Difference between revisions of "CSC103: DT's Notes 1"
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− | Now, if we | + | Now, if we observe the first table, we should recognize the table for the '''and''' operator! So it is true: arithmetic on bits can actually be done as a logic operation. But is it true of the '''S''' bit? We do not recognize the truth table of a known operator. But remember the ice cream example; we probably come up with a logic expression that matches this table. An easy way to come up with this expression is to express it in English first and then translate it into a logic expression: |
::'''S''' is true in two cases: when '''a''' is true and '''b''' is false, or when '''a''' is false and '''b''' is true. | ::'''S''' is true in two cases: when '''a''' is true and '''b''' is false, or when '''a''' is false and '''b''' is true. |
Revision as of 08:29, 25 March 2014
--© D. Thiebaut 08:10, 30 January 2012 (EST)
Last revised --D. Thiebaut (talk) 08:05, 9 October 2013 (EDT)