Mixing
Why can we multiply?
Clash Royale CLAN TAG#URR8PPP
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My friend and I were discussing some mathematical philosophy and how the numbers systems were created when we reached a question. Why can we multiply two different number like this?:
Say we had to multiply 13*34. One may break this up like (10+3)*(30+4). Applying distributive property here will give us the answer of 442. We can also choose to multiply this as (6+7)(22+12). Intuitively, we can hypothesize that this should give us the same answer as 13*34. How can we prove that our answer will be equal regardless of how we break up the numbers?
Thanks
algebra-precalculus arithmetic
edited 36 mins ago
Henning Makholm
233k16298531
asked 43 mins ago
Dude156
32212
add a comment |Â
up vote
3
down vote
favorite
My friend and I were discussing some mathematical philosophy and how the numbers systems were created when we reached a question. Why can we multiply two different number like this?:
Say we had to multiply 13*34. One may break this up like (10+3)*(30+4). Applying distributive property here will give us the answer of 442. We can also choose to multiply this as (6+7)(22+12). Intuitively, we can hypothesize that this should give us the same answer as 13*34. How can we prove that our answer will be equal regardless of how we break up the numbers?
Thanks
algebra-precalculus arithmetic
edited 36 mins ago
Henning Makholm
233k16298531
asked 43 mins ago
Dude156
32212
3
This will depend wildly on how you have defined multiplication in the first place, and which axioms you're willing you let your proof rely on.
– Henning Makholm
35 mins ago
Is this a stupid question? Or is it ok to ask this.... I honestly can't tell
– Dude156
34 mins ago
1
You can do this because the distributive property works and is inviolate. Now why the universe (or our brains) work this way? Who knows?
– fleablood
27 mins ago
2
It's not a dumb question. Investigating structures where this isn't the case is a part of abstract algebra. If you take your question seriously you will be able to find out exactly what rules (in this algebraic structure) allow you to make these leaps.
– Mason
16 mins ago
Thanks for the comforting reply Mr. Mason. I've been thinking about these questions and I think I am going insane sometimes lol.
– Dude156
2 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
My friend and I were discussing some mathematical philosophy and how the numbers systems were created when we reached a question. Why can we multiply two different number like this?:
Say we had to multiply 13*34. One may break this up like (10+3)*(30+4). Applying distributive property here will give us the answer of 442. We can also choose to multiply this as (6+7)(22+12). Intuitively, we can hypothesize that this should give us the same answer as 13*34. How can we prove that our answer will be equal regardless of how we break up the numbers?
Thanks
algebra-precalculus arithmetic
edited 36 mins ago
Henning Makholm
233k16298531
asked 43 mins ago
Dude156
32212
My friend and I were discussing some mathematical philosophy and how the numbers systems were created when we reached a question. Why can we multiply two different number like this?:
Say we had to multiply 13*34. One may break this up like (10+3)*(30+4). Applying distributive property here will give us the answer of 442. We can also choose to multiply this as (6+7)(22+12). Intuitively, we can hypothesize that this should give us the same answer as 13*34. How can we prove that our answer will be equal regardless of how we break up the numbers?
Thanks
algebra-precalculus arithmetic
algebra-precalculus arithmetic
edited 36 mins ago
Henning Makholm
233k16298531
asked 43 mins ago
Dude156
32212
edited 36 mins ago
Henning Makholm
233k16298531
asked 43 mins ago
Dude156
32212
edited 36 mins ago
Henning Makholm
233k16298531
edited 36 mins ago
Henning Makholm
233k16298531
edited 36 mins ago
Henning Makholm
233k16298531
233k16298531
asked 43 mins ago
Dude156
32212
asked 43 mins ago
Dude156
32212
asked 43 mins ago
Dude156
32212
32212
3
This will depend wildly on how you have defined multiplication in the first place, and which axioms you're willing you let your proof rely on.
– Henning Makholm
35 mins ago
Is this a stupid question? Or is it ok to ask this.... I honestly can't tell
– Dude156
34 mins ago
1
You can do this because the distributive property works and is inviolate. Now why the universe (or our brains) work this way? Who knows?
– fleablood
27 mins ago
2
It's not a dumb question. Investigating structures where this isn't the case is a part of abstract algebra. If you take your question seriously you will be able to find out exactly what rules (in this algebraic structure) allow you to make these leaps.
– Mason
16 mins ago
Thanks for the comforting reply Mr. Mason. I've been thinking about these questions and I think I am going insane sometimes lol.
– Dude156
2 mins ago
add a comment |Â
3
This will depend wildly on how you have defined multiplication in the first place, and which axioms you're willing you let your proof rely on.
– Henning Makholm
35 mins ago
Is this a stupid question? Or is it ok to ask this.... I honestly can't tell
– Dude156
34 mins ago
1
You can do this because the distributive property works and is inviolate. Now why the universe (or our brains) work this way? Who knows?
– fleablood
27 mins ago
2
It's not a dumb question. Investigating structures where this isn't the case is a part of abstract algebra. If you take your question seriously you will be able to find out exactly what rules (in this algebraic structure) allow you to make these leaps.
– Mason
16 mins ago
Thanks for the comforting reply Mr. Mason. I've been thinking about these questions and I think I am going insane sometimes lol.
– Dude156
2 mins ago
3
3
This will depend wildly on how you have defined multiplication in the first place, and which axioms you're willing you let your proof rely on.
– Henning Makholm
35 mins ago
This will depend wildly on how you have defined multiplication in the first place, and which axioms you're willing you let your proof rely on.
– Henning Makholm
35 mins ago
Is this a stupid question? Or is it ok to ask this.... I honestly can't tell
– Dude156
34 mins ago
Is this a stupid question? Or is it ok to ask this.... I honestly can't tell
– Dude156
34 mins ago
1
1
You can do this because the distributive property works and is inviolate. Now why the universe (or our brains) work this way? Who knows?
– fleablood
27 mins ago
You can do this because the distributive property works and is inviolate. Now why the universe (or our brains) work this way? Who knows?
– fleablood
27 mins ago
2
2
It's not a dumb question. Investigating structures where this isn't the case is a part of abstract algebra. If you take your question seriously you will be able to find out exactly what rules (in this algebraic structure) allow you to make these leaps.
– Mason
16 mins ago
It's not a dumb question. Investigating structures where this isn't the case is a part of abstract algebra. If you take your question seriously you will be able to find out exactly what rules (in this algebraic structure) allow you to make these leaps.
– Mason
16 mins ago
Thanks for the comforting reply Mr. Mason. I've been thinking about these questions and I think I am going insane sometimes lol.
– Dude156
2 mins ago
Thanks for the comforting reply Mr. Mason. I've been thinking about these questions and I think I am going insane sometimes lol.
– Dude156
2 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
There are two ways I see to answer this question. One is from an axiomatic standpoint, where numbers are merely symbols on paper that are required to follow certain rules. The other uses the interpretation of multiplication as computing area. The former would take a good 5-10 pages to build up from the Peano axioms.
For the latter, you can draw a rectangle 13 units by 34 units. Break one side into 10 units and 3 units, and the other side into 30 units and 4 units. At this point, you should see that this decomposes the rectangle into four pieces, corresponding to the four terms you get from the distributive rule. The various was to compute $13 cdot 34$ are all just ways to decompose the rectangle, and at the end of the day they all compute the same number: the area of the rectangle.
answered 33 mins ago
RghtHndSd
4,92621431
But would that be a proof? Or would it just be intuition? Idk, i've seen multiplication defined wildly differently. Some folks use multiplication as repeated addition, while others use it as area. What is the correct interpretation?
– Dude156
32 mins ago
1
@Dude156: Defining multiplication as repeated addition is the axiomatic viewpoint I was referring to. And one can prove the distributive property from this definition, without any reference to area. However you should also see the definition of area, where the sides are whole numbers, as repeated addition. What the area interpretation offers is a way to see geometrically what's going on, and in my opinion, it greatly clarifies what's going on with the distributive property.
– RghtHndSd
25 mins ago
There is no one "correct" interpretation. But repeated addition only works for multiplying by integers. And integer area is repeated addition ....
– fleablood
23 mins ago
add a comment |Â
up vote
1
down vote
Well, that is literally what the distributive law tells you. It tells you that $$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd$$ and so whenever you break up the two factors of a product as a sum, you can use the "pieces" to compute the product.
So what you are really asking for is a proof of the distributive law itself. What constitutes a "proof" of such a basic fact depends heavily on what definitions of "numbers" and the operations on them that you are using (for some definitions, the distributive law is simply an axiom that you assume). But here is an intuitive explanation that works for natural numbers (and this can be turned into a rigorous proof if you define arithmetic of natural numbers in terms of cardinalities of sets).
We want to prove that $(a+b)c=ac+bc$. What does a product $xy$ of natural numbers mean? Well, it means you draw a grid of dots with $x$ rows and $y$ columns, and count up the total number of dots. So, to compute $(a+b)c$, you draw a grid with $a+b$ rows and $c$ columns. Now we observe that we can split such a grid into two pieces: the top $a$ rows and the bottom $b$ rows. The top $a$ rows form a grid with $a$ rows and $c$ columns, so they have $ac$ dots. The bottom $b$ rows form a grid with $b$ rows and $c$ columns, so they have $bc$ dots. In total, then, we have $ac+bc$ dots, so $(a+b)c=ac+bc$.
(In the computation of $(a+b)(c+d)$ above I used the distributive law in two different versions, one with the sum on the left side of the product and one with the sum on the right side of the product. You can prove the version with the sum on the right side of the product in the same way (you just split up the columns instead of the rows), or you can deduce it from the other version using commutativity of multiplication.)
answered 28 mins ago
Eric Wofsey
172k12198318
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
There are two ways I see to answer this question. One is from an axiomatic standpoint, where numbers are merely symbols on paper that are required to follow certain rules. The other uses the interpretation of multiplication as computing area. The former would take a good 5-10 pages to build up from the Peano axioms.
For the latter, you can draw a rectangle 13 units by 34 units. Break one side into 10 units and 3 units, and the other side into 30 units and 4 units. At this point, you should see that this decomposes the rectangle into four pieces, corresponding to the four terms you get from the distributive rule. The various was to compute $13 cdot 34$ are all just ways to decompose the rectangle, and at the end of the day they all compute the same number: the area of the rectangle.
answered 33 mins ago
RghtHndSd
4,92621431
But would that be a proof? Or would it just be intuition? Idk, i've seen multiplication defined wildly differently. Some folks use multiplication as repeated addition, while others use it as area. What is the correct interpretation?
– Dude156
32 mins ago
1
@Dude156: Defining multiplication as repeated addition is the axiomatic viewpoint I was referring to. And one can prove the distributive property from this definition, without any reference to area. However you should also see the definition of area, where the sides are whole numbers, as repeated addition. What the area interpretation offers is a way to see geometrically what's going on, and in my opinion, it greatly clarifies what's going on with the distributive property.
– RghtHndSd
25 mins ago
There is no one "correct" interpretation. But repeated addition only works for multiplying by integers. And integer area is repeated addition ....
– fleablood
23 mins ago
add a comment |Â
up vote
4
down vote
There are two ways I see to answer this question. One is from an axiomatic standpoint, where numbers are merely symbols on paper that are required to follow certain rules. The other uses the interpretation of multiplication as computing area. The former would take a good 5-10 pages to build up from the Peano axioms.
For the latter, you can draw a rectangle 13 units by 34 units. Break one side into 10 units and 3 units, and the other side into 30 units and 4 units. At this point, you should see that this decomposes the rectangle into four pieces, corresponding to the four terms you get from the distributive rule. The various was to compute $13 cdot 34$ are all just ways to decompose the rectangle, and at the end of the day they all compute the same number: the area of the rectangle.
answered 33 mins ago
RghtHndSd
4,92621431
But would that be a proof? Or would it just be intuition? Idk, i've seen multiplication defined wildly differently. Some folks use multiplication as repeated addition, while others use it as area. What is the correct interpretation?
– Dude156
32 mins ago
1
@Dude156: Defining multiplication as repeated addition is the axiomatic viewpoint I was referring to. And one can prove the distributive property from this definition, without any reference to area. However you should also see the definition of area, where the sides are whole numbers, as repeated addition. What the area interpretation offers is a way to see geometrically what's going on, and in my opinion, it greatly clarifies what's going on with the distributive property.
– RghtHndSd
25 mins ago
There is no one "correct" interpretation. But repeated addition only works for multiplying by integers. And integer area is repeated addition ....
– fleablood
23 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
There are two ways I see to answer this question. One is from an axiomatic standpoint, where numbers are merely symbols on paper that are required to follow certain rules. The other uses the interpretation of multiplication as computing area. The former would take a good 5-10 pages to build up from the Peano axioms.
For the latter, you can draw a rectangle 13 units by 34 units. Break one side into 10 units and 3 units, and the other side into 30 units and 4 units. At this point, you should see that this decomposes the rectangle into four pieces, corresponding to the four terms you get from the distributive rule. The various was to compute $13 cdot 34$ are all just ways to decompose the rectangle, and at the end of the day they all compute the same number: the area of the rectangle.
answered 33 mins ago
RghtHndSd
4,92621431
There are two ways I see to answer this question. One is from an axiomatic standpoint, where numbers are merely symbols on paper that are required to follow certain rules. The other uses the interpretation of multiplication as computing area. The former would take a good 5-10 pages to build up from the Peano axioms.
For the latter, you can draw a rectangle 13 units by 34 units. Break one side into 10 units and 3 units, and the other side into 30 units and 4 units. At this point, you should see that this decomposes the rectangle into four pieces, corresponding to the four terms you get from the distributive rule. The various was to compute $13 cdot 34$ are all just ways to decompose the rectangle, and at the end of the day they all compute the same number: the area of the rectangle.
answered 33 mins ago
RghtHndSd
4,92621431
answered 33 mins ago
RghtHndSd
4,92621431
answered 33 mins ago
RghtHndSd
4,92621431
answered 33 mins ago
RghtHndSd
4,92621431
4,92621431
But would that be a proof? Or would it just be intuition? Idk, i've seen multiplication defined wildly differently. Some folks use multiplication as repeated addition, while others use it as area. What is the correct interpretation?
– Dude156
32 mins ago
1
@Dude156: Defining multiplication as repeated addition is the axiomatic viewpoint I was referring to. And one can prove the distributive property from this definition, without any reference to area. However you should also see the definition of area, where the sides are whole numbers, as repeated addition. What the area interpretation offers is a way to see geometrically what's going on, and in my opinion, it greatly clarifies what's going on with the distributive property.
– RghtHndSd
25 mins ago
There is no one "correct" interpretation. But repeated addition only works for multiplying by integers. And integer area is repeated addition ....
– fleablood
23 mins ago
add a comment |Â
But would that be a proof? Or would it just be intuition? Idk, i've seen multiplication defined wildly differently. Some folks use multiplication as repeated addition, while others use it as area. What is the correct interpretation?
– Dude156
32 mins ago
1
@Dude156: Defining multiplication as repeated addition is the axiomatic viewpoint I was referring to. And one can prove the distributive property from this definition, without any reference to area. However you should also see the definition of area, where the sides are whole numbers, as repeated addition. What the area interpretation offers is a way to see geometrically what's going on, and in my opinion, it greatly clarifies what's going on with the distributive property.
– RghtHndSd
25 mins ago
There is no one "correct" interpretation. But repeated addition only works for multiplying by integers. And integer area is repeated addition ....
– fleablood
23 mins ago
But would that be a proof? Or would it just be intuition? Idk, i've seen multiplication defined wildly differently. Some folks use multiplication as repeated addition, while others use it as area. What is the correct interpretation?
– Dude156
32 mins ago
But would that be a proof? Or would it just be intuition? Idk, i've seen multiplication defined wildly differently. Some folks use multiplication as repeated addition, while others use it as area. What is the correct interpretation?
– Dude156
32 mins ago
1
1
@Dude156: Defining multiplication as repeated addition is the axiomatic viewpoint I was referring to. And one can prove the distributive property from this definition, without any reference to area. However you should also see the definition of area, where the sides are whole numbers, as repeated addition. What the area interpretation offers is a way to see geometrically what's going on, and in my opinion, it greatly clarifies what's going on with the distributive property.
– RghtHndSd
25 mins ago
@Dude156: Defining multiplication as repeated addition is the axiomatic viewpoint I was referring to. And one can prove the distributive property from this definition, without any reference to area. However you should also see the definition of area, where the sides are whole numbers, as repeated addition. What the area interpretation offers is a way to see geometrically what's going on, and in my opinion, it greatly clarifies what's going on with the distributive property.
– RghtHndSd
25 mins ago
There is no one "correct" interpretation. But repeated addition only works for multiplying by integers. And integer area is repeated addition ....
– fleablood
23 mins ago
There is no one "correct" interpretation. But repeated addition only works for multiplying by integers. And integer area is repeated addition ....
– fleablood
23 mins ago
add a comment |Â
up vote
1
down vote
Well, that is literally what the distributive law tells you. It tells you that $$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd$$ and so whenever you break up the two factors of a product as a sum, you can use the "pieces" to compute the product.
So what you are really asking for is a proof of the distributive law itself. What constitutes a "proof" of such a basic fact depends heavily on what definitions of "numbers" and the operations on them that you are using (for some definitions, the distributive law is simply an axiom that you assume). But here is an intuitive explanation that works for natural numbers (and this can be turned into a rigorous proof if you define arithmetic of natural numbers in terms of cardinalities of sets).
We want to prove that $(a+b)c=ac+bc$. What does a product $xy$ of natural numbers mean? Well, it means you draw a grid of dots with $x$ rows and $y$ columns, and count up the total number of dots. So, to compute $(a+b)c$, you draw a grid with $a+b$ rows and $c$ columns. Now we observe that we can split such a grid into two pieces: the top $a$ rows and the bottom $b$ rows. The top $a$ rows form a grid with $a$ rows and $c$ columns, so they have $ac$ dots. The bottom $b$ rows form a grid with $b$ rows and $c$ columns, so they have $bc$ dots. In total, then, we have $ac+bc$ dots, so $(a+b)c=ac+bc$.
(In the computation of $(a+b)(c+d)$ above I used the distributive law in two different versions, one with the sum on the left side of the product and one with the sum on the right side of the product. You can prove the version with the sum on the right side of the product in the same way (you just split up the columns instead of the rows), or you can deduce it from the other version using commutativity of multiplication.)
answered 28 mins ago
Eric Wofsey
172k12198318
add a comment |Â
up vote
1
down vote
Well, that is literally what the distributive law tells you. It tells you that $$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd$$ and so whenever you break up the two factors of a product as a sum, you can use the "pieces" to compute the product.
So what you are really asking for is a proof of the distributive law itself. What constitutes a "proof" of such a basic fact depends heavily on what definitions of "numbers" and the operations on them that you are using (for some definitions, the distributive law is simply an axiom that you assume). But here is an intuitive explanation that works for natural numbers (and this can be turned into a rigorous proof if you define arithmetic of natural numbers in terms of cardinalities of sets).
We want to prove that $(a+b)c=ac+bc$. What does a product $xy$ of natural numbers mean? Well, it means you draw a grid of dots with $x$ rows and $y$ columns, and count up the total number of dots. So, to compute $(a+b)c$, you draw a grid with $a+b$ rows and $c$ columns. Now we observe that we can split such a grid into two pieces: the top $a$ rows and the bottom $b$ rows. The top $a$ rows form a grid with $a$ rows and $c$ columns, so they have $ac$ dots. The bottom $b$ rows form a grid with $b$ rows and $c$ columns, so they have $bc$ dots. In total, then, we have $ac+bc$ dots, so $(a+b)c=ac+bc$.
(In the computation of $(a+b)(c+d)$ above I used the distributive law in two different versions, one with the sum on the left side of the product and one with the sum on the right side of the product. You can prove the version with the sum on the right side of the product in the same way (you just split up the columns instead of the rows), or you can deduce it from the other version using commutativity of multiplication.)
answered 28 mins ago
Eric Wofsey
172k12198318
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Well, that is literally what the distributive law tells you. It tells you that $$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd$$ and so whenever you break up the two factors of a product as a sum, you can use the "pieces" to compute the product.
So what you are really asking for is a proof of the distributive law itself. What constitutes a "proof" of such a basic fact depends heavily on what definitions of "numbers" and the operations on them that you are using (for some definitions, the distributive law is simply an axiom that you assume). But here is an intuitive explanation that works for natural numbers (and this can be turned into a rigorous proof if you define arithmetic of natural numbers in terms of cardinalities of sets).
We want to prove that $(a+b)c=ac+bc$. What does a product $xy$ of natural numbers mean? Well, it means you draw a grid of dots with $x$ rows and $y$ columns, and count up the total number of dots. So, to compute $(a+b)c$, you draw a grid with $a+b$ rows and $c$ columns. Now we observe that we can split such a grid into two pieces: the top $a$ rows and the bottom $b$ rows. The top $a$ rows form a grid with $a$ rows and $c$ columns, so they have $ac$ dots. The bottom $b$ rows form a grid with $b$ rows and $c$ columns, so they have $bc$ dots. In total, then, we have $ac+bc$ dots, so $(a+b)c=ac+bc$.
(In the computation of $(a+b)(c+d)$ above I used the distributive law in two different versions, one with the sum on the left side of the product and one with the sum on the right side of the product. You can prove the version with the sum on the right side of the product in the same way (you just split up the columns instead of the rows), or you can deduce it from the other version using commutativity of multiplication.)
answered 28 mins ago
Eric Wofsey
172k12198318
Well, that is literally what the distributive law tells you. It tells you that $$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd$$ and so whenever you break up the two factors of a product as a sum, you can use the "pieces" to compute the product.
So what you are really asking for is a proof of the distributive law itself. What constitutes a "proof" of such a basic fact depends heavily on what definitions of "numbers" and the operations on them that you are using (for some definitions, the distributive law is simply an axiom that you assume). But here is an intuitive explanation that works for natural numbers (and this can be turned into a rigorous proof if you define arithmetic of natural numbers in terms of cardinalities of sets).
We want to prove that $(a+b)c=ac+bc$. What does a product $xy$ of natural numbers mean? Well, it means you draw a grid of dots with $x$ rows and $y$ columns, and count up the total number of dots. So, to compute $(a+b)c$, you draw a grid with $a+b$ rows and $c$ columns. Now we observe that we can split such a grid into two pieces: the top $a$ rows and the bottom $b$ rows. The top $a$ rows form a grid with $a$ rows and $c$ columns, so they have $ac$ dots. The bottom $b$ rows form a grid with $b$ rows and $c$ columns, so they have $bc$ dots. In total, then, we have $ac+bc$ dots, so $(a+b)c=ac+bc$.
(In the computation of $(a+b)(c+d)$ above I used the distributive law in two different versions, one with the sum on the left side of the product and one with the sum on the right side of the product. You can prove the version with the sum on the right side of the product in the same way (you just split up the columns instead of the rows), or you can deduce it from the other version using commutativity of multiplication.)
answered 28 mins ago
Eric Wofsey
172k12198318
answered 28 mins ago
Eric Wofsey
172k12198318
answered 28 mins ago
Eric Wofsey
172k12198318
answered 28 mins ago
Eric Wofsey
172k12198318
172k12198318
add a comment |Â
add a comment |Â
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3
This will depend wildly on how you have defined multiplication in the first place, and which axioms you're willing you let your proof rely on.
– Henning Makholm
35 mins ago
Is this a stupid question? Or is it ok to ask this.... I honestly can't tell
– Dude156
34 mins ago
1
You can do this because the distributive property works and is inviolate. Now why the universe (or our brains) work this way? Who knows?
– fleablood
27 mins ago
2
It's not a dumb question. Investigating structures where this isn't the case is a part of abstract algebra. If you take your question seriously you will be able to find out exactly what rules (in this algebraic structure) allow you to make these leaps.
– Mason
16 mins ago
Thanks for the comforting reply Mr. Mason. I've been thinking about these questions and I think I am going insane sometimes lol.
– Dude156
2 mins ago