Problem 2 Recall the toBinary algorithm from Tutorial 7 problem 3, which finds the binary representation of a given integer n > 0. For example, toBinary (13) = (1, 1, 0, 1), and 1-2³ +1.2² +0-2¹ +1-2⁰ = 8 +4+0+1 = 13. toBinary (n) : 1: if n ≤ 1 then return (n) 2: 3: 4: 5: 6: else (nd,...,no) I = parity (n) return (nd,..., no, x) toBinary ([n/2]) In Tutorial 7 we showed by induction that for any n ≥ 0, if toBinary (n) d>0, then we have on,2² = n. = (nd,...,no) for some Note that toBinary takes in any integer n ≥ 0 and spits out a binary number, which is an element of {0,1}* (that is, an element of Uzo{0,1}). In other words, toBinary is a function from the domain of Z2º to the codomain of {0, 1}*. Use the result from Tutorial 7 to prove that toBinary : Z20 ➜ {0,1}* is one-to-one via a proof by contrapositive.
Problem 2 Recall the toBinary algorithm from Tutorial 7 problem 3, which finds the binary representation of a given integer n > 0. For example, toBinary (13) = (1, 1, 0, 1), and 1-2³ +1.2² +0-2¹ +1-2⁰ = 8 +4+0+1 = 13. toBinary (n) : 1: if n ≤ 1 then return (n) 2: 3: 4: 5: 6: else (nd,...,no) I = parity (n) return (nd,..., no, x) toBinary ([n/2]) In Tutorial 7 we showed by induction that for any n ≥ 0, if toBinary (n) d>0, then we have on,2² = n. = (nd,...,no) for some Note that toBinary takes in any integer n ≥ 0 and spits out a binary number, which is an element of {0,1}* (that is, an element of Uzo{0,1}). In other words, toBinary is a function from the domain of Z2º to the codomain of {0, 1}*. Use the result from Tutorial 7 to prove that toBinary : Z20 ➜ {0,1}* is one-to-one via a proof by contrapositive.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![Problem 2
Recall the toBinary algorithm from Tutorial 7 problem 3, which finds the binary representation of
a given integer n > 0. For example, toBinary (13) = (1, 1, 0, 1), and 1-2³ +1.2² +0-2¹ +1-2⁰ =
8 +4+0+1 = 13.
toBinary (n) :
1: if n ≤ 1 then
return (n)
2:
3:
4:
5:
6:
else
(nd,...,no)
I = parity (n)
return (nd,..., no, x)
toBinary ([n/2])
In Tutorial 7 we showed by induction that for any n ≥ 0, if toBinary (n)
d>0, then we have on,2² = n.
= (nd,...,no) for some
Note that toBinary takes in any integer n ≥ 0 and spits out a binary number, which is an element
of {0,1}* (that is, an element of Uzo{0,1}). In other words, toBinary is a function from the
domain of Z2º to the codomain of {0, 1}*.
Use the result from Tutorial 7 to prove that toBinary : Z20 ➜ {0,1}* is one-to-one via a proof
by contrapositive.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20fe110a-296a-48a1-a7d8-d1b56fc22b3f%2Fa2d49195-ae83-42aa-ae48-f4b01d5c2ada%2F796x72_processed.png&w=3840&q=75)
Transcribed Image Text:Problem 2
Recall the toBinary algorithm from Tutorial 7 problem 3, which finds the binary representation of
a given integer n > 0. For example, toBinary (13) = (1, 1, 0, 1), and 1-2³ +1.2² +0-2¹ +1-2⁰ =
8 +4+0+1 = 13.
toBinary (n) :
1: if n ≤ 1 then
return (n)
2:
3:
4:
5:
6:
else
(nd,...,no)
I = parity (n)
return (nd,..., no, x)
toBinary ([n/2])
In Tutorial 7 we showed by induction that for any n ≥ 0, if toBinary (n)
d>0, then we have on,2² = n.
= (nd,...,no) for some
Note that toBinary takes in any integer n ≥ 0 and spits out a binary number, which is an element
of {0,1}* (that is, an element of Uzo{0,1}). In other words, toBinary is a function from the
domain of Z2º to the codomain of {0, 1}*.
Use the result from Tutorial 7 to prove that toBinary : Z20 ➜ {0,1}* is one-to-one via a proof
by contrapositive.
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