I have an equation      y = ab + cb'   How would I implement this with only NOR Gates?  Thanks
122326
I'm having some trouble understanding how I can convert a boolean expression to a NOR-gate only expression.   What I'm working with looks like \$T = A B' C + A' B C' + A B\$. How would I go into implementing it with NOR gates?
74422
How do I generalize the equation to be able to plug in any result for $\phi(n)=12$ and find any possible integer that works?
13583
I am trying to create a logic circuit using only NAND Gates for this expression: (NOT Q AND P) OR R   This question has really gotten me stuck! Can somebody please help?
92873
Please guide me:  $y' + ay +b = 0$ ($a$ not zero) is supposed to be separable and has solution  $y = ce^{-ax} - \frac ba$  Here is my start to this problem:  $\frac{dy}{dx} + ay = -b$ is as far as I can go with this. How should I go about separating $x$ and
61024
My proof: By the way of contradiction, suppose that we found $(x,y) \in \mathbb N^*$ satisfies:$$\frac{x+y}{xy}=5$$ So : $$x+y=5xy \\ \iff x=5xy-y \iff x=y(5x-1)$$ Since $x\in \mathbb N^* $ so the minimum value of $5x$ is $5&gt;1 \implies y\mid x$ By the same method we have: $$y=x(5y-1) \implies x\mid y$$ $x\mid y, y\mid x \implies x=y$, Now we can write the original equation as : $$\frac{2x}{x^2}=5 \iff 2=5x \implies x\notin \mathbb N^* $$ Therefore there is no solution to this equation: $$\frac{x+y}{xy} =5$$ Does this considered a proof?
31703
If $A$ and $B$ are $n\times n$ matrices, and $AB$ is invertible then $A$ and $B$ are invertible.    I started out by writing that since $AB$ is invertible, then for the equation $ABx=b$ has a unique solution for any $b$, $x=(AB)^{-1}b$.  Then I don't know how to break up the composition of AB to prove they are each individually similar.
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A similar question was asked in .  A paper by Roberto Di Cosmo and Thomas Dufour () asserts in footnote 2 (p.3) that it has been shown "that there are no nontrivial equations for $⟨\mathbb{N}, \mathrm{Ack}(n, \_, \_)⟩$ if $n &gt; 2.$" References make it clear that $\mathrm{Ack}(n,x,y)=H_{n+1}(x,y)$ in the , so the assertion is as follows:     $\text{There are no nontrivial equations for the binary function }(\_\uparrow^y\_)\text{ if }y&gt;1\\ \text{ (i.e., for hyperoperations above exponentiation)}\tag{*}.$   (For this, the above paper cites Charles F. Martin, "Axiomatic bases for equational theories of natural numbers", Notices of the Am. Math. Soc., 19(7):778, 1972, which I have not found online.)  Q.1: What, precisely, is meant by a "nontrivial equation" here?  (The defining recursion $$x\uparrow^{y+1}(z+1)=x\uparrow^y (x\uparrow^{y+1}z),$$ which certainly holds for $y&gt;1$, is evidently considered trivial.)  Q.2: Is the following a valid proof based on assertion $(*)$? ...  Claim:     If $x,y,z$ are integers $\gt 1$, then the value of $\ x\uparrow^y z\ $ uniquely determines $x,y,z$; i.e.,  $$x\uparrow^y z = x'\uparrow^{y'} z' \quad\implies\quad x=x',\ y=y',\ z=z'.\quad \tag{1}$$  (The purpose of restricting the domain to integers $&gt;1$ is to remove obvious counterexamples that arise from equations like $x\uparrow (a\cdot z) = (x\uparrow a)\uparrow z\ \ $ or $\ \ 2\uparrow^y 2=4\uparrow^{y'} 1\ \ $ or $\ \ 1\uparrow^y z=1\uparrow^{y'} z'.$)  Proof:  Suppose, for the sake of contradiction, that there is a counterexample to $(1)$, in which the LHS of the implication holds but not the RHS. Then the LHS is a nontrivial equation for some hyperoperation(s) beyond exponentiation, which is impossible as per $(*)$. QED
56990
The problem is as stated in the title. I've tried factorization, but it doesn't seem to yield anything. Any help is greatly appreciated. Also please don't use modular arithmetic, thank you.
69512
I am trying to give a non-algebraic proof for this equality:  $$\dbinom{a}{b}\dbinom{b}{c}=\dbinom{a}{a-c}\dbinom{a-c}{a-b}$$  So far, I could only use the identity $\dbinom{x}{y}=\dbinom{x}{x-y}$.  What I got is $\dbinom{a}{a-c}\dbinom{a-c}{a-b}=\dbinom{a}{c}\dbinom{a-c}{b-c}$. But I don't think I have made any progress from there. It seems reasonable if we think about the identity, but I just find it difficult to give a proper proof for it.  Thanks for the help.
84986
Basically, I want to turn a continuous "1" into a short "1" pulse then "0". I'm new to logic gates, apologies if it's too obvious.
114077
How can we show that the assumption $a,b,c&gt;0$ and $a+b+c=1$ implies   $$\sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}~?$$
120379
I've scoured the math stackexchange and the philosophy one for some guidance on how to go about this while using the Fitch System.  Anyone can attempt it here;    I can reach the conclusion, but it's not in the correct "layer" of the proof. I'm stuck with the answer as a subproof.  (Note that the fitch system lacks contradiction elimination (RAA) as a method for getting out of assumptions).
83554
I'm really struggling with "simple" proofs, and I could only come up with this. I googled for the solution and know how to do it now, but I first want to know if there's anything wrong with my solution? $$ (AB)^{-1}=B^{-1}*A^{-1}. $$ For this equation to hold, $B$ must be invertible and so must $A$. //
67892
Let $ a, b, c $ be sides of a triangle and $ ab+bc+ca=1 $. Show $$(a+1)(b+1)(c+1)&lt;4 $$  I tried Ravi substitution and got a close bound, but don't know how to make it all the way to $4 $. I am looking for a non-calculus solution (no Lagrange multipliers).  Do you know how to do it?
69459
Prove that if $a,b,c$ are non-negative real numbers, then   $(ab + bc + ca)^2 \geq 3abc(a+b+c)$.   I tried to compute from $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$.
46158
If $a,b,c$ and $d$ satisfy the equations:$$a+7b+3c+5d=0$$$$8a+4b+6c+2d=-16$$$$2a+6b+4c+8d=16$$$$5a+3b+7c+d=-16$$ then $(a+d)(b+c)$ equals $-16$.   I can't understand why $(a+d)(b+c)$ equals $-16$?
3104
I'm stuck on what how to continue. I know I'm missing a few steps but this is what I have so far. Thank you in advance!   ¬Cube(b) → Small(b) Small(c) → (Small(d) ∨ Small(e))  Small(d) → ¬Small(c) Cube(b) → ¬Small(e)   Small(c)   ¬Small(b)   ¬Cube(b) Small(b) (→ Elim 1,7) ⊥ (⊥intro 6, 8)    10.Cube(b) (¬Intro 7-9)   ¬Small(e) (→ Elim 4,10)     Goal: Small(c) → Small(b)