Nah but it can be expressed as a * 1/a+b
ummmm wrote:
Question for all the math buffs: if you have a/(a+b), can you re-write this as 1 + a/b?
+1
do what you want wrote:
You can rewrite it that way if you wish, but it is not correct.
Fixed that for you. And no.
ummmm wrote:
Question for all the people who finished middle school:if you have a/(a+b), can you re-write this as 1 + a/b?
Disregard all of this gobbledygook.
Nope wrote:
a/(a+b) = a/a(1+b/a) = 1/(1+b/a)
Plug in some actual values to double check. Best to use numbers that are wildy different (ex. a=112 and b = 17). If you use numbers like 2 and 3, you'll get a lot of false positives when double checking. Also, good resource is Mathway http://www.mathway.com/Algebra
a/(a+b) = 112/(112+17) = 112/129
1/(1+b/a) = 1/(1+17/112) = 1/(112/112+17/112) = 1/(129/112) = 112/129
1+a/b = 1+112/17 = 7 and some change = not equivalent
I guess in this scenario zero doesn't count as a real number. Sad!
Bad Wigins wrote:
not only can't you do that in general, it's not equivalent for any a and b in the set of real numbers.
[...]
b^2 + a^2 = 0
I guess in this scenario division by zero counts as a defined operation. Sad!
here lies subzero wrote:Bad Wigins wrote:I guess in this scenario zero doesn't count as a real number. Sad!
not only can't you do that in general, it's not equivalent for any a and b in the set of real numbers.
[...]
b^2 + a^2 = 0
YES!!!
ummmm wrote:
Question for all the math buffs: if you have a/(a+b), can you re-write this as 1 + a/b?