# Is 'O' in BODMAS really for 'of'?

We were trying to solve a BODMAS question (with many solutions like always) when I came across this link.

https://www.mathsisfun.com/operation-order-bodmas.html

For the entire life, I have studied (and think that you might have also studied) that O in BODMAS is for 'of' meaning multiplication when * is not specified but bracket form is mentioned, like: 2(2+2) where 'of' makes it 2*(2+2). However, this website is claiming that 'O' means order, i.e., equations containing powers and roots. Is this really true? Seems like the world is shaking.

Also, can you please tell me what should be the answer of the following equation according to you?

3*{8/2(2+2)}+2

Note: Please bear with me if have been studying wrong my entire life.

• See Order of operations : BODMAS : it is only mnemonic; nothing really "deep" here. See Pedmas, where "E" stands for "Exponents". – Mauro ALLEGRANZA Nov 8 '19 at 9:57
• If $*$ is exponentiation, 2*(2+2) will be $2^{(2+2)}=2^4$. This is not about order of operations (we have enough parentheses to disambiguate the formula) but about the way to express exponentiation. Mult is $\times$, e.g. $2 \times 2$, while Exp is $a^b$. – Mauro ALLEGRANZA Nov 8 '19 at 10:02
• For your last question, see math.stackexchange.com/questions/33215/what-is-48%c3%b7293. It's ambiguous, and that's it. – Arthur Nov 8 '19 at 10:03
• Well, I learnt it as "BEDMAS" with "E" standing for "Exponents" in my computer textbook, which was the order if operations carried out by the computer. – AryanSonwatikar Nov 8 '19 at 10:03
• The answer regarding 3*{8/2(2+2)}+2 is very simple: Never write an expression this way that bears the risk of being ambiguous – Hagen von Eitzen Nov 8 '19 at 10:52

Through out my life I have used o for 'of' , meaning the bracket operation Solution :-

3×{8÷2(2+2)} +2 .......( If l am interpreting same you want to ask)


3×{8÷2(4)}+2

3×{8÷8}+2

3×1+2

3+2