# $10+10\times 0$ equals $0$ or $10$

I thought $10+10\times 0$ equals $0$ because: $$10+10 = 20$$

And $$20\times 0 = 0$$

I know about BEDMAS and came up with conclusion it should be $$0$$ not $$10$$

But as per this, answer is $10$, are they right?

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en.wikipedia.org/wiki/Order_of_operations – user2468 Jul 3 '12 at 18:51
In mathematics, they are right. In some computer languages or calculators, computations may be done in some other order, and of course $(10+10)\times 0 = 0$. – GEdgar Jul 3 '12 at 18:52
If you know that the order of operation has multiplication before addition (that's the "M" and the "A" in "BEDMAS" [boy, do I hate dumb acronyms that don't teach anything]), then why are you doing the Addition of $10+10$ before the Multiplication $10\times 0$? You say you know "about BEDMAS"; do you understand what it is trying to tell you? – Arturo Magidin Jul 3 '12 at 18:53
anything is right. There is no such rule which one has to be done first . If u want to say that something has to be done first then you have to use signs like () {}, [] etc . – Theorem Jul 3 '12 at 19:31
@DougSpoonwood: I didn't ask you anything, and you often misrepresent and misunderstand what I write, whether out of willful ignorance, your personal agenday, or something else. I care not. You can take it as read that I'm never asking you anything. So don't answer me. Don't ping me. Don't direct your post-mortem equine flagelations in my direction. Discuss whatever you want with other people, not with me. "You like to shut off discussion" is another misrepresentation on your part. I just don't want to discuss antyhing with you. And as such, I don't want to hear you either. – Arturo Magidin Jul 4 '12 at 22:40

There is a "precedence of operations": some operations should be done before other operations in the absence of indications to the contrary. This is a convention used, among other things, to make writing things like polynomials simpler and to require fewer parentheses.

Multiplication has higher precedence than addition. That means that to perform $10+10\times 0$, you should first do the product, then do the sum. So first you do $10\times 0$, then you add the answer of that to $10$. The answer is $10$.

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Realized it later on but thanks for confirmation – Blaster Jul 3 '12 at 19:02

Multiplication has a higher precedence than addition so the multiplication is performed first.

$$10+10\times0 = 10+(10\times0) = 10$$

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Remember order of basic operations: multiplication/division have precedence over sum/substraction, so

$$10+10\cdot 0= 10+0=10$$

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I assume there is a typo: $10+0\not=0$. ;) – Shaktal Jul 3 '12 at 18:50
Yes, there was...and a rather huge one. Thanks to Peter. – DonAntonio Jul 3 '12 at 20:34

To elucidate what you said above in the original post, consider that $20\times0=0$, and consider also that $10+10=20$. If we have two equations like that, with one number $n$ on one side of an equation by itself with $m$ on the other side of the equation, and $n$ also appearing in the middle of a formula elsewhere, we should then have the ability to replace $n$ by $m$ in the middle of the formula elsewhere. The rule of replacement basically says this. In other words, $20\times0=10+10\times0$, since $20=10+10$, and $20\times0=0$, we replace $20$ by $10+10$ in $20\times0=0$ and obtain $10+10\times0=0$, right? But, BEDMAS says that $10+10\times0=10$. Why the difference?

The catch here lies in that the $10+10\times0$ obtained in the first instance does NOT mean the same thing as $10+10\times0$, given by fiat, in the second instance. Technically speaking, neither $20\times0=0$ and $10+10=20$ ends up as quite correct enough that you can use the rule of replacement as happened above. More precisely, $20\times0=0$ abbreviates $(20\times0)=0$, and $10+10=20$ abbreviates $(10+10)=20$. Keeping that in mind then we can see that using the rule of replacement here leads us to $((10+10)\times0)=0$ or more shortly $(10+10)\times0=0$. BEDMAS says that $10+10\times0$ means $(10+(10\times0))$ or more shortly $10+(10\times0)$, which differs from $(10+10)\times0$. So, the problem here arises, because the infix notation you've used makes it necessary to keep parentheses in formulas if you wish the rule of replacement as mechanically as you did.

If you do express all formulas with complete parenthesis in infix notation, then BEDMAS becomes unnecessary. If you wish to drop parentheses and use the rule of replacement mechanically as you did, then you'll need to write formulas in either Polish notation or Reverse Polish notation, or fully parenthesized infix notation, instead of partially parenthesized, "normal" infix notation. If you wish to keep BEDMAS and like conventions around, and write in normal infix notation, then you have to refrain from applying the rule of replacement as mechanically as you did. Conventional mathematicians and authors of our era generally appear to prefer the latter. I don't claim to understand why they appear to prefer a notation that makes such a simple logical rule, in some cases at least, harder to use than needed.

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