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10.23 Proof

  1. Prove that for \(n\in\N\), \(n(n+1)\) is even

    If \(n\) is even \(\Rightarrow\) \(n=2k\) for \(k\in \Z\). Then \(n(n+1)=2k(2k+1)=2(k(k+1))\)​

    Thus by definition, \(n(n+1)\) is even.

    If \(n\) is odd \(\Rightarrow\) \(n=2k+1\) for \(k\in\Z\). Then \(n(n+1)=(2k+1)(2k+1+1)=(2k+1)2(k+1)=2(2k+1)(k+1)\)​

    Thus by definition, \(n(n+1)\) is even.

  2. Prove that if \(x^2\leq 1\Rightarrow x^2-7x>-10\)​

    Proof

    Since \(x^2\leq 1\), then \(-1\leq x\leq 1\Rightarrow x<2\Rightarrow (x-2)(x-5)=x^2-7x+10\)​

    Since \(x<2\) and \(x<5\), then \(x-2<0\) and \(x-5<0\). Then \(x^2-7x>-10\)​

  3. A circle has it's center at \(P=(2,4)\). Prove that \(P_1=(-1,5)\) and \(P_2=(5,1)\) are not both on the circle

    Suppose \(P_1,P_2\) are both in the circle, then \(PP_1=PP_2\)​

    Then \(PP_1=\sqrt{\left(2-\left(-1\right)\right)^2+\left(4-5\right)^2}=\sqrt{9+1}=\sqrt{10}\)​

    \(PP_2=\sqrt{\left(2-5\right)^2+\left(4-1\right)^2}=\sqrt{9+9}=3\sqrt2\)

    Obviously, \(PP_1\neq PP_2\) which is a contradiction!

  4. Prove that \(\forall\varepsilon>0,\exists\delta>0,\forall x,|x-1|<\delta\Rightarrow|x^2-1|<\varepsilon\)

    Analysis

    First, we can do some transformation.

    \(-\delta<x-1<\delta\Rightarrow-\delta+1<x<\delta+1\Rightarrow\left(-\delta+1\right)^2<x^2<\left(\delta+1\right)^2\Rightarrow\left(-\delta+1\right)^2-1<x^2-1<\left(\delta+1\right)^2-1\)

    Thus we can let \(\varepsilon=(\delta+1)^2-1\), then \(\delta=\sqrt{\varepsilon+1}-1\).

    We can try some example which is \(\varepsilon=0.02,0.1,0.9,1,5,10,50\)

    Then we can find if \(\varepsilon<1\), the smaller \(\varepsilon\) is, the closer the ratio of \(\varepsilon\) to \(\delta\) gets to 2, but it remains greater than 2.

    If \(\varepsilon\geq 1\), the bigger \(\varepsilon\) is, the ratio of \(\varepsilon\) to \(\delta\) gets to positive infinity

    the smaller \(\varepsilon\) is, the closer the ratio of \(\varepsilon\) to \(\delta\) will slight smaller than 2.5

    Thus we can think about the relation of \(\varepsilon\) and \(\delta\) can be \(\delta=\min\{1,\frac{\varepsilon}{3}\}\)

    We will find if to 2,不会大于2.5，三合理 大于一 to 无穷所以 想到

    Another way is that we expand the condition and get \(x+1<2+\delta<3\).

    And \(|x^2-1|=\left|x+1\right|\left|x-1\right|<\delta\left|x+1\right|<3\delta\)

    And we can get the thread.

    Proof

    We can consider 2 cases.

    If \(\delta =1\), then \(|x-1|<1\Rightarrow0<x<2\Rightarrow\varepsilon>3\)

    If \(\delta=\frac{\varepsilon}{3}\), then \(|x-1|<\frac{\varepsilon}{3}\Rightarrow1-\frac{\varepsilon}{3}<x<1+\frac{\varepsilon}{3}\Rightarrow\left(1-\frac{\varepsilon}{3}\right)^2<x^2<\left(1+\frac{\varepsilon}{3}\right)^2\Rightarrow\left(1-\frac{\varepsilon}{3}\right)^2-1<x^2-1<\left(1+\frac{\varepsilon}{3}\right)^2-1\)

    Then we can let \(\left(1+\frac{\varepsilon}{3}\right)^2-1\leq\varepsilon\Rightarrow\varepsilon\in\left\lbrack0,3\right\rbrack\)

    Thus we prove it.

    Analysis

    Since \(\delta\) represents a very subtle element, then it tells us we need to consider \(x\rightarrow 1\)

    \(|x^2-1|=|x+1||x-1|<\delta\cdot|x+1|\)

    In order to bound \(|x^2-1|\) to \(\varepsilon\), we need to bound \(|x+1|\)

    Then let \(|x+1|<c\), also \(\delta\cdot c=\varepsilon\), thus \(\delta=\frac{\varepsilon}{c}\)

    Then \(-\delta+2<x+1<\delta+2\)

    So if we choose \(\delta=min=\left\lbrace c,\frac{\varepsilon}{c+2}\right\rbrace\), is ok

    Another way,

    I will show that given any \(\varepsilon >0\), \(\delta=\min\{1,\frac{\varepsilon}{3}\}\) works

    If \(|x-1|<\delta\Rightarrow -\delta<x-1<\delta\Rightarrow -\delta+1<x<\delta+1\Rightarrow0<-\delta+2<x+1<\delta+2\leq 1+2=3\)​

    Now, \(|x^2-1|=|x-1||x+1|<3\cdot|x-1|<3\cdot\frac{\varepsilon}{3}=\varepsilon\)​

    image​

    Since we need to find a \(\delta\) that for any arbitrary \(x\), which means \(\delta\) is fixed and \(x\) is not.

    Thus the above operation is not valid.

  5. Prove that there exists a \(3\) digit number in \(\N\) less than \(400\), with all it's digit different such that the sum of the digits is \(17\) and the product is \(108\)​

    We have \(a+b+c=17\) and \(abc=108\) and \(100a+10b+c<400\)

    Which means \(a=1,2,3\) (3 cases)

    1. \(a=1\)​

      \(\begin{cases}b+c=16\\ bc=108\end{cases}\) no integer solution! 2. \(a=2\)​

      \(\begin{cases}b+c=15\\ bc=54\end{cases}\) the solution is \(\begin{cases}c=6\\ b=9\end{cases}\) or \(\begin{cases}c=9\\ b=6\end{cases}\) 3. \(a=3\)​

      \(\begin{cases}b+c=14\\ bc=36\end{cases}\)no integer solution!

    Thus the number is \(269\) or \(296\)​

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