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So, as you see in the title the question is about $(3+5)(3+5)$

I find two ways to solve it:

$1)$ the first way: $(3+5)(3+5) =(8)(8) =8\cdot8 =64$
In this way of solving it, I just respected the rule of solving that what is insides de parenthesis in priority.

$2)$ the second way: $(3+5)(3+5) =(3+5)^2$
I use the identity: $(a+b)^2=a^2+2ab+b^2$ to get $(3+5)^2=3^2+2\cdot3\cdot5+5^2 =9+30+25 =64$

As you see here I didn't solve that what is inside de the parenthesis in priority while I could.

My question is: Is the second method wrong because I didn't respect the parenthesis priority?

Thank you for reading.

-Edit: I add a way to solve it to make it clearer for me:

-3) So I just multiply the expessions in parenthesis one by one without using an identity to get $$(3+5)(3+5) =(3*3)+(3*5)+(5*3)+(5*5) =9+15+15+25 =64$$

Here I disrespected the parenthesis priority in a more visible way. Is it still correct?

-Edit: I opened a new question here still to clarify my point: Two ways for $\frac{(1-1)(1-1)}{(1-1)}$ two results

lazare
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    Lazare, don’t worry, the second method is correct too. – Angelo Oct 14 '24 at 17:53
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    The second way is also correct. When we say "parentheses take priority", this just means that you should think of $(3+5)(3+5)$ as meaning the product of $3+5$ with itself, and not, say, $3$ plus $5(3+5)$. The phrase "parentheses take priority" should not be understood as an instruction manual for the order in which you should simplify an expression. And $(a+b)^2$ equals $a^2+2ab+b^2$ for all values of $a$ and $b$, so it's fine to replace $(3+5)^2$ with $3^2+2(3)(5)+5^2$. – Joe Oct 14 '24 at 17:53
  • This is weard beceause the identity is naturaly never used for some a and b that you can obviously add – lazare Oct 14 '24 at 17:58
  • One more of the deep mysteries of arithmetic. – A rural reader Oct 14 '24 at 18:06
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    There are many correct ways to compute it. You could also sum up $8$ times $3+5$, or $3+3+3+3+3+3+3+3+5+5+5+5+5+5+5+5=64$. – Dietrich Burde Oct 14 '24 at 18:11
  • @Joe: The fact that operator priority is not the same thing as order of evaluation is important in computer programming, where functions can have "side effects", and y = f(x) * (g(x) + h(x)); doesn't necessarily have a well-defined order in which the calls to f, g, and h are made. It's for this reason that I hate the phrase “order of operations” to refer to operator priority/precedence. – Dan Oct 14 '24 at 18:28
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    P.S. We typically say "evaluate/compute (3+5)(3+5)" rather than "solve (3+5)(3+5)". – ryang Oct 14 '24 at 18:29
  • I edited my post – lazare Oct 14 '24 at 18:30
  • If you have two paranthesis that get multiplied, yes it is correct. You just have to multiply all the different pairs and add them up: $(a+b+c+d)(e+f)=ae+be+ce+de+af+bf+cf+df$. – user408858 Oct 14 '24 at 20:16
  • If you have three paranthesis, you have to pick all the different triplets and add them up: $(a+b+c)(e+f)(g+h)=aeg+beg+ceg+afg+bfg+cfg+aeh+beh+ceh+afh+bfh+cfh$ – user408858 Oct 14 '24 at 20:18
  • And if you have four, five, six, "..." paranthesis you multiply the different fourlets, fivelets, sixlets, "..."lets and add them up. The same logic applies. – user408858 Oct 14 '24 at 20:19
  • lazare - what age are you / what level of maths are you? Have you learned how numbers (integers, then rational numbers, then real numbers) are formed from axioms? And finally, where do you know about the identity $(a+b)^2=a^2+2ab+b^2$ from? – Adam Rubinson Oct 14 '24 at 20:57
  • Adam, I know how the numbers are formed from axiom. And I know the binomial theorem and understand its demonstration by induction. My study level is just before hightschool where I left school but after I learn a lot on internet. – lazare Oct 14 '24 at 21:05
  • Well then you should know that there is no axiom where you have to "respect parenthesis priority". BIDMAS/BODMAS/PEDMAS are not axioms when constructing the set of natural, rational or real numbers. But what is an axiom is the distributive axiom, which can be used to show that your identity is true, like in user408858's answer. Then all you are using is like an axiom of substitution (although whether or not substitution is taken as an axiom depends on the way you have formally defined things). – Adam Rubinson Oct 14 '24 at 21:33
  • We almost always use the first way of computing something like $(3+5)(3+5)$ because most people find it faster to compute. If it were easier to compute the other way we would do it that way. But if the second way were not perfectly correct then $(a+b)(a+b)=a^2+2ab+b^2$ would not be correct either, because it's supposed to work for all values of $a$ and $b$, including when $a=3$ and $b=5$. – David K Oct 14 '24 at 21:57

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I'd like to point out some very general information, which may be useful in other contexts beyond just seeing that your second computation is perfectly OK.

The precedence rules, like "parentheses have priority" are intended to define the meanings of expressions. They have to be very specific, because an expression should have just one meaning. But these rules are not intended to limit how one might compute the values of those expressions, and there might be many ways to compute the same expression.

In most cases, the meaning of the expression will provide a way to compute it, but that need not be the best way, or even a reasonable way. For example, I would not compute $(2024\cdot1776)/2024$ by first performing the multiplication in the parentheses and then performing the division, even though that's what the expression means. Many of the formulas and rules that you learn can serve to simplify computations (without damaging the meanings). For example, one of those rules can be applied to $(2024\cdot1776)/2024$ to cancel the 2024s and read off the answer $1776$ without having to do any "real" work.

Andreas Blass
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  • I edited my post. – lazare Oct 14 '24 at 18:31
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    @lazare: You are missing the point. What this and the other answers want to tell you is that, as long as you are not making a mathematically incorrect simplification, it is correct. All the three ways in the question are correct. Read this answer in particular once again. Parentheses do not determine the order of computation. – ultralegend5385 Oct 14 '24 at 18:36
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No, it is not wrong. It is simply using the algebraic laws. Let's imagine first, that $(a+b)=c$. (In your example $c=8$.) Then you find, as you write, either

$$(a+b)(a+b)=c^2$$

or

$$ \begin{aligned} (a+b)(a+b) &=(a+b)c\\ &=ac+bc\\ &=a(a+b)+b(a+b)\\ &=aa+ab+ba+bb\\ &=a^2+2ab+b^2 \end{aligned} $$

(Of course you can do this more rigorously by applying the associative, commutative and distributive laws precisely. Though, I hope you get the picture, why this identity is true.)

user408858
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Your "second method" is correct too. What the parentheses are telling you is that (as @Joe comments) you should think of each $(3+5)$ as a single number.

The distributive law is what connects the two methods. It says $$ a(b+c) = ab + ac , $$ which you can think of as two ways to evaluate the same expression. Use whichever is most convenient.

Yoy prove the identity $$ (a+b)^2 = a*2 + 2ab + b^2 $$ by applying the distributive law twice.

Ethan Bolker
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Both methods are fine and in general all methods which agree with algebraic axioms and derived rules, for example we can also use

$$(3+5)(3+5)=(8)\cdot (1+1+1+1+1+1+1+1) =8+8+8+8+8+8+8+8=64$$

The parenthesis are used to avoid confusion with, for example

$$3+5 \cdot 3+5 =3+15+5=23$$

but as noticed we can invoke distributive law and so

$$(3+5)(3+5)=3(3+5)+5(3+5)=24+40=64$$

which is pefectly fine with algebraic rules.

user
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