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--- math104-f21:hw1 [2021/08/28 07:54]
pzhou
+++ math104-f21:hw1 [2026/02/21 14:41] (current)
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-====== HW 1 ======
+====== HW 1 (with solution) ======
 Due Tuesday (Aug 31) 6pm. 2 points each.
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 . Someone claims that he has found a smallest positive rational number, but would not tell you which number it is, can you prove that this is impossible? (Optional extra question: can you prove that there is no smallest rational number among all rational numbers that are larger than $\sqrt{2}$?)
-. Prove that, if $r$ is a rational number, $x$ is an irrational number, then $r + x$ and $rx$ are irrational.
+. Prove that, if $r$ is a **non-zero** rational number, $x$ is an irrational number, then $r + x$ and $rx$ are irrational. (UPDATE: if $r=0$, of course $rx=0$)
 . Prove that there is no rational number whose square is $20$.
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+====== Solution ======
+. Say this number is $p$, and $p= m/n$ for some co-prime $m,n$, then $m/(n+1)$ would be a smaller rational than $p$ and still is positive. (One can also use other method to get a smaller prime, say $p/2$).
+For the optional problem, following Rudin, we can let $p' = p - (p^2-2)/(p+2)$, then one can check that $\sqrt{2} < p' < p$.
+. If $r + x$ is rational, say $r+x = s$, then $x = s - r$. However, we know arithemetic operations preserves rational numbers, so $s-r$ is rational, contradicting with $x$ being irrational. Thus $r+x$ is irrational. Similarly, if $rx = s$ is rational, then $x = s/r$ would be rational, contradicting with $x$ being irrational.
+. Suppose $p^2 = 20$, then $(p/2)^2=5$. We claim that there is no rational number whose square is $5$. The proof is similar to that of no rational's square is $2$, as in Rudin.
+. Let $P(n)$ be the statement that $f(n)=7^n - 6n - 1$ is divisible by $36$. Then $P(1)$ is true, since $f(1)=0$. Suppose $P(n)$ is true for $n \leq m$, we now prove that $P(n)$ is true for $n=m+1$. We have
+$$ f(m+1) - f(m) = 7^{m+1} - 7^{m} - 6 = 6 (7^m - 1) $$
+Hence $36 \mid f(m+1) $ is equivalent to $36 \mid f(m+1) - f(m)$ (since we know $36 \mid f(m)$ by induction), and is equivalent to $36 \mid 6(7^m-1)$ and is equivalent to $6 \mid 7^m - 1$. Now, one can either use the binomial formula
+$$ 7^m = (1+6)^m = 1 + {m \choose 1} 6 + {m \choose 2} 6^2 + \cdots + 6^m $$
+and get $6 \mid 7^m -1$. Or, one can use induction again to prove that $6 \mid 7^m -1 $ for all positive $m$. Let $Q(m)$ be the statement that $6 \mid 7^m -1 $, then $Q(1)$ is true. Suppose $Q(m)$ is true, ie., $6\mid 7^m-1$, then $(7^{m+1} -1) - (7^m-1) = 6 \cdot 7^m$ is divisible by $6$, hence $6 \mid 7^{m+1}- 1$, that is, $Q(m+1)$ is true. Thus $Q(m)$ is true for all positive integer $m$ .

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