Actually it’s (axb), since a(b+c)=(ab+ac). This is where a lot of people go wrong in the order of operations questions you see on socials - removing the brackets too soon. 1/ab=1/(axb) NOT 1/axb. If a=2 and b=3 then 1/ab=1/(2x3)=1/6, but 1/axb=1/2x3=3/2. Note that this also means it gets solved in the Brackets step of order of operations, NOT the “Multiplication” step (another common mistake).
If there is no operator though, it is assumed multiplication
It’s not “multiplication”, it’s a Product, a single number written as a product of factors. If a=2 and b=3 then ab is 6 written as the product of 2 and 3. ab=(2)(3)=(2x3)=6. axb=ab, 2x3=6, axb=2x3, ab=6.
“I’m unsure why that is.”
To show it’s a single number, not 2 separate numbers to be multiplied. Think of things like F=ma. You have to show that ma is a single number (equal to the Force), not 2 separate numbers multiplied., mxa. If you were doing something like dividing by the Force, then you have to have 1/ma=1/(mxa), NOT 1/mxa.
Actually it’s (axb), since a(b+c)=(ab+ac). This is where a lot of people go wrong in the order of operations questions you see on socials - removing the brackets too soon. 1/ab=1/(axb) NOT 1/axb. If a=2 and b=3 then 1/ab=1/(2x3)=1/6, but 1/axb=1/2x3=3/2. Note that this also means it gets solved in the Brackets step of order of operations, NOT the “Multiplication” step (another common mistake).
It’s not “multiplication”, it’s a Product, a single number written as a product of factors. If a=2 and b=3 then ab is 6 written as the product of 2 and 3. ab=(2)(3)=(2x3)=6. axb=ab, 2x3=6, axb=2x3, ab=6.
To show it’s a single number, not 2 separate numbers to be multiplied. Think of things like F=ma. You have to show that ma is a single number (equal to the Force), not 2 separate numbers multiplied., mxa. If you were doing something like dividing by the Force, then you have to have 1/ma=1/(mxa), NOT 1/mxa.