algebraic expressions ?

[1123581321]

I have recently been studying algebra and came across this which didn't make sense - a(squared) + 1 / a + 1 and stated that it could not be simplified any further due to lack of common factors... however, can't the denominator 'a' be taken away from the numerator 'a(squared)', therefore just leaving 'a' with the two 1s being cancelled. Or is simplification (of algebraic expressions) different to solving them..

[ydoaPs]

[math]\frac{a^2+1}{a+1}[/math] is fully simplified. You can only cross out terms that cancel. The ones don't cancel because they're being added instead of multiplied.

Edited February 13, 2011 by ydoaPs

[Cap'n Refsmmat]

Indeed. You can't do this:

 

[math]\frac{a^2 + 1}{a + 1} \neq \frac{a^2}{a}[/math]

 

Or this:

 

[math]\frac{a^2 + 1}{a + 1} \neq \frac{a + 1}{1}[/math]

You can only do something like this:

 

[math]\frac{a(a^2 + 1)}{a(a+1)} = \frac{a^2 + 1}{a+1}[/math]

You can only cross out factors that are multiplied. Division is the opposite of multiplication, so [math]a \times a \div a = a[/math]. It's not the case that [math]a \times a \div (a+1) = a[/math].

[1123581321]

why is it that you can only cancel multiplied factors ?

[Cap'n Refsmmat]

"Cancelling" is really division. I mean, here's a concrete example:

 

[math]\frac{4 \times (2 + 1)}{4 \times (7 - 2)} = \frac{2 + 1}{7 - 2}[/math]

You can do that because [imath]\frac{4}{4} = 1[/imath], and they divide out. On the other hand:

 

[math]\frac{4 + 2}{1 + 4} \neq \frac{2}{1}[/math]

You can see that just be doing the addition. Because division is the opposite of multiplication, you can "cancel" (divide) things that are multiplied. But you can't "cancel" things that are added, because you're not undoing any multiplication.

[1123581321]

ok, and can you also explain what: 1 over a letter means. e.g. 1/a or 1/a(squared) or something etc.. ?

[ydoaPs]

It is simply 1 divided by that variable.

[1123581321]

So 1/a is just a. yeh sorry, thats pretty obvious. But what about: a^-1 = 1/a & a^-m = 1/a^m ? how does that work..

[Cap'n Refsmmat]

[imath]\frac{1}{a} \neq a[/imath], unless [math]a = 1[/math]. No, [imath]\frac{1}{a} = 1 \div a[/imath], which is different.

 

As for the second bit: Perhaps this can be best explained another way. This makes sense, right:

 

[math]\frac{a^5}{a^2} = a^3[/math]

because:

 

[math]\frac{a^5}{a^2} = \frac{a\times a\times a\times a\times a}{a \times a}[/math]

and you cancel stuff out. So it works something like this:

 

[math]\frac{a^m}{a^n} = a^{m - n}[/math]

When you divide exponents, you subtract. Now, we also know that [imath]a^0 = 1[/imath], so you can do this:

 

[math]a^{-m} = a^{0 - m} = \frac{a^0}{a^m} = \frac{1}{a^m}[/math]

since that's just the same thing I did before, but in reverse.

 

Is that making sense, or is it too complicated?

[1123581321]

No, that seems to have hit the jackpot. Thanks.