SDNet Brain Trust - enlighten me on 6 / 2 (1+2)

[AniThyng]

I'm going to go with 9 because of order of precedence rules and also because C will evaluate it as such.

But it seems opinions differ.

Have at it!

[bilateralrope]

6/2 * (1+2)
6/2 * (3)
3*3
9

How do those people favoring incorrect answers try to justify their answers ?

[AniThyng]

It seems that the most common justification (or at least the only one that comes close to making sense to me) is that they presume that 2( is somehow higher precedence than 2*( so they evaluate it as

6/2(1+2)
6/2(3)
6/6
1

or

given b = (1+2)

does

6/2b = 6b/2 ?

I had jumped to this "intuitively" also at first but after being forced to write it down I came around.

[Terralthra]

It's an ambiguous expression, because it's not possible to set standard type using horizontal fractions. If it were, whether it's "6/[2*(1+2)]" or "6/2*(1+2)". With the assumption that what is typed is what is meant, 9 is the correct answer. If it were set with a horizontal fraction instead of the (ambiguous) division operator, it would be immediately obvious which is meant.

[Feil]

Operations with equivalent primacy evaluate left to right. Putting parentheses around 6/2 would add clarity, but not change meaning.

[Highlord Laan]

6/2 * (1+2)
6/2 * (3)
3*3
9

How do those people favoring incorrect answers try to justify their answers ?

Being out of practice. I did catch myself after my first mistake though.

[The Vortex Empire]

Judging by the comments on these things that go around Facebook every now and then, people get it wrong because they don't know whether or not to use Order of Operations. Even though you always use it.

[slebetman]

The correct answer is that in algebra, the expression is ambiguous. The left-to-right rule is from calculators and programming languages but not algebra per se. As such the answer is that the expression is technically a form of syntax error.

The correct way to write the expression is either to add additional parenthesis to clarify what you mean of to write it in proper algebraic form:

Code: Select all

      6
__________

 2 ( 1 + 2)

or

Code: Select all

6
_    ( 1 + 2)

2

[Terralthra]

That's what I was trying to say, but I didn't use code blocks to make it clear. Thanks for the reinforcement.

[Batman]

The way I was taught math, it works the way bilateralrope did it. You calculate the stuff in parentheses then figure the result into the equation.

[fuzzymillipede]

It's an ambiguous expression, because it's not possible to set standard type using horizontal fractions. If it were, whether it's "6/[2*(1+2)]" or "6/2*(1+2)". With the assumption that what is typed is what is meant, 9 is the correct answer. If it were set with a horizontal fraction instead of the (ambiguous) division operator, it would be immediately obvious which is meant.

It's not ambiguous. If you want to specify a horizontal fraction using the standard division symbol, you need to enclose the denominator in a set of parenthesis.

Code: Select all

      6
__________

2 ( 1 + 2)

translates to

Code: Select all

6 / (2 ( 1 + 2))

In no case does it translate to

Code: Select all

6 / 2 ( 1 + 2)

, which is evaluated differently.

[ArmorPierce]

Where people seem to be making the mistake is applying the order of operations, or as people remember it, PEMDAS. They see Multiplication comes before Division so assume that you have to multiply first when in actuality multiplication and Division (and Addition and subtraction) have the same precedence and in those cases you work the problem for that left to right.

[Blayne]

I think you can also prove it through mathematical induction.

6/2 * (1+2) = 9

(3*2)/2 * (1+2) = (3)^2

2(2k+1)/2 * (2k+1) = (2k+1)^2

If k is true, then k+1 is true.

2(2(k+1)+1)/2 * (2(k+1)+1) = (2(k+1)+1)^2

Simply: (2k+3)(2k+3)=(2k+3)^2

I think I've skipped a step in the induction process of where we add (2k+3)(2k+3) to both sides of our original equation but it should suffice to show that LHS = RHS in this instance no matter for what value of k; also I'm not 100% if that's what you do when you are not proving a sequence/series.

[Surlethe]

Blayne, how is induction relevant to this arithmetic question? You use induction to prove a family of propositions indexed by the natural numbers, not to evaluate a single arithmetic expression. Actually you seem to be proving that (2k+1)(2k+1) = (2k+1)^2 for all k ... you don't need induction to tell you that.

[Blayne]

I thought it was valid use of induction but I am mistaken.