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# 6÷2(1+2)=?

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posted on May, 2 2011 @ 09:36 AM

Believe what you want. I think this notation is poor and this thread demonstrates why I think so. Again write out one third times three.

1 / 3 * 3 = 1 not one ninth. which is the equivalent of what many are doing.

1 / 3 * (3 + 0) = 1 or do some of you insist on the answer being 0.111111111 or 1/9?

edit on 2-5-2011 by MegaMind because: (no reason given)

edit on 2-5-2011 by MegaMind because: (no reason given)

edit on 2-5-2011 by MegaMind because: (no reason given)

posted on May, 2 2011 @ 09:50 AM

Originally posted by MegaMind

1 / 3 * 3 = 1 not one ninth. which is the equivalent of what many are doing.

1 / 3 * 3 != 1 / 3 (3)

Many people here are treating A(B+C) as A*(B+C), which is wrong.

1/3*3 = 1 (as you stated)

not to be confused with

1/3(3) = 0.111

This is shown and proven (to me at least) on my Casio fx-82TL scientific calculator.

posted on May, 2 2011 @ 09:51 AM

Wrong. A * (B + C) = A(B + C)

I don't know what else to say. There are conventions for math notation. Most accept this notation the way I interpret it also. People have shown examples online here.

edit on 2-5-2011 by MegaMind because: (no reason given)

edit on 2-5-2011 by MegaMind because: (no reason given)

posted on May, 2 2011 @ 09:55 AM

Originally posted by MegaMind

Wrong. A * (B + C) = A(B + C)

BTW don't write any computer programs while thinking the way you do - it simply won't work.

I use this way of thinking for solving this maths equation, not for programming. I understand the limitations of the compilers

People who type this into google and tell me adamantly that the answer is nine don't understand the limitations, though.

posted on May, 2 2011 @ 09:58 AM

Originally posted by GobbledokTChipeater

Originally posted by MegaMind

Wrong. A * (B + C) = A(B + C)

BTW don't write any computer programs while thinking the way you do - it simply won't work.

I use this way of thinking for solving this maths equation, not for programming. I understand the limitations of the compilers

People who type this into google and tell me adamantly that the answer is nine don't understand the limitations, though.

Good luck with the math.

posted on May, 2 2011 @ 09:58 AM

Originally posted by MegaMind

1 / 3 * (3 + 0) = 1 or do some of you insist on the answer being 0.111111111 or 1/9?

Yes the way it is written, the answer is 1.

When you remove the multiplication sign though, the answer changes. I know you will tell me this is not the case, but I would rather believe my very good scientific calculator than you, sorry.

posted on May, 2 2011 @ 10:01 AM

Let me ask you what is 1/3 (X + Y) distributed?

or

(X + Y)/3

or

(X + Y)
--------
3

X/3 + Y/3 correct?

edit on 2-5-2011 by MegaMind because: (no reason given)

posted on May, 2 2011 @ 10:01 AM
Guys, what is 15y ÷ 5y ?
Did you get 3 or 3y^2 ?

If your answer were 9 in previous question, you should get 3y^2 because
15y ÷ 5y = 15 x y ÷ 5 x y
= 15y ÷ 5 x y
= 3y x y
= 3y^2

What if you got answer 1 for previous question?
15y ÷ 5y = 15y/5y
= 3

So, back to the question 6÷2(1+2)=?

Let x = (1+2) = 3 -------------eq 1

6÷2x = 3/x

Put eq 1 into 3/x, you get
3/3 = 1

posted on May, 2 2011 @ 10:03 AM

Originally posted by MegaMind

Good luck with the math.

I have written many programs for many embedded systems, some of those using maths (wow, yeh I know). And guess what, they actually work!

Math within programs is not scientific calculator maths.

For example (I work with 8, 16 and 32 bit PIC's) if, within the program, you do (for example) 22/7, it will tell me the answer is 3.

Clearly not the correct answer but if you understand the limitations of what you're working with (be it a microcontroller or a compiler), then it causes no problems.

posted on May, 2 2011 @ 10:05 AM

Originally posted by Red_xi

Originally posted by solargeddon

Originally posted by Red_xi

WhoTF decided to put the extra bracket in, that isn't in the original sum, the sum is... 6/2(1+2)

There is no extra bracket used

Look at my calculators and tell me where you see an extra bracket?

If you go back to my first reply, you will see I refernce the OP, not you, so cool your heels.

Second I haven't even read the conclusion you came to, because SHOCK HORROR ! I haven't read the entire thread, just the gist, understandable given that I am not a troll !

Nothing wrong with a bit of healthy curiosity, oh and while I'm on it, don't know what country youre from, but perhaps this is a cultural issue, as from where I come from this is how the sum is solved.

Sorry, I'm not trying to bash you, was never my intention.

I guess your in the UK due to you mentioning GCSE's

I'm in England and this is definitely how it is taught.

Any way I'll be quiet now, I've said my piece

edit on 2-5-2011 by Red_xi because: (no reason given)

Nice to bash with a fellow Brit

Youre alright

Maths, bank holiday, and hormones, nasty mix lol (thats me, not you)

Interestingly enough I did it in my head like quite a few others did, then used google which came up with the same, but calculators (scientific) come up with a different answer, perhaps neither is right or wrong, but then that has to be impossible, because maths is logic based, and cannot be both things at once.

I don't know I agree with you, said my bit and now I'm moving on

Have a great day

posted on May, 2 2011 @ 10:07 AM

Originally posted by MegaMind

Let me ask you what is 1/3 (X + Y) distributed?

x/3 + y/3 correct?

or

(X + Y)/3

or

(X + Y)

1 / (3X + 3Y).

Now you mention it this way, I distinctly remember this problem from my uni text books. If I can find it tomorrow, I will scan the relevant pages up and post.

posted on May, 2 2011 @ 10:09 AM

Originally posted by GobbledokTChipeater

Originally posted by MegaMind

Let me ask you what is 1/3 (X + Y) distributed?

x/3 + y/3 correct?

or

(X + Y)/3

or

(X + Y)

1 / (3X + 3Y).

Now you mention it this way, I distinctly remember this problem from my uni text books. If I can find it tomorrow, I will scan the relevant pages up and post.

You are violating the rules with that.

A (B + C) = AB + AC

A = 1/3

therefore,

1/3B + 1/3C

Basic algebra

same as (X + Y)/3 or (B + C)/3
edit on 2-5-2011 by MegaMind because: (no reason given)

edit on 2-5-2011 by MegaMind because: (no reason given)

posted on May, 2 2011 @ 10:13 AM

Originally posted by MegaMind

Originally posted by GobbledokTChipeater

Originally posted by MegaMind

Let me ask you what is 1/3 (X + Y) distributed?

x/3 + y/3 correct?

or

(X + Y)/3

or

(X + Y)

1 / (3X + 3Y).

Now you mention it this way, I distinctly remember this problem from my uni text books. If I can find it tomorrow, I will scan the relevant pages up and post.

You are violating the rules with that.

A (B + C) = AB + AC

A = 1/3

therefore,

1/3B + 1/3C

Basic algebra

same as (X + Y)/3 or (B + C)/3
edit on 2-5-2011 by MegaMind because: (no reason given)

edit on 2-5-2011 by MegaMind because: (no reason given)

...and we go around in circles. I think we should agree to disagree.

You keep asking me the same question structured differently, and the answer remains the same.

If A = 1/3 then it should be in brackets, otherwise it is taken as A = 3.

You keep thinking it means (1/3)(X+Y). I keep thinking it means 1/(3(X+Y))

You think you are right, I think I am right. My calculator agrees with me.

edit on 2/5/11 by GobbledokTChipeater because: (no reason given)

posted on May, 2 2011 @ 10:14 AM

You are hopelessly lost I'm outta here.

posted on May, 2 2011 @ 10:15 AM

Originally posted by IamBoon

I have a \$200 calculator that says 9 is wrong. My cheap one says the answer is 9 though.

The thing about \$200 calculators is that they can be tricky and often times have sub keys that are suppose to be used instead of primary keys. Cheap calculators are generally more reliable because they're more user friendly.

That aside; using my built in calculator, the problem is coming up as 9.
edit on 2-5-2011 by Mactire because: (no reason given)

posted on May, 2 2011 @ 10:15 AM

In order to get the correct answer of 3.14285714285714 you would need to cast 22 and 7 as doubles instead of ints.

posted on May, 2 2011 @ 10:22 AM

edit on 2-5-2011 by knightseifer because: (no reason given)

posted on May, 2 2011 @ 10:28 AM

Originally posted by knightseifer

Mate, 1/(3B+3C) cant be written as 1/3B + 1/3C

No kidding, but I didn't write that did I?

1/3(B + C) = B/3 + C/3 or 1/3B + 1/3C

A ( B + C) = AB + BC

"A" can be any damn thing you want it to be!!!

You people pass algebra???
edit on 2-5-2011 by MegaMind because: (no reason given)

edit on 2-5-2011 by MegaMind because: (no reason given)

posted on May, 2 2011 @ 10:31 AM

Originally posted by MegaMind

Originally posted by knightseifer

Mate, 1/(3B+3C) cant be written as 1/3B + 1/3C

No kidding but I didn't write that did I?

1/3(B + C) = B/3 + C/3
edit on 2-5-2011 by MegaMind because: (no reason given)

Original equation should be in focus... there are really only 2 ways to do this and it seems to revolve around seeing the problem as a fraction or not. Order of Ops do not change but perception of the problem does with using the 6 as the numerator and the rest of the equation as the denominator.

posted on May, 2 2011 @ 10:33 AM

Originally posted by MegaMind

"A" can be any damn thing you want it to be!!!

Yes and if you want it to be 1/3, then you should write (1/3). Otherwise A is taken to mean the number directly before the brackets.

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