All Questions
Tagged with real-numbers proof-writing
127
questions
1
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4
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8k
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Prove $(-x)y=-(xy)$ using axioms of real numbers
Working on proof writing, and I need to prove
$$(-x)y=-(xy)$$
using the axioms of the real numbers. I know that this is equivalent to saying that the additive inverse of $xy$ is $(-x)y$ but I am ...
user124862
1
vote
1
answer
54
views
My proof of: $|x - y| < \varepsilon \Leftrightarrow y - \varepsilon < x < y + \varepsilon$
Is it reasonable to prove the following (trivial) theorem?
If yes, is there a better way to do it?
Let $x, y \in \mathbb{R}$.
Let $\varepsilon \in \mathbb{R}$ with $\varepsilon > 0$.
$\textbf{...
- 399
0
votes
1
answer
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$xy \le xz$ if both $y \le z$ and $0 \le x$. (very easy proof exercise)
As an exercise, I tried to prove the following theorem.
Please share your thoughts about what I wrote.
(The proof only uses the utensils which are listed below.)
Theorem
Let $x,y,z \in \mathbb{R}$.
...
- 1,321
3
votes
1
answer
2k
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$x^n y^n = (xy)^n$, proof exercise
As an exercise, I tried to prove the following theorem.
Please share your thoughts about what I wrote.
(The proof only uses the utensils which are listed below.)
Theorem
\begin{equation*}
x^n y^n ...
- 1,321
3
votes
2
answers
1k
views
$x,y$ are real, $x<y+\varepsilon$ with $\varepsilon>0$. How to prove $x \le y$? [closed]
$x,y \in \mathbb R$ are such that $x \lt y + \epsilon$ for any $\epsilon \gt 0$ Then prove $x \le y$.
1
vote
4
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120
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Proof: For every $x \in \mathbb R\,\exists n\in\mathbb Z,\,r\in[0,1):x=n+r.$
I still do not understand how to approach proofs. Any help would be appreciated.
For every real number $x$, there exists an integer $n$ and a real number $r\in [0,1)$ such that $x = n + r$.
Hint: ...
- 35
1
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2
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An upper bound $u$ is the supremum of $A$ if and only if for all $\epsilon > 0$ there is an $a \in A$ such that $u-\epsilon < a$
Problem:
Let $u$ be an upper bound of non-empty set $A$ in $\mathbb{R}$.
Prove that $u$ is the supremum of $A$
if and only if for all $\epsilon > 0$ there is an $a \in A$ such that
$u-\epsilon &...
- 6,684
2
votes
3
answers
75
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If $x>y$, then $\lfloor x\rfloor\ge \lfloor y\rfloor$, formal proof
For x ∈ ℝ, define by: ⌊x⌋ ∈ ℤ ∧ ⌊x⌋ ≤ x ∧ (∀z ∈ ℤ, z ≤ x ⇒ z ≤ ⌊x⌋).
Claim 1.1: ∀x ∈ ℝ, ∀y ∈ ℝ, x > y ⇒ ⌊x⌋ ≥ ⌊y⌋.
Assume, x, y ∈ ℝ # Domain assumption
...
- 93
1
vote
1
answer
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There exists a positive real number $u$ such that $u^3 = 3$
Modify the Theorem that states There exists a positive real number x such that $x^2 = 2$.
Show that there exists a positive real number $u$ such that $u^3 = 3$.
So far, I have come up with the ...
- 81
0
votes
1
answer
236
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Identify a countable union of nested intervals using the Archimedean principle [closed]
$\displaystyle\bigcup_{n=2}^\infty \left[\frac1n,3-\frac 2n\right]=(0,3)$. I can't prove using limits. I have to use the Archimedean principle and I don't know how to go about doing that..
- 41
2
votes
3
answers
85
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$a > b+1 \Rightarrow a>x>b$?
If I have $a,b \in \mathbb R$ such that $$a > b+1 $$
It is assured that $\exists\space x \in \mathbb Z: a>x>b$
Does this property have some special name?
How can this be proved?
This idea ...
- 6,834
0
votes
1
answer
43
views
Proving of Inequalities
How to prove:
If $a>0$ and $b>0$, and $a^2>b^2$, then $a>b.$
I've tried different methods but I really can't prove this one. Thank you for your help!
2
votes
2
answers
704
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How to show that an infinite decimal is equal to a unique real number?
I don't understand how the proof above shows that two distinct real numbers correspond to different infinite decimal.
All I got out of the explanation is given any two distinct real numbers $a$ and $...
- 3,600
0
votes
1
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73
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Help with 2 questions my professor gave us
I was wondering how to solve these two proofs my professor put on the blackboard today. He said they were pretty easy but i'm still unsure how to prove them. ANy help would be greatly appreciated!
(i)...
- 107
0
votes
3
answers
72
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Help solving a proof
My professor put this up on the blackboard and I was wondering how to solve it.
Let $x,y \in \mathbb{R}$. Then |$x$|< |$y$| if and only if $x^2 < y^2$.
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