B Basic Algebra Question: 3x = 15

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Not really. If the coefficient is zero, then its reciprocal is undefined.
It may sometimes be easier to notice that the coefficient is zero if we aim to multiply by its reciprocal instead of directly dividing by the coefficient. In the first false proof example from the referenced math humor page:
Given that a and b are integers such that a = b + 1,Prove: 1 = 0
1. a = b + 11. Given
2. (a-b)a = (a-b)(b+1)2. Multiplication Prop. of =
3. a2 - ab = ab + a - b2 - b3. Distributive Property
4. a2 - ab -a = ab + a -a - b2 - b4. Subtraction Prop. of =
5. a(a - b - 1) = b(a - b - 1)5. Distributive Property
6. a = b6. Division Property of =
7. b + 1 = b7. Transitive Property of = (Steps 1, 7)
8. Therefore, 1 = 08. Subtraction Prop. of =
Statement '5.' is vacuously true, but given statement '1.' as true, statement '6.' cannot be true. The false proof works partly by obscuring the zero divisor in the bilateral division operation employed to justify the transition from statement '5.' to statement '6.'. I think it is harder to fail to notice that the divisor is zero if the division operation is seen as multiplying by ##1/(a-b-1)## instead of as simply canceling the ##(a-b-1)##, whatever that quantity might be, from both sides.
 
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mathwonk

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wow. i wonder how we would explain how to solve 1.X = 5?, i.e. divide by 1, or multiply by 1^-1.
 

symbolipoint

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wow. i wonder how we would explain how to solve 1.X = 5?, i.e. divide by 1, or multiply by 1^-1.
Is the low-positioned dot intended to be the multiplication operation symbol? The person would be expected to recognize the Identity Element 1, and understand that X is 5.

EDIT: Another way to say this is, the person would logically recognize the occurrence of the identity element 1, and understand that X is 5.
 
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The false proof works partly by obscuring the zero dividend in the bilateral division operation employed to justify the transition from statement '5.' to statement '6.'.
Except that from statement 1, a = b + 1, which is equivalent to a - b = 1, we can see that a - b - 1 = 0. This means that we are dividing by 0 going from step 5 to step 6.
 
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wow. i wonder how we would explain how to solve 1.X = 5?, i.e. divide by 1, or multiply by 1^-1.
No need to divide by anything, as 1 is the multiplicative identity: ##1 \cdot a = a## for any a.
 
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Except that from statement 1, a = b + 1, which is equivalent to a - b = 1, we can see that a - b - 1 = 0. This means that we are dividing by 0 going from step 5 to step 6.
Well, we can also start by recognizing that 1 ≠ 0, and not bother finding which step in the false proof is in error. In the false proof example, in which "we are dividing by 0 going from step 5 to step 6" is quod erat obscurandum (what was to be obscured), multiplying by the reciprocal instead of just canceling out the unevaluated parenthetical expressions, can give us pause to examine their content, and so be more apt to not fail to notice that it sums to zero.
 
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Well, we can also start by recognizing that 1 ≠ 0, and not bother finding which step in the false proof is in error.
Well, clearly 1 ≠ 0, but sorting out this contradiction requires finding the exact step that is invalid. I don't think we're in disagreement, since both dividing by zero or multiplying by the reciprocal of something that is zero are invalid operations.

For the record, when I taught algebra, I avoided the use of the term "cancel," as too often students equated cancalling things with just crossing them off, such as "cancelling" the 5s in this expression, ##\frac {2 + 1} 2##, obtaining 1.
 

DrClaude

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I would have never thought that the question "How do you solve ##3x=15##?" would bring about so much discussion.

In addition to the discussion of multiplying by 1/3 versus dividing by 3, I would like to know if anyone else visualizes the operations as I do. I understand (I think :smile:) the formal mathematics of the transformation from ##3x = 15## to ##x = 15/3##, and I am careful, for example, when discussing it with my son to always talk about it in terms of applying the same operation to both sides of the equation, but this is not how I picture it when I do algebra. For me, the symbols move from one side of the equation to the other, as in
algebra001.jpg

Likewise, if we had ##x + 3 = 15##, I would "see" the 3 going from the lhs to the rhs and acquiring a minus sing in the process.
 

pinball1970

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Same thing.
Study of Basic Algebra
Thanks.
I was asking why chose to multiply rather than divide when the unknown is already being multiplied by something.
Same thing.
Study of Basic Algebra
Yes I get that I was asking why choose to multiply over division.
It looks like things have moved on a bit since then.
 
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Really @DrClaude? You would call the fraction bar in ##\frac {15} 3## a minus sign? 😉
 

pinball1970

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I would have never thought that the question "How do you solve ##3x=15##?" would bring about so much discussion.

In addition to the discussion of multiplying by 1/3 versus dividing by 3, I would like to know if anyone else visualizes the operations as I do. I understand (I think :smile:) the formal mathematics of the transformation from ##3x = 15## to ##x = 15/3##, and I am careful, for example, when discussing it with my son to always talk about it in terms of applying the same operation to both sides of the equation, but this is not how I picture it when I do algebra. For me, the symbols move from one side of the equation to the other, as in
View attachment 242858
Likewise, if we had ##x + 3 = 15##, I would "see" the 3 going from the lhs to the rhs and acquiring a minus sing in the process.
Yes that is exactly how I was taught and I have used it ever since.
That is why I asked about multiplication over division approach.
Posts #19&21 explain from a maths and programming side but it would take practice to train my brain to look at simple equations this.
 

pinball1970

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I think the basic subtlety is that we like to define algebraic fields in terms of addition and multiplication with additive and multiplicative inverses.
It's related so hopefully the mods will allow.
Rules of indices turns multiplication into addition and division to subtraction is this part of what you mention above?
 
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It's related so hopefully the mods will allow.
Rules of indices turns multiplication into addition and division to subtraction is this part of what you mention above?
No, I think you’re referring to logarithms here whereas I’m talking about the properties of a number field like addition and multiplication over the real numbers aka the axiomatic approach.


See the Axiomatic section and the properties of addition and multiplication over a real number field.
 

symbolipoint

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No, I think you’re referring to logarithms here whereas I’m talking about the properties of a number field like addition and multiplication over the real numbers aka the axiomatic approach.


See the Axiomatic section and the properties of addition and multiplication over a real number field.
Not sure what you mean in first part, but second part yes for sure.
 

symbolipoint

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#33, yes, that is how I too think of it most of the time.
 
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Here's the context of pinball's and jedi's remarks:
Rules of indices turns multiplication into addition and division to subtraction is this part of what you mention above?
No, I think you’re referring to logarithms here
Not sure what you mean in first part,
On this side of the "pond" we call 'em exponents rather than indexes/indices.
I'm pretty sure that pinball's comment above has to do with these properties:
##\log_b(m \cdot n) = \log_b(m) + \log_b(n)##
##\log_b(\frac m n) = \log_b (m) - \log_b (n)##
 

jbriggs444

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Here's the context of pinball's and jedi's remarks:


On this side of the "pond" we call 'em exponents rather than indexes/indices.
I'm pretty sure that pinball's comment above has to do with these properties:
##\log_b(m \cdot n) = \log_b(m) + \log_b(n)##
##\log_b(\frac m n) = \log_b (m) - \log_b (n)##
I think he had more the laws of exponents in mind. e.g. ##b^{m+n}=b^m \cdot b^n##

Of course, it is six of one, half dozen of the other.
 

PeroK

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Back to the original question. I would try to find the answer by a direct knowledge of multiplication. For

##3x= 15##

I know that ##3 × 5 =15##, so ##x=5##.

I would generally try this with less obvious equations, including a bit of trial and error, although eventually I'd be forced into long division or a calculator.
 

symbolipoint

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Back to the original question. I would try to find the answer by a direct knowledge of multiplication. For

##3x= 15##

I know that ##3 × 5 =15##, so ##x=5##.

I would generally try this with less obvious equations, including a bit of trial and error, although eventually I'd be forced into long division or a calculator.
One would really not need a calculator for something as simple as that.
 
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This thread is beginning to sound like the Road Not Taken by Frost:


...
Two roads diverged in a wood, and I—
I took the one less traveled by,
(the one with a calculator)
And that has made all the difference.
 
Do you have to divide both sides by 3
No, I can start with nothing, add 3 x times until I arrive at 15. Then I just have to remember how many times I added 3. This is very fundamental starting from definitions.

x = 15/3 follows from the definition of division.

Of course you can just divide both sides by 3.
 

mathwonk

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I love this! So we should first transform 3x=15 to log(3x) = log(15), hence log(3) + log(x) = log(15), hence
log(x) = log(15)-log(3) = log(15/3) = log(5), then exponentiate, getting x = 5. wonderful. this for people who only know how to solve additive equations.
 

symbolipoint

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I love this! So we should first transform 3x=15 to log(3x) = log(15), hence log(3) + log(x) = log(15), hence
log(x) = log(15)-log(3) = log(15/3) = log(5), then exponentiate, getting x = 5. wonderful. this for people who only know how to solve additive equations.
The original question of the topic is really just pre-algebra or basic Algebra 1 level stuff. Not so complicated.
 

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