# I have a month to learn college algebra **URL:** \ **Category:** Miscellaneous and Personal Stuff I Must Share\ **Created:** [June 21, 2008, 2:17am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656 "2008-06-21T02:17:45Z")\ **Posts on this page:** 20\ **Page:** 5
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 26, 2008, 11:56pm UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/81 "2008-06-26T23:56:02Z")
I don’t understand. To my eyes, it looks like Thudlow Boink is saying, in the text I quoted, that I should start with the 5 x instead of the 10[sup]n[/sup], which - again, to my unskilled eyes - looks like putting multiplication in front of exponent in the order of operations. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 12:11am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/82 "2008-06-27T00:11:57Z")
No, he is not saying that you should start with the “5 x” instead of the “10[sup]n[/sup]”; perhaps you are misreading his “you used 10 as your starting point”. He is saying that to calculate 10[sup]n[/sup], you start from 1 (_not_ 10) and multiply by 10 n times. To calculate 10[sup]-n[/sup], you start from 1 (_not_ 10) and divide by 10 n times. Thus: 5 x 10[sup]1[/sup] = 5 x (1 x 10) = 5 x 10 = 50 5 x 10[sup]0[/sup] = 5 x (1) = 5 x 1 = 5 5 x 10[sup]-1[/sup] = 5 x (1 / 10) = 5 x 0.1 = 0.5 5 x 10[sup]-2[/sup] = 5 x ((1 / 10) / 10) = 5 x 0.01 = 0.05 5 x 10[sup]-3[/sup] = 5 x (((1/10)/10)/10) = 5 x 0.001 = 0.005 ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 12:36am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/83 "2008-06-27T00:36:59Z")
[QUOTE=Indistinguishable] No, he is not saying that you should start with the “5 x” instead of the “10[sup]n[/sup]”; perhaps you are misreading his “you used 10 as your starting point”. He is saying that to calculate 10[sup]n[/sup], you start from 1 (_not_ 10) and multiply by 10 n times. To calculate 10[sup]-n[/sup], you start from 1 (_not_ 10) and divide by 10 n times. Thus: 5 x 10[sup]1[/sup] = 5 x (1 x 10) = 5 x 10 = 50 5 x 10[sup]0[/sup] = 5 x (1) = 5 x 1 = 5 5 x 10[sup]-1[/sup] = 5 x (1 / 10) = 5 x 0.1 = 0.5 5 x 10[sup]-2[/sup] = 5 x ((1 / 10) / 10) = 5 x 0.01 = 0.05 5 x 10[sup]-3[/sup] = 5 x (((1/10)/10)/10) = 5 x 0.001 = 0.005 [/QUOTE] Brilliant, thanks. I get it now. So is that always the rule with negative exponents? Divide by 1, [base] times? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 12:47am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/84 "2008-06-27T00:47:19Z")
In b[sup]n[/sup], n is the “exponent” or “power”, and b is the “base”. The rule, concerning integer exponents at least, is to start with 1, and multiply by the base b, doing so n many times. [If n is negative, then consider this to mean _dividing_ by the base b the corresponding positive number of times] So: 2[sup]3[/sup] = 1 x 2 x 2 x 2 = 8 2[sup]2[/sup] = 1 x 2 x 2 = 4 2[sup]1[/sup] = 1 x 2 = 2 2[sup]0[/sup] = 1 = 1 2[sup]-1[/sup] = 1 / 2 = 0.5 2[sup]-2[/sup] = (1 / 2) / 2 = 0.25 2[sup]-3[/sup] = ((1 / 2) / 2) / 2 = 0.125 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [June 27, 2008, 12:51am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/85 "2008-06-27T00:51:49Z")
The general rule for exponents is that, if a and b are positive integers, x[sup]a[/sup] \* x[sup]b[/sup] = x[sup]a + b[/sup] (try a few examples to verify this). This is a very nice rule, and we’d like it to hold in general, so that influences how we define exponentiation with exponents that are not positive integers. For instance, with b = 0, we want that x[sup]a[/sup] \* x[sup]0[/sup] = x[sup]a[/sup], which is fine as long as x[sup]0[/sup] = 1. Likewise, we want that x[sup]a[/sup] \* x[sup]-a[/sup] = x[sup]0[/sup] = 1, so it follows that multiplying by x[sup]-a[/sup] must be the same as dividing by x[sup]a[/sup]. Make sense? ETA: If you really want to make sure you understand this rule, tell me how we should define x[sup]1/2[/sup]. ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 1:31am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/86 "2008-06-27T01:31:14Z")
[QUOTE=Indistinguishable] In b[sup]n[/sup], n is the “exponent” or “power”, and b is the “base”. The rule, concerning integer exponents at least, is to start with 1, and multiply by the base b, doing so n many times. [If n is negative, then consider this to mean _dividing_ by the base b the corresponding positive number of times] So: 2[sup]3[/sup] = 1 x 2 x 2 x 2 = 8 2[sup]2[/sup] = 1 x 2 x 2 = 4 2[sup]1[/sup] = 1 x 2 = 2 2[sup]0[/sup] = 1 = 1 2[sup]-1[/sup] = 1 / 2 = 0.5 2[sup]-2[/sup] = (1 / 2) / 2 = 0.25 2[sup]-3[/sup] = ((1 / 2) / 2) / 2 = 0.125 [/QUOTE] Excellent explanation, thank you. > [@ultrafilter](#): > > The general rule for exponents is that, if a and b are positive integers, xa \* xb = xa + b (try a few examples to verify this). This is a very nice rule, and we’d like it to hold in general, so that influences how we define exponentiation with exponents that are not positive integers. For instance, with b = 0, we want that xa \* x0 = xa, which is fine as long as x0 = 1. Likewise, we want that xa \* x-a = x0 = 1, so it follows that multiplying by x-a must be the same as dividing by xa. Make sense? > > ETA: If you really want to make sure you understand this rule, tell me how we should define x1/2. Also a very nice explanation, thank you. Is it 1/2? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 1:34am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/87 "2008-06-27T01:34:06Z")
That’s odd… how come my superscripts were maintained through quoting but **ultrafilter** ’s were not? On edit: Oh, nevermind, you must’ve copied and pasted **ultrafilter** ’s while "Reply"ing to mine. Yet another problem with the way sup tags work… ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 2:10am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/88 "2008-06-27T02:10:47Z")
[QUOTE=Indistinguishable] That’s odd… how come my superscripts were maintained through quoting but **ultrafilter** ’s were not? On edit: Oh, nevermind, you must’ve copied and pasted **ultrafilter** ’s while "Reply"ing to mine. Yet another problem with the way sup tags work… [/QUOTE] Yep. Did I get his question right? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 2:17am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/89 "2008-06-27T02:17:08Z")
If you are saying that x[sup]1/2[/sup] should equal 1/2 regardless of what x is, then, no, that is not correct. Remember what **ultrafilter** said, you want x[sup]a[/sup] times x[sup]b[/sup] to equal x[sup]a+b[/sup]. So, think about what this means for x[sup]1/2[/sup] in particular. (It’s a little tricky, for a minor technical reason that may become apparent once you realize what the condition imposed is, but all in due time…) ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 2:33am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/90 "2008-06-27T02:33:43Z")
[QUOTE=Indistinguishable] If you are saying that x[sup]1/2[/sup] should equal 1/2 regardless of what x is, then, no, that is not correct. Remember what **ultrafilter** said, you want x[sup]a[/sup] times x[sup]b[/sup] to equal x[sup]a+b[/sup]. So, think about what this means for x[sup]1/2[/sup] in particular. (It’s a little tricky, for a minor technical reason that may become apparent once you realize what the condition imposed is, but all in due time…) [/QUOTE] Oh, I was thinking x=1 for some reason. I know why: I’m mathtarded. I have no idea what x[sup]1/2[/sup] is. ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 2:35am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/91 "2008-06-27T02:35:05Z")
Maybe I’m shooting in the dark here, but is the answer X÷2? ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [June 27, 2008, 2:37am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/92 "2008-06-27T02:37:36Z")
Remember that x[sup]a[/sup] \* x[sup]b[/sup] = x[sup]a + b[/sup]. Consider the case when a = b = 1/2. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 2:38am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/93 "2008-06-27T02:38:28Z")
[QUOTE=Cisco] Oh, I was thinking x=1 for some reason. I know why: I’m mathtarded. [/QUOTE] Aw, don’t say that. Nobody’s intrinsically mathtarded or math-adept; it’s just a matter of practice and experience. Here, I’ll walk you through it: We want x[sup]a[/sup] \* x[sup]b[/sup] to equal x[sup]a+b[/sup], for all values of a and b, since it’s such a nice law that works out for the usual integer cases. In particular, what does that condition say when a and b are both 1/2? ETA: Oh, **ultrafilter** beat me to it ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 2:46am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/94 "2008-06-27T02:46:00Z")
Is it x=1? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 2:47am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/95 "2008-06-27T02:47:33Z")
We’re not solving for the value of x here; we’re trying to see what x[sup]1/2[/sup] should be, in terms of x. Let’s make it more concrete. We want 3[sup]1/2[/sup] \* 3[sup]1/2[/sup] to equal 3[sup]1/2 + 1/2[/sup]. What does that tell us that 3[sup]1/2[/sup] should equal? ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 2:50am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/96 "2008-06-27T02:50:55Z")
0? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 2:54am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/97 "2008-06-27T02:54:28Z")
Well, you haven’t explained your reasoning (the question mark suggests perhaps a random guess?), but let’s see… We want 3[sup]1/2[/sup] \* 3[sup]1/2[/sup] to equal 3[sup]1/2 + 1/2[/sup]. If 3[sup]1/2[/sup] were 0, this would mean that 0 \* 0 would have to equal 3[sup]1/2 + 1/2[/sup]. Is that the case? [Incidentally, in case it’s not clear, the asterisk \*\*\*\*\* stands for multiplication, whereas we’ve generally been using **x** not to stand for multiplication but as the name of a variable] ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 2:59am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/98 "2008-06-27T02:59:38Z")
I was thinking 0[sup]1/2[/sup]\*0[sup]1/2[/sup] = 0[sup]1/2[/sup]+[sup]1/2[/sup] . . . I guess. I don’t even think I understand the question. ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 3:29am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/99 "2008-06-27T03:29:30Z")
Is it -3[sup]1/2[/sup]? ---
**Author:** ![OpalCat](https://avatars.discourse-cdn.com/v4/letter/o/8edcca/32.png) [@OpalCat](https://boards.straightdope.com/u/OpalCat)\ **Post date:** [June 27, 2008, 3:38am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/100 "2008-06-27T03:38:45Z")
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