Starting Out With Java: From Control Structures Through Objects, Student Value Edition (7th Edition)
7th Edition
ISBN: 9780134802817
Author: Tony Gaddis
Publisher: PEARSON
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Textbook Question
Chapter 1, Problem 11SA
What is the difference between machine language code and byte code?
Expert Solution & Answer
Explanation of Solution
Machine Language:
- Machine language is also called as object code.
- The machine language varies from operating system (OS) to another.
Byte code:
Java program is the high-level language. So, java compiler is used to compile the java program (high-level language) into java bytecode (standard machine code).
Difference between machine language code and byte code:
| Machine Language code | Byte code |
| Machine language is also called as object code. | The Java compiler generates the source code into “Bytecode”. |
| It is executed directly by CPU. It is in binary format. | The JVM (Java Virtual Machine) interpreter is designed to execute the bytecode program produced by the compiler. |
| It is difficult for humans to read and write. But, it is easily understood by computers. | The interpreter takes the bytecode to convert the machine code and executes it. |
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By dragging statements from the left column to the right column below, give a proof by induction of the following statement:
an
=
= 9" - 1 is a solution to the recurrence relation an = 9an-18 with ao
=
: 0.
The correct proof will use 8 of the statements below.
Statements to choose from:
Note that a₁ =
9a0 + 8.
Now assume that P(n) is true for all n ≥ 0.
Your Proof: Put chosen statements in order in this
column and press the Submit Answers button.
Let P(n) be the predicate, "a = 9″ – 1".
απ = 90 − 1 =
Note that
Let P(n) be the predicate, "an
9" - 1 is a
solution to the recurrence relation an = 9an-1 +8
with ao = 0."
-
Now assume that P(k + 1) is true.
Thus P(k) is true for all k.
Thus P(k+1) is true.
Then ak+1
=
9ak +8, so P(k + 1) is true.
= 1 − 1 = 0, as required.
Then = 9k — 1.
ak
Now assume that P(k) is true for an arbitrary
integer k ≥ 1.
By the recurrence relation, we have
ak+1 =
ak+1 =
= 9ak + 8 = 9(9k − 1) + 8
This simplifies to
9k+19+8 = 9k+1 − 1
Then 9k+1 − 1 = 9(9*…
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Chapter 1 Solutions
Starting Out With Java: From Control Structures Through Objects, Student Value Edition (7th Edition)
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