Using distributive property to prove that the product of two odd numbers is odd
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I've been reading Algebra The Easy Way, and there's a problem at the end of the chapter to prove that the product of two odd numbers is an odd number.
In this problem, an even number is defined as two multiplied by a natural number, so (2 * m). An odd number is defined as an even number plus one, so (2 * n + 1).
Therefore, two odd numbers multiplied by each other is
(2 * m + 1) * ( 2 * n + 1).
The answer says to use the distributive property to convert this to
(2 * m + 1) * 2 * n + (2 * m + 1)
My question is how does the distributive property allow (2 * m + 1) * ( 2 * n + 1) to be converted to (2 * m + 1) * 2 * n + (2 * m + 1)? Is this a typo in the book, or is it actually possible to do this conversion using the distributive property?
algebra-precalculus
asked Nov 13 at 22:16
Toby Artisan
1032
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up vote
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down vote
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I've been reading Algebra The Easy Way, and there's a problem at the end of the chapter to prove that the product of two odd numbers is an odd number.
In this problem, an even number is defined as two multiplied by a natural number, so (2 * m). An odd number is defined as an even number plus one, so (2 * n + 1).
Therefore, two odd numbers multiplied by each other is
(2 * m + 1) * ( 2 * n + 1).
The answer says to use the distributive property to convert this to
(2 * m + 1) * 2 * n + (2 * m + 1)
My question is how does the distributive property allow (2 * m + 1) * ( 2 * n + 1) to be converted to (2 * m + 1) * 2 * n + (2 * m + 1)? Is this a typo in the book, or is it actually possible to do this conversion using the distributive property?
algebra-precalculus
asked Nov 13 at 22:16
Toby Artisan
1032
1
Because $2m + 1 = 1 cdot (2m + 1)$.
– T. Bongers
Nov 13 at 22:20
Yes. (2*n+1)=(2*n)+(1), so (2*m+1) can distribute.
– herb steinberg
Nov 13 at 22:45
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've been reading Algebra The Easy Way, and there's a problem at the end of the chapter to prove that the product of two odd numbers is an odd number.
In this problem, an even number is defined as two multiplied by a natural number, so (2 * m). An odd number is defined as an even number plus one, so (2 * n + 1).
Therefore, two odd numbers multiplied by each other is
(2 * m + 1) * ( 2 * n + 1).
The answer says to use the distributive property to convert this to
(2 * m + 1) * 2 * n + (2 * m + 1)
My question is how does the distributive property allow (2 * m + 1) * ( 2 * n + 1) to be converted to (2 * m + 1) * 2 * n + (2 * m + 1)? Is this a typo in the book, or is it actually possible to do this conversion using the distributive property?
algebra-precalculus
asked Nov 13 at 22:16
Toby Artisan
1032
I've been reading Algebra The Easy Way, and there's a problem at the end of the chapter to prove that the product of two odd numbers is an odd number.
In this problem, an even number is defined as two multiplied by a natural number, so (2 * m). An odd number is defined as an even number plus one, so (2 * n + 1).
Therefore, two odd numbers multiplied by each other is
(2 * m + 1) * ( 2 * n + 1).
The answer says to use the distributive property to convert this to
(2 * m + 1) * 2 * n + (2 * m + 1)
My question is how does the distributive property allow (2 * m + 1) * ( 2 * n + 1) to be converted to (2 * m + 1) * 2 * n + (2 * m + 1)? Is this a typo in the book, or is it actually possible to do this conversion using the distributive property?
algebra-precalculus
algebra-precalculus
asked Nov 13 at 22:16
Toby Artisan
1032
asked Nov 13 at 22:16
Toby Artisan
1032
asked Nov 13 at 22:16
Toby Artisan
1032
asked Nov 13 at 22:16
Toby Artisan
1032
asked Nov 13 at 22:16
Toby Artisan
1032
1032
1
Because $2m + 1 = 1 cdot (2m + 1)$.
– T. Bongers
Nov 13 at 22:20
Yes. (2*n+1)=(2*n)+(1), so (2*m+1) can distribute.
– herb steinberg
Nov 13 at 22:45
add a comment |
1
Because $2m + 1 = 1 cdot (2m + 1)$.
– T. Bongers
Nov 13 at 22:20
Yes. (2*n+1)=(2*n)+(1), so (2*m+1) can distribute.
– herb steinberg
Nov 13 at 22:45
1
1
Because $2m + 1 = 1 cdot (2m + 1)$.
– T. Bongers
Nov 13 at 22:20
Because $2m + 1 = 1 cdot (2m + 1)$.
– T. Bongers
Nov 13 at 22:20
Yes. (2*n+1)=(2*n)+(1), so (2*m+1) can distribute.
– herb steinberg
Nov 13 at 22:45
Yes. (2*n+1)=(2*n)+(1), so (2*m+1) can distribute.
– herb steinberg
Nov 13 at 22:45
add a comment |
1 Answer
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1
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To answer your specific question, let $nu = (2m+1)$. Then, with substitution, $$(2m+1)(2n+1) = \
nu(2n+1) = \ nu2n + nu
$$Does that last line make sense? If not, please have a look over distributive multiplication. Going on, replace $nu$ above with $(2m+1)$ and you get the answer in the book.
The overall proof may actually be easier to use FOIL, to obtain $4mn + 2m + 2n + 1$ then factor out a 2 to get $$2(2mn + m + n) + 1 = \2(mathrm{some integer})+1$$Since you've established that an odd number has the form, $2i+1$ for some integer $i$, and the above expansion has that form, then the above expansion is odd. Since the above expansion was created as a product of two odd numbers, then the product is an odd number.
answered Nov 14 at 0:35
Russ
37919
Using "v" as a substitution makes it perfectly clear. I also like your suggestion of using FOIL. Thank you.
– Toby Artisan
Nov 14 at 14:59
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
To answer your specific question, let $nu = (2m+1)$. Then, with substitution, $$(2m+1)(2n+1) = \
nu(2n+1) = \ nu2n + nu
$$Does that last line make sense? If not, please have a look over distributive multiplication. Going on, replace $nu$ above with $(2m+1)$ and you get the answer in the book.
The overall proof may actually be easier to use FOIL, to obtain $4mn + 2m + 2n + 1$ then factor out a 2 to get $$2(2mn + m + n) + 1 = \2(mathrm{some integer})+1$$Since you've established that an odd number has the form, $2i+1$ for some integer $i$, and the above expansion has that form, then the above expansion is odd. Since the above expansion was created as a product of two odd numbers, then the product is an odd number.
answered Nov 14 at 0:35
Russ
37919
Using "v" as a substitution makes it perfectly clear. I also like your suggestion of using FOIL. Thank you.
– Toby Artisan
Nov 14 at 14:59
add a comment |
up vote
1
down vote
accepted
To answer your specific question, let $nu = (2m+1)$. Then, with substitution, $$(2m+1)(2n+1) = \
nu(2n+1) = \ nu2n + nu
$$Does that last line make sense? If not, please have a look over distributive multiplication. Going on, replace $nu$ above with $(2m+1)$ and you get the answer in the book.
The overall proof may actually be easier to use FOIL, to obtain $4mn + 2m + 2n + 1$ then factor out a 2 to get $$2(2mn + m + n) + 1 = \2(mathrm{some integer})+1$$Since you've established that an odd number has the form, $2i+1$ for some integer $i$, and the above expansion has that form, then the above expansion is odd. Since the above expansion was created as a product of two odd numbers, then the product is an odd number.
answered Nov 14 at 0:35
Russ
37919
Using "v" as a substitution makes it perfectly clear. I also like your suggestion of using FOIL. Thank you.
– Toby Artisan
Nov 14 at 14:59
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To answer your specific question, let $nu = (2m+1)$. Then, with substitution, $$(2m+1)(2n+1) = \
nu(2n+1) = \ nu2n + nu
$$Does that last line make sense? If not, please have a look over distributive multiplication. Going on, replace $nu$ above with $(2m+1)$ and you get the answer in the book.
The overall proof may actually be easier to use FOIL, to obtain $4mn + 2m + 2n + 1$ then factor out a 2 to get $$2(2mn + m + n) + 1 = \2(mathrm{some integer})+1$$Since you've established that an odd number has the form, $2i+1$ for some integer $i$, and the above expansion has that form, then the above expansion is odd. Since the above expansion was created as a product of two odd numbers, then the product is an odd number.
answered Nov 14 at 0:35
Russ
37919
To answer your specific question, let $nu = (2m+1)$. Then, with substitution, $$(2m+1)(2n+1) = \
nu(2n+1) = \ nu2n + nu
$$Does that last line make sense? If not, please have a look over distributive multiplication. Going on, replace $nu$ above with $(2m+1)$ and you get the answer in the book.
The overall proof may actually be easier to use FOIL, to obtain $4mn + 2m + 2n + 1$ then factor out a 2 to get $$2(2mn + m + n) + 1 = \2(mathrm{some integer})+1$$Since you've established that an odd number has the form, $2i+1$ for some integer $i$, and the above expansion has that form, then the above expansion is odd. Since the above expansion was created as a product of two odd numbers, then the product is an odd number.
answered Nov 14 at 0:35
Russ
37919
answered Nov 14 at 0:35
Russ
37919
answered Nov 14 at 0:35
Russ
37919
answered Nov 14 at 0:35
Russ
37919
37919
Using "v" as a substitution makes it perfectly clear. I also like your suggestion of using FOIL. Thank you.
– Toby Artisan
Nov 14 at 14:59
add a comment |
Using "v" as a substitution makes it perfectly clear. I also like your suggestion of using FOIL. Thank you.
– Toby Artisan
Nov 14 at 14:59
Using "v" as a substitution makes it perfectly clear. I also like your suggestion of using FOIL. Thank you.
– Toby Artisan
Nov 14 at 14:59
Using "v" as a substitution makes it perfectly clear. I also like your suggestion of using FOIL. Thank you.
– Toby Artisan
Nov 14 at 14:59
add a comment |
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Because $2m + 1 = 1 cdot (2m + 1)$.
– T. Bongers
Nov 13 at 22:20
Yes. (2*n+1)=(2*n)+(1), so (2*m+1) can distribute.
– herb steinberg
Nov 13 at 22:45