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4

The issue here is in the step where you say $$ 2(a-b)=(a-b) $$ $$ 2=1 $$ Since $a=b$, $a-b=0$, and therefore in that step you are dividing by zero. This proof is about as accurate as saying: $$ 0=0 $$ $$ (2)(0)=(1)(0) $$ $$ 2=1 $$


3

It’s not the proof itself that will be used throughout the book: it’s the format of the proof, in which the justification for each equality is given in curly braces immediately after the equals sign. For instance, the third step, $$\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha=2\sin\alpha\cos^2\alpha+\cos2\alpha\sin\alpha\;,$$ is justified by the reason ...


0

There is one more problem. The correct formulation is $$\forall x\in \mathcal P(\mathbb N): x\ne\emptyset \Rightarrow \exists y\in x\ \forall z\in x: y\le z$$ Note the $x\ne \emptyset$ vs. $x\notin\emptyset$ and $y\le z$ vs. $y<z$.


0

Yes, this is the Well-Ordering Principle: Every nonempty subset of the natural numbers contains a least element. The inequality should indeed be changed to be $y \leq z$.


0

I'll start with some examples with actual numbers... Is $10$ a divisor of $20\times23=460$? Yes. Note that $10\mid20$. This gives us a general rule: If $d\mid a$, then $d\mid ab$. Is $10$ a divisor of $580-110=470$? Yes. Note that $10\mid570$ and $10\mid110$. This gives us a general rule: If $d\mid a$ and $d\mid b$, then $d\mid a-b$. (P.S. Those are not ...


2

Suppose that $d\mid n$ and $d\mid kn+b$; then there are integers $\ell$ and $m$ such that $n=d\ell$ and $kn+b=dm$. Substitute the first equation into the second to find that $k(d\ell)+b=dm$ and hence that $b=dm-dk\ell=d(m-k\ell)$, so that $d\mid b$. Now apply this with $n=2a+1$, $k=2a-1$, and $b=2$. More generally, you can prove similarly that if $d$ ...


0

You have to prove it on both sides, but you can make the assumption that you prove it for earlier cases. So to prove for $n+1$ you may assume it holds for $n$.


1

The goal was to prove that $f$ is a well defined function, i.e. it assigns exactly one value to any element of its domain. $\Bbb Z/6\Bbb Z$ contains six elements. Note that $[6]=[0]$ therein (and $[7]=[1]$ and $[8]=[2]$ and $[12]=[0]$ again, and $[-1]=[5]$, and so on..). So, we take one element of this quotient set, which is one equivalence class. But we ...


0

I have a problem with the step where you say $6n^{3} -4n^{2} + 3n + 2 \geq 6n^3 + n^3$. This just isn't true in the long run. You can cut to the chase a little bit faster: $6n^{3} -4n^{2} + 3n + 2 \geq 5n^{3} - n^{2} + n + 1$ for all $n \geq 0$ if and only if $n^{3} - 3n^{2} + 2n + 1 \geq 0$ for all $n \geq 0$ Observe that this is true for $n = 0, 1$. ...


2

Hint. See if you can fill in the gaps in the following proof. The zero polynomial in $P_2$ is $0+0x+0x^2$. This is in $S$ because ... Let $p=a+bx+cx^2$ and $q=d+ex+fx^2$ be polynomials in $S$. This means that $a+b+c=0$ and ... Now adding the two polynomials and simplifying gives $$p+q=\cdots\ .$$ The sum of the coefficients of $p+q$ is ... and so $p+q$ ...


1

The elements of $\mathbb{Z}/4\mathbb{Z}$ are all of the form $[a]$, where $a\in\mathbb{Z}$. However, $[a]$ can also be represented as $[b]$, with $a\ne b$; for instance, $$ [0]=[4]=[8]=\dots $$ Defining $f([a])=[3a+1]$ requires showing that, if $[a]=[b]$, then $[3a+1]=[3b+1]$, so the image of $[a]$ doesn't depend on the particular element we use to ...



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