Fill in the blanks in the following proof by contraposition that for every integer n , if 5 / n 2 then 5 / n . Proof (by contraposition): [The contrapositive is: For every integer n, if 5| nthen 5| n 2 .] Suppose n is any integer such that __(a)__ [We must show that __(b)__] By definition of divisibility, n = _ _ ( c ) _ for some integer k . By substitution, n 2 = _ _ ( d ) _ _ = 5 ( 5 k 2 ) . But 5 k 2 is an integer because it is a product of integers. Hence n 2 = 5 ⋅ (an integer), and so __(e)___ [as was to be shown].
Fill in the blanks in the following proof by contraposition that for every integer n , if 5 / n 2 then 5 / n . Proof (by contraposition): [The contrapositive is: For every integer n, if 5| nthen 5| n 2 .] Suppose n is any integer such that __(a)__ [We must show that __(b)__] By definition of divisibility, n = _ _ ( c ) _ for some integer k . By substitution, n 2 = _ _ ( d ) _ _ = 5 ( 5 k 2 ) . But 5 k 2 is an integer because it is a product of integers. Hence n 2 = 5 ⋅ (an integer), and so __(e)___ [as was to be shown].
Solution Summary: The author explains that 5|n is an integer because it is a product of integers.
Fill in the blanks in the following proof by contraposition that for every integer n, if
5
/
n
2
then
5
/
n
.
Proof (by contraposition): [The contrapositive is: For every integer n, if 5|nthen 5|
n
2
.] Suppose n is any integer such that __(a)__ [We must show that __(b)__] By definition of divisibility,
n
=
_
_
(
c
)
_
for some integer k. By substitution,
n
2
=
_
_
(
d
)
_
_
=
5
(
5
k
2
)
.
But
5
k
2
is an integer because it is a product of integers. Hence
n
2
=
5
⋅
(an integer), and so __(e)___ [as was to be shown].
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RELATIONS-DOMAIN, RANGE AND CO-DOMAIN (RELATIONS AND FUNCTIONS CBSE/ ISC MATHS); Author: Neha Agrawal Mathematically Inclined;https://www.youtube.com/watch?v=u4IQh46VoU4;License: Standard YouTube License, CC-BY