How can we prove by contradiction and fundamental theorem of arithmetic that $3430000^\frac14$ is irrational?
My try:By calculating that number,it looks obvious but don't know how to prove?Thank you
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1It is equal to $10 \times 7^{\frac{3}{4}}$ ... – Donald Splutterwit Mar 31 '17 at 01:00
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1How can you judge a number is irrational by calculation? You will need infinitely many decimals to guarantee that it doesn't terminate or repeat, right? You will have to proceed by contradiction in this case. Look up the famous proof that $\sqrt 2$ is irrational, try to mimic it. Also, the above comment provides a simplification. – Sarvesh Ravichandran Iyer Mar 31 '17 at 01:02
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Donald kindly chops the large beast down into $10$ and $7^{\frac{3}{4}}$. We need to show that $7^{\frac{3}{4}}$ is irrational. To do so, we suppose it is not. Then, we can write this number as a reduced fraction $\frac{a}{b}$, where $a, b$ are relatively prime integers. Now, $\frac{a^4}{b^4} = 7^3 = 343$. This implies that $a^4 = 343b^4$. However, since $7$ divides $a$ (why?), $7^4$ divides $a^4$. Dividing by $343$ gives us that $7$ divides $b^4$ and this is only possible if $7$ divides $b$. Therefore, no such reduced fraction exists. Hence, $7^{\frac{3}{4}}$ is irrational.
Yunus Syed
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$a^4 = 343b^4$ and $343 = 7^3$. If $7$ does not divide $a$, is there a way for $343$ to divide $a^4$. – Yunus Syed Mar 31 '17 at 15:19
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Yes, I understand. But other readers might not have gone through this line of reasoning before. So, you might consider adding some commentary on this point. -Mark – Mark Viola Mar 31 '17 at 15:21
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If $\,7\nmid a \,$ then $(7^3 a)^{1/4} = \frac{c}d\Rightarrow\, 7^3 ad^4\! = c^4.$ $\,7$ occurs to odd power on LHS vs. even power on RHS, contra uniqueness of prime factorizations.
Bill Dubuque
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