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I could be over thinking or tired... But I am to embarrassed to ask my prof. this probably very simple algebra rule I am ignorant of... Also this is just a snip-it from a inductive proof example.

Say you have something like this

$${\frac{4^k - 1}{3}} +{\frac{3*4^k}{3}} =$$ $${\frac{4*4^k-1}{3}}$$ My question is what algebra rules or rules of fractions allow this? In other words what computation is going on? I can agree that the 3 should cancel out $${\frac{3*4^k}{3} =} {{4^k}}$$ But wouldn't that mean it be something like this $${\frac{(4^k-1)*4^k}{3}}$$ or more accurately $${\frac{(4^k-1)}{3}{* 4^k}}$$ and not this? $${\frac{4*4^k-1}{3}}$$

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  • $\begingroup$ I'm not sure which part you are asking about. But $u+3u=4u$ So $4^x+3 \cdot 4^x=4 \cdot 4^x$. $\endgroup$ – randomgirl Mar 30 '15 at 1:04
  • $\begingroup$ Wait are you talking about how do you combine fractions? $\endgroup$ – randomgirl Mar 30 '15 at 1:05
  • $\begingroup$ $\frac{u(x)}{3}+c=\frac{u(x)}{3}+\frac{3c}{3}=\frac{u(x)+3c}{3}$. We are able to do this because $c=\frac{n}{n}c, n \neq 0$. Like you know since $\frac{n}{n}=1 \text{ where } n \neq 0$ $\endgroup$ – randomgirl Mar 30 '15 at 1:06
  • $\begingroup$ @randomgirl I guess It would be about exponent rules. And so 10*11^x = 11? $\endgroup$ – T.Malo Mar 30 '15 at 1:08
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    $\begingroup$ Its the Distributive law $a(b + c) = ab + ac$ $$ 4^{k}\cdot4 = 4^{k}\cdot(1 + 3) = 4^{k} + 3\cdot4^{k} $$ $\endgroup$ – user222031 Mar 30 '15 at 1:10
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${\frac{4^k - 1}{3}} +{\frac{3\cdot4^k}{3}} \\ \text{ these fractions have the same denominator so we can write the fraction as one now } \\ \frac{4^k-1+3 \cdot 4^k}{3} \\ \text{ recall } u+3u=4u \\ \text{ so we have } \\ \frac{4^k(1+3)-1}{3}=\frac{4^k (4)-1}{3}=\frac{4 \cdot 4^k-1}{3}=\frac{4^{k+1}-1}{3} \\ $

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  • $\begingroup$ Oh my God! I need to take a break from math - yes I see it now. Thanks. I will make this the answer. $\endgroup$ – T.Malo Mar 30 '15 at 1:17
  • $\begingroup$ I was starting to lose my mind thinking that $${4^x + 3*4^x = 4*4^x}$$ meant $${3⋅4^x} {=} {4}$$ $\endgroup$ – T.Malo Mar 30 '15 at 1:19

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