I want to prove that if $a$ and $b$ are positive integers, the following equality is true:  $$\left\lceil \frac{a}{b}\right\rceil = \left\lfloor\frac{a+b-1}{b}\right\rfloor$$  What have I thought about?  Then $a$ = $qb + r$, where $q$ is the quotient and $r$ the remainder  \begin{align} b-1 &amp;\ge r \ge 0 \\ qb + (b-1) &amp;\ge qb+r \ge qb \\ qb + (b-1) &amp;\ge a\ge qb \\ \left\lfloor\frac{a}{b} \right\rfloor +1-\frac1{b} &amp;\ge \frac{a}{b} \ge \left\lfloor\frac{a}{b}\right\rfloor \\ \left\lfloor \frac{a+b}{b} \right\rfloor -\frac1{b} &amp;\ge \frac{a}{b} \ge \left\lfloor\frac{a}{b}\right\rfloor \end{align}
123477
Mathematically, why is this true?  $$\left\lceil\frac{a}{b}\right\rceil= \left\lfloor\frac{a+b-1}{b}\right\rfloor$$  Assume $a$ and $b$ are positive integers.  Is this also true if $a$ and $b$ are real numbers?
63872
I am currently working on this problem from Hardy's Course of Pure Mathematics and have gotten stuck near the end. I was wondering if someone could help me determine what to go next.  Question  If $a_1, a_2, ...,a_n$ are all positive and $S_n=a_1+a_2+...+a_n$ then:  $(1+a_1)(1+a_2)...(1+a_n) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}$  My Attempt  Proof by induction:  I had first shown that it was true for $n=1$ and $n=2$.  Now suppose that it is true for n. Then it must also be true for $n+1$  $(1+a_1)(1+a_2)...(1+a_n)(1+a_{n+1}) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!} + \dfrac{(S_{n})^{n+1}}{(n+1)!}$  Define: $(1+a_1)(1+a_2)...(1+a_n)=x$ and $1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}=y$  Then: $x+xa_{n+1} \le y+\dfrac{(S_{n})^{n+1}}{(n+1)!}$  By the induction assumption, we know that $x \le y$.  What I can't figure out  From this point, what I think the next natural step would be is to show that  $xa_{n+1} \le \dfrac{(S_{n})^{n+1}}{(n+1)!}$  I know this rearranges to:  $xa_{n+1} \le \dfrac{(S_{n})^{n}}{n!} \times \dfrac{S_n}{(n+1)} $  And feel there may be something I can do here, but haven't been able to figure anything out.
97590
I'm trying to understand properties of the greatest integer function and I am struggling to find the value of $\lfloor x+y \rfloor$  where $x \in \mathbb{R}$, $y \in \mathbb{Z}$, and prove that it is correct value.  I don't really know how to prove this, but I have been dividing it into different cases. I think that it equals $\lfloor x \rfloor + y$ when $x,y$ are both positive but not sure how to prove it. Depending on if one or both $x$ and $y$ are negative, and their ultimate sum, I get different values. I am having trouble determining when exactly this happens though and then proving the results. Any help would be great, thanks!
37833
Given $A$ and $B$ positive-definite matrices and $Q$ unitary matrix, prove that if $A = BQ$, then $A=B$. $Q$ is unitary, so $QQ^*=I$ If $A$ and $B$ are positive-definite, than $A=A^*$ and $B=B^*$. $A^*=(BQ)^*=Q^*B^*$ $A^2=AA=AA^*=(BQ)(Q^*B^*)=B(QQ^*)B^*=BIB^*=B^2$ $A^2=B^2$. I don't know how to use the fact that $A$ and $B$ are positive-definite to finish the proof.
123308
I figured we'd have to show $a=qb$ from $2^a-1=r(2^b-1)$ for $q,r\in\mathbb{Z}$ for the first implication and vice versa, how should I proceed from here?
74884
Why is it that $$ab=ba$$ for positive integers $a,b$? Is there an intuitive explanation or we just need to accept it as a given fact?
124253
Let $a$ and $b$ be positive integers. Prove that: If $a^2$ divides $b^2$, then $a$ divides $b$.   Context: the lecturer wrote this up in my notes without proving it, but I can't seem to figure out why it's true. Would appreciate a solution.
47264
Let $a$ and $b$ be positive integers such that $a | b^2, b^2 | a^3, a^3 | b^4, b^4 | a^5, \cdots $ Prove that $a = b$.  My way is as follows: Let $A=v_p(a), B=v_p(b)$ be the exact power of a prime $p$ dividing $a$ and $b$ respectively. Then by assumption, $(2n-1)A \leq 2nB, n=1,2,3,\cdots.$ Hence $A \leq (2n/(2n-1))B$. Taking limit as $n$ tends to infinite yields $A \leq B.$ Likewise, $2nB \leq (2n+1)A, n=1,2,3,\cdots,$ by a similar argument, we have $B \leq A.$ So that $A=B$ and we are done. But I am rather unsatisfied with the proof because it involves limit. I wanna ask if there is any other alternative proof, thank in advances.
40198
We have to prove $b|a$ and $b|c \Rightarrow b|ka+lc$ for all $k,l \in \mathbb{Z}$.  I thought it would be enough to say that $b$ can be expressed both as $b=ka$ and $b=lc$. Now we can reason that since $ka+lc=2b$ and $b|2b$, it directly follows that $b|ka+lc$?  But I'm not sure if that works for any value of $k$ and $l$ (namely $k$ and $l$ are defined through quotient between $a$ and $c$, respectively).  What am I missing?
18846
Determine all possible values of $\frac{a^2+b^2+1}{ab-1}$ where $a,b$ are positive integers.  I am quite certain one should use a Vieta jumping argument, but I cannot complete it.  Let $\frac{a^2+b^2+1}{ab-1} = k \in \mathbb{Z}$ with $a+b$ being minimal. Clearly $k &gt; 0$ and we have $a^2 - kab + b^2 + k + 1 = 0$. Without loss of generality $a\geq b$. We see that $x_1 = a$ and $x_2 = kb-a = \frac{b^2+k+1}{a}$ are the roots of $x^2 - kbx + b^2 + k + 1 = 0$. The expression $kb-a$ shows they are both integers and $\frac{b^2+k+1}{a}$ they are both positive. To reach a contradiction, it suffices to have $b + \frac{b^2+k+1}{a} &lt; a + b$, i.e. $b^2 + k + 1 &lt; a^2$. Hmm, if we can perhaps somehow assure $k\leq a$, we will just consider pairs $(a,b)$ with $a\neq b$ and $a+b$ minimal $-$ that's how we can rule out the case $a\neq b$ and the rest will be very easy. So any idea how to complete this argument, if it indeed works?  Any help appreciated!
85477
For all positive integers $n,r,k$ that $n=rk$, how to prove the inequality $$(r-1)!^k\le \frac{c(n,k)}{S(n,k)}\le (n-k)!$$ $c(n,k)$ is unsigned Stirling numbers of the first kind and $S(n,k)$ is Stirling numbers of the second kind.  Any help would be appreciated!
72479
Find all triples of positive integers (a, b, c) such that  $$ \left( 1 + \frac{1}{a} \right)\left( 1 + \frac{1}{b} \right)\left( 1 + \frac{1}{c} \right) = 2 $$  I tried to expand, yielding $ a+b+c+ab+bc+ca+1 = abc $, but I couldn't go anywhere from here.   This problem clearly involves using the fact that $a, b, c$ are positive integers, since there are three variables and only one equation. It would be great if someone also addressed this condition and maybe tried to solve the problem for other requirements, like when $a, b, c$ are positive or negative integers.  Finally - can anyone post or link other such problems which involves a mix of algebra manipulation and number theory. I would really appreciate it.  (Here is the full paper: )
39825
If $p,q$ are positive quantities and $0 \leq m\leq 1$ then Prove that $$(p+q)^m \leq p^m+q^m$$    Trial: For $m=0$, $(p+q)^0=1 &lt; 2= p^0+q^0$  and for $m=1$, $(p+q)^1=p+q =p^1+q^1$.   So, For $m=0,1$ the inequality is true.How I show that the inequality is also true for $0 &lt; m &lt; 1$.   Please help.
44581
Given that $a,b,c$ are distinct positive real numbers, prove that $(a + b +c)\big( \frac1{a}+ \frac1{b} + \frac1{c}\big)&gt;9$   This is how I tried doing it:  Let $p= a + b + c,$ and $q=\frac1{a}+ \frac1{b} + \frac1{c}$.  Using AM>GM for $p, q$, I get:  $$\frac{p+q}{2} &gt; {(pq)}^{1/2}$$ $$\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} &lt; \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2$$ And for any $x\in \mathbb{R}, \space \space x+\frac1{x}≥2.$ Thus,   $$(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)&lt;9, $$ which is the opposite of what had to be proven. What did I do wrong?
109324
Can anyone find a hole in my proof or tell me how to make it stronger or simpler? The &quot;lemma above&quot; referenced in the fourth paragraph is a small but solid proof showing that $r$ is strictly less than $b$ in the division algorithm if it's written $a=bq+r$.  $\mathbf {Theorem}$ If $m$ and $n$ are positive integers that are relatively prime, then there are integers $a$ and $b$ such that $am+bn=1$. $\mathbf {Proof}$ Let $D = \{d \in \Bbb Z : am+bn&gt;0\}$. Then there is a least element $l$ in D. Since $m$ and $n$ are relatively prime, we need only show that $l$ divides $m$ and $n$. Suppose $l$ does not divide $m$. Then there are integers $q$ and $r$ such that $m=ql+r$. Rewritten, we have that $r=ql-m=q(am+bn)-m=qam-m+qbn=m(qa-1)+n(qb)$. Thus, $r \in D$. By the lemma above, $r&lt;l$, contradicting that $l$ is the least element of $D$. We conclude then that $l$ divides $m$. Switching $m$ for $n$ gives the same result for $n$. Since $m$ and $n$ are relatively prime, $l=1$ and it follows that there are integers $a$ and $b$ such that $am+bn=1$.
15359
Show that if $a$ and $b$ are positive integers, then $$\left(a +\frac12\right)^n + \left(b+\frac{1}{2}\right)^n$$is an integer for only finitely many positive integers $n$.   I tried hard but nothing seems to work. :(
93582
Let $a,b,$ and $c$ be positive numbers with $a+b+c = 1$. Prove that $$\left (\dfrac{1}{a}+1 \right)\left (\dfrac{1}{b}+1 \right)\left (\dfrac{1}{c}+1 \right) \geq 64.$$   Attempt  Expanding the LHS we obtain $\left (\dfrac{1+a}{a} \right)\left (\dfrac{1+b}{b} \right)\left (\dfrac{1+c}{c} \right)$. We are given that $a+b+c = 1$, so substituting that in we get $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right)$. Then do I say $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right) \geq 64$ and see if I can get a true statement from this?