# Why can't you add terms with different exponents?

When evaluating algebraic expressions,

1) you can add together like terms. $3x^5 + 6x^5 = 9x^5$, but you cannot add together different terms: $2x^4 + 3x^5$, because these have different exponents.

2) you can multiple different terms: $2x^4 \cdot 3x^5 = 6x^9$. When you multiple terms, the exponents are added together.

Why can't you add terms with different exponents?

Someone said it's because of the properties of algebra:

Commutative property: $a + b = b + a$ and $ab = ba$.

Associative property: $a + (b + c) = b + (c + a)$ and $a \cdot (b \cdot c) = b \cdot (a \cdot c)$.

Distributive property: $x(a+b) = xa + xb$.

So how do these properties suggest that you cannot add terms when exponents are different, but you can multiply terms with different exponents?

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The right question to ask yourself may be "why do these properties suggest that you can add terms in the other situation?" –  Hurkyl Apr 14 '13 at 23:19

If you pay for two bananas and then three bananas you have paid for five bananas ($2x^2 + 3x^2 \equiv 5x^2$). But if you pay for one banana and one peach, what have you paid for except one banana and one peach?! ($x^2+x^3 \equiv x^2+x^3$).

Even in primary school (elementary school in the US), children are taught the rule of BODMAS (or BIDMAS). We evaluate Brackets first, then Orders (or Indices). Then we do Division, Multiplication, Addition and finally Subtraction.

How can we hope to combine terms of the form $x^2$ and $x^3$? An expression of the form $(x \times x) + (x \times x \times x)$ has only one meaning, even if we take the brackets away, we still have to multiply first.

\begin{array}{ccc} 2x^2 + 3x^2 &\equiv& 2(x \times x) + 3(x \times x) \\ &\equiv& [(x \times x) + (x \times x)] + [(x \times x) + (x \times x) + (x \times x)] \\ &\equiv& (x \times x) + (x \times x) + (x \times x) + (x \times x) + (x \times x) \\ &\equiv& 5(x \times x) \\ &\equiv& 5x^2 \end{array}

We can even verify this by experimentation: $2\times 4^2 + 3\times 4^2 = 32 + 48 = 80$ while $5 \times 4^2 = 80$. However, let's try to do this with terms of a different order, say $x^2$ and $x^3$:

\begin{array}{ccc} x^2 + x^3 &\equiv& (x \times x) + (x \times x \times x) \\ &\equiv& \ldots\ldots? \end{array}

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You can add different exponents together. Your example of $2x^4 + 3x^5$ illustrates this perfectly. However, you cannot simplify this expression further. In the case of $3x^5 + 6x^5$, this equals $(3+6)x^5$ by the distributive property which simplifies to $9x^5$.

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First of all, what you call communicative is actually called commutative, and the fact that $ab=ba$ doesn't have anything to do with this. In non-commutative rings it doesn't all of a sudden become possible to add monomials of different degrees.

When it comes to addition (or multiplication) of polynomials, it is just a matter of definition. The addition is simply defined to be as it is and if you've got several monomials of different degree, you can't add them. Unless I'm utterly unaware of it, there isn't anything deep going on there. The main point is to consider polynomials as formal objects and not as their corresponding maps, unless there's a very good reason to do so.

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I could be wrong, but I really doubt the OP knows what a "ring" is, or what it means to consider polynomials "as formal objects and not as their corresponding maps." –  Jesse Madnick Apr 14 '13 at 22:53
You may very well be right, I'm afraid. It's a bit hard for me to imagine asking things about polynomials when you don't know about rings, especially if you mention certain algebraic properties. I will try to refrain from answering questions like this in future, thanks for the feedback! –  HSN Apr 14 '13 at 22:59

What happens if you try?

$$3x^2+6x^2=(3+6)x^2=9x^2$$

$$3x^2+6x^3=(3+6x)x^2=?$$

If you factor out as many $x$'s as possible, the only case where you'll be left with only numbers in the bracket (which you can, of course, add) is when the exponents are the same. You can simplify $3+6$ into $9$, but you can't simplify $3+6x$ into anything, so in this case it doesn't get you anywhere.

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