arithmetic) nothing is specified regarding that.

In clause 5.2 on page 18, it is clear that when the result is exact (which is the case here as confirmed in clause 6.1) then we should have the preferred exponent of the operation.

In clause 5.4.1. on page 21, the division(x, y) has a preferred exponent Q(x) -Q(y).  That leads one to ask what is the "exponent" of infinity?

In clause 3.5.2 on page 11, point b) explains that remaining bits in G and T are ignored for infinities but it does not explicitly say that the exponent is treated as if it were zero as the previous point of the same clause states clearly for NaNs.

So, is the correct approach to treat the exponent of infinity as zero similar to NaNs? Shall we explicitly mention that in the following revision of the standard?

Hossam A. H. Fahmy

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