How does the step from $$=\biggr(\frac{7}{4}\biggr)^{k-2}\biggr(\frac{11}{4}\biggl)$$ To $$=\biggr(\frac{7}{4}\biggr)^{k-2}\biggr(\frac{7}{4}\biggl)^2$$ work given $(\frac{7}{4})^2 \neq (\frac{11}{4})$
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It is $\biggr(\frac{7}{4}\biggr)^{k-2}\biggr(\frac{11}{4}\biggl)<\biggr(\frac{7}{4}\biggr)^{k-2}\biggr(\frac{7}{4}\biggl)^2$ – CY Aries Jun 10 '17 at 10:59
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Wait so when the writer uses the $=$ instead of $<$ for step 2 and 3 its not just short hand it means an actual conversion? – Sonny Da Silva-Peters Jun 10 '17 at 11:01
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The writer uses "$=$" because the expressions are equal. – CY Aries Jun 10 '17 at 11:03
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$$\left(\frac{7}{4}\right)^{k-2}\cdot \frac{11}{4}=\left(\frac{7}{4}\right)^k\left(\frac{4}{7}\right)^2\cdot \frac{11}{4}=\left(\frac{7}{4}\right)^k\frac{44}{49}$$
Dr. Sonnhard Graubner
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how is this $$\left(\frac{7}{4}\right)^k\frac{44}{49}$$ the same as $$\left(\frac{7}{4}\right)^{k-2}\biggl(\frac{7}{4}\biggl)^2 = \biggl(\frac{7}{4}\biggl)^k$$ am i making a mistake in one of these? – Sonny Da Silva-Peters Jun 10 '17 at 11:11
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$$\left(\frac{7}{4}\right)^{k-2}\cdot \left(\frac{7}{4}\right)^2=\left(\frac{7}{4}\right)^{k-2+2}=\left(\frac{7}{4}\right)^k$$ – Dr. Sonnhard Graubner Jun 10 '17 at 11:15
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But didnt your original answer come out to $$\biggl(\frac{7}{4}\biggl)^k\biggl(\frac{44}{49}\biggl)$$ is it due to the fact that you are proving its greater than that you can ignore the $44/49$?? – Sonny Da Silva-Peters Jun 10 '17 at 11:19
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yes and $$\left(\frac{7}{4}\right)^{-2}=\left(\frac{4}{7}\right)^2$$ – Dr. Sonnhard Graubner Jun 10 '17 at 11:23
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$$=\biggr(\frac{7}{4}\biggr)^{k-2}\biggr(\frac{11}{4}\biggl)$$ To $$=\biggr(\frac{7}{4}\biggr)^{k-2}\biggr(\frac{7}{4}\biggl)^2$$ – Sonny Da Silva-Peters Jun 10 '17 at 11:23
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The expression $\displaystyle {(\frac 74)}^{k-2}(\frac {11}4) <{(\frac 74)}^{k-2}(\frac {7}4)^2$ means the left hand term is less than the right hand term. It is not equality. It is an inequality. The expression $a < b$ means that $a$ is less than $b$. The expression $a > b$ means that $a$ is greater than $b$.
And the inequality holds because $\displaystyle (\frac {11}4) < (\frac {7}4)^2$. Do you see now?
Can you see how to prove that $\displaystyle (\frac {11}4) < (\frac {7}4)^2$, or do you need help with this part too?
Deepak
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ohhhhhhhhhhhh okay so when $<$ is used with no visible left part then the left part is assumed to be the line above it? – Sonny Da Silva-Peters Jun 10 '17 at 11:27
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1@SonnyDasilvapeters Yes. Just the same as when you break a line and start the next line with '='. It's just typography, not mathematics. – Deepak Jun 10 '17 at 11:28
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I am so sorry i took so much time figuring that out, i'm new to trying to read more mathematical books, thank you so much – Sonny Da Silva-Peters Jun 10 '17 at 11:29
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