How do you determine if #7/2# is a monomial?

2 Answers
Nov 27, 2016


It is a monomial. It is one term.


A monomial is an expression with one term.

Terms are separated by + and - signs.

The term that is given can be regarded as a single number, it has an exact position on the number line.

#7/2 = (3 1/2) = 3.5#

It is indeed a monomial.

The only possible confusion is if you consider the number to be
#3 + 1/2# which appears to be 2 terms.
This is overcome by either using brackets to indicate that it represents one term #(3 1/2)#,

Or by writing it as #7div2 =3.5#


It is a monomial.


A monomial, by definition, contains just one term.

According to regents prep:

"A monomial is the product of non-negative integer powers of variables. Consequently, a monomial has NO variable in its denominator. It has one term. (mono implies one)."

There have to be no negative exponents, and no fractional exponents.

In the case of your fraction, it could be seen as the product of a non-negative integer power, because:

#(7/2)x^0 = 7/2*1=7/2#

The integer power, zero , is not a negative, so it meets that requirement.

The denominator, 2, is not a variable, so it meets that requirement. Also consider that the denominator could be 1 if you put your fraction into decimal form, which is 3.5. In either case, it's not a variable, it's an integer.

Lastly, unlike a binomial (the prefix bi- means two) or polynomials (many), your number, #7/2# or 3.5 has only one term, meaning it's not being added or subtracted by or with anything else.

This all means that any number that isn't attached to a variable is a monomial (that means it's not being divided by or near a variable within the same term). And some individual terms could be monomials, if they satisfy the rules above.

Just to provide more clarity, here are a few examples that are not monomials:

#7/(2x)# #-># It has a variable in its denominator.

#3.5^-2# #-># It has a negative exponent.

#3.5^-(7/2)# #-># It has a fractional exponent.

Notice that those examples above all have just one term, but they don't satisfy the other rules.

Good luck!