|
- How do I know when to use let and suppose in a proof?
I often use 'suppose' when my goal is to derive a contradiction, and 'let' when I instantiate a variable when I'm not going to derive a contradiction I'm not sure if this is standard
- elementary set theory - Suppose $A$ and $B$ are sets. Prove that $A . . .
Suppose A is subset of B Let X belongs to A then by hypothesis, X will belong to B Hence X belong to A and X belong to B implies that X belongs to A intersection B Accordingly A is subset of A intersection B But we know that A intersection B is always subset of A Hence A intersection B is equal to A On the other hand, suppose A intersection B is equal to A Then in particular, A is
- Suppose $v_1,. . . ,v_m$ is linearly independent in $V$ and $w \in V . . .
You'll need to complete a few actions and gain 15 reputation points before being able to upvote Upvoting indicates when questions and answers are useful What's reputation and how do I get it? Instead, you can save this post to reference later
- Suppose $n$ is an even positive integer and $H$ is a subgroup of . . .
Suppose $n$ is an even positive integer and $H$ is a subgroup of $\mathbb Z n\mathbb Z$ Prove that either every element of $H$ is even or exactly half of its elements are even
- Suppose $V$ is finite-dimensional and $T_1,T_2∈L(V,W)$. Prove that . . .
You'll need to complete a few actions and gain 15 reputation points before being able to upvote Upvoting indicates when questions and answers are useful What's reputation and how do I get it? Instead, you can save this post to reference later
- Suppose $CA=I_n$ (the $n \times n$ identity matrix. Show that the . . .
Does this answer your question? Linear Algebra - Suppose $CA=I_n$ Show that the equation $Ax = 0$ has only the trivial solution
- Suppose $f: [a,b]\to\mathbb {R}$ is Riemann integrable. Prove that . . .
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges,
- Let $a,b,c\in\mathbb {Z}$. Suppose $a$ and $b$ are not both zero, and . . .
"Should I have proved the claim "if x∈Z is a divisor of both a and b, then −x is also" or is it ok to assume the reader would find this obvious" In my opinion I'd have said less BTW it is not that $\gcd (a,b)$ is positive because they are both non zero It is finite because they are both non-zero
|
|
|