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Proof verification of $f(x) = alpha x - E(alpha x) < 1, ;forall xinmathbb Z, ;alpha in mathbb R setminus mathbb Q $
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I'm wondering whether the below is a valid proof for:
$f(x) = alpha x - E(alpha x) < 1, ;forall xinmathbb Z, ;alpha in mathbb R setminus mathbb Q $ where $E(x)$ is integer part of $x$
I've used two steps.
Step 1
Put $mathbb I = mathbb R setminus mathbb Q$ which denotes irrational numbers.
Let $alpha = k + r, ; k in mathbb Z,; rin(0; 1),; rin mathbb I$. Then the function may be rewritten as:
$$
beginalign f(x) & = alpha x - E(alpha x) \
& = (k+r)x - E((k+r)x) \
&= kx + rx - E(kx + rx) \
&= rx - E(rx) tag 1
endalign $$
Since $x$ may be arbitrary we need step 2.
Step 2
Let $rx = a + b, ; ain mathbb Z, ; b in mathbb I, b in [0, 1)$. Rewrite $(1)$:
$$
$$
beginalign rx - E(rx) & = a + b - E(a + b) \
& = a + b - a - E(b) \
&= b - E(b) \
&= b - 0 \
&= b tag 2
endalign $$
$$
But $b in [0, 1)$ and therefore $f(x) in [0, 1)$ which proves $f(x) < 1$
algebra-precalculus proof-verification
asked Aug 3 at 13:26
roman
3721212
add a comment |Â
up vote
0
down vote
favorite
I'm wondering whether the below is a valid proof for:
$f(x) = alpha x - E(alpha x) < 1, ;forall xinmathbb Z, ;alpha in mathbb R setminus mathbb Q $ where $E(x)$ is integer part of $x$
I've used two steps.
Step 1
Put $mathbb I = mathbb R setminus mathbb Q$ which denotes irrational numbers.
Let $alpha = k + r, ; k in mathbb Z,; rin(0; 1),; rin mathbb I$. Then the function may be rewritten as:
$$
beginalign f(x) & = alpha x - E(alpha x) \
& = (k+r)x - E((k+r)x) \
&= kx + rx - E(kx + rx) \
&= rx - E(rx) tag 1
endalign $$
Since $x$ may be arbitrary we need step 2.
Step 2
Let $rx = a + b, ; ain mathbb Z, ; b in mathbb I, b in [0, 1)$. Rewrite $(1)$:
$$
$$
beginalign rx - E(rx) & = a + b - E(a + b) \
& = a + b - a - E(b) \
&= b - E(b) \
&= b - 0 \
&= b tag 2
endalign $$
$$
But $b in [0, 1)$ and therefore $f(x) in [0, 1)$ which proves $f(x) < 1$
algebra-precalculus proof-verification
asked Aug 3 at 13:26
roman
3721212
The ideas are generally fine, but there is one factual inaccuracy (and the proof can be shortened significantly by focusing only on Case 2). It is not always true that $alpha x$ can be written as $a+b$ where $ain mathbb Z$ and $b in (0,1)$: the exception is when $x=0$. But you don't need $b>0$, so you can relax this assertion (and then greatly simplify the proof).
– Erick Wong
Aug 3 at 13:35
Your proof is fine, but if $yin mathbbR$ is any real number, then $0leq y-E(y)<1$. Why? Well, you've proved this kinda multiple times: If $y=n+r$ with $nin mathbbZ$ and $rin mathbbIcap [0,1)$, then $y-E(y)=r$.
– Hamed
Aug 3 at 13:40
@ErickWong Thank you, that's a good point, I've included $0$ in the interval for $b$ which makes $ain mathbb Z,; b in [0, 1)$ a valid statement for $x = 0$
– roman
Aug 3 at 13:47
@roman Notice now that Step 2 works equally well for $alpha x$ as it does for $rx$, making Step 1 redundant.
– Erick Wong
Aug 3 at 15:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm wondering whether the below is a valid proof for:
$f(x) = alpha x - E(alpha x) < 1, ;forall xinmathbb Z, ;alpha in mathbb R setminus mathbb Q $ where $E(x)$ is integer part of $x$
I've used two steps.
Step 1
Put $mathbb I = mathbb R setminus mathbb Q$ which denotes irrational numbers.
Let $alpha = k + r, ; k in mathbb Z,; rin(0; 1),; rin mathbb I$. Then the function may be rewritten as:
$$
beginalign f(x) & = alpha x - E(alpha x) \
& = (k+r)x - E((k+r)x) \
&= kx + rx - E(kx + rx) \
&= rx - E(rx) tag 1
endalign $$
Since $x$ may be arbitrary we need step 2.
Step 2
Let $rx = a + b, ; ain mathbb Z, ; b in mathbb I, b in [0, 1)$. Rewrite $(1)$:
$$
$$
beginalign rx - E(rx) & = a + b - E(a + b) \
& = a + b - a - E(b) \
&= b - E(b) \
&= b - 0 \
&= b tag 2
endalign $$
$$
But $b in [0, 1)$ and therefore $f(x) in [0, 1)$ which proves $f(x) < 1$
algebra-precalculus proof-verification
asked Aug 3 at 13:26
roman
3721212
I'm wondering whether the below is a valid proof for:
$f(x) = alpha x - E(alpha x) < 1, ;forall xinmathbb Z, ;alpha in mathbb R setminus mathbb Q $ where $E(x)$ is integer part of $x$
I've used two steps.
Step 1
Put $mathbb I = mathbb R setminus mathbb Q$ which denotes irrational numbers.
Let $alpha = k + r, ; k in mathbb Z,; rin(0; 1),; rin mathbb I$. Then the function may be rewritten as:
$$
beginalign f(x) & = alpha x - E(alpha x) \
& = (k+r)x - E((k+r)x) \
&= kx + rx - E(kx + rx) \
&= rx - E(rx) tag 1
endalign $$
Since $x$ may be arbitrary we need step 2.
Step 2
Let $rx = a + b, ; ain mathbb Z, ; b in mathbb I, b in [0, 1)$. Rewrite $(1)$:
$$
$$
beginalign rx - E(rx) & = a + b - E(a + b) \
& = a + b - a - E(b) \
&= b - E(b) \
&= b - 0 \
&= b tag 2
endalign $$
$$
But $b in [0, 1)$ and therefore $f(x) in [0, 1)$ which proves $f(x) < 1$
algebra-precalculus proof-verification
asked Aug 3 at 13:26
roman
3721212
edited Aug 3 at 13:44
asked Aug 3 at 13:26
roman
3721212
asked Aug 3 at 13:26
roman
3721212
asked Aug 3 at 13:26
roman
3721212
3721212
The ideas are generally fine, but there is one factual inaccuracy (and the proof can be shortened significantly by focusing only on Case 2). It is not always true that $alpha x$ can be written as $a+b$ where $ain mathbb Z$ and $b in (0,1)$: the exception is when $x=0$. But you don't need $b>0$, so you can relax this assertion (and then greatly simplify the proof).
– Erick Wong
Aug 3 at 13:35
Your proof is fine, but if $yin mathbbR$ is any real number, then $0leq y-E(y)<1$. Why? Well, you've proved this kinda multiple times: If $y=n+r$ with $nin mathbbZ$ and $rin mathbbIcap [0,1)$, then $y-E(y)=r$.
– Hamed
Aug 3 at 13:40
@ErickWong Thank you, that's a good point, I've included $0$ in the interval for $b$ which makes $ain mathbb Z,; b in [0, 1)$ a valid statement for $x = 0$
– roman
Aug 3 at 13:47
@roman Notice now that Step 2 works equally well for $alpha x$ as it does for $rx$, making Step 1 redundant.
– Erick Wong
Aug 3 at 15:39
add a comment |Â
The ideas are generally fine, but there is one factual inaccuracy (and the proof can be shortened significantly by focusing only on Case 2). It is not always true that $alpha x$ can be written as $a+b$ where $ain mathbb Z$ and $b in (0,1)$: the exception is when $x=0$. But you don't need $b>0$, so you can relax this assertion (and then greatly simplify the proof).
– Erick Wong
Aug 3 at 13:35
Your proof is fine, but if $yin mathbbR$ is any real number, then $0leq y-E(y)<1$. Why? Well, you've proved this kinda multiple times: If $y=n+r$ with $nin mathbbZ$ and $rin mathbbIcap [0,1)$, then $y-E(y)=r$.
– Hamed
Aug 3 at 13:40
@ErickWong Thank you, that's a good point, I've included $0$ in the interval for $b$ which makes $ain mathbb Z,; b in [0, 1)$ a valid statement for $x = 0$
– roman
Aug 3 at 13:47
@roman Notice now that Step 2 works equally well for $alpha x$ as it does for $rx$, making Step 1 redundant.
– Erick Wong
Aug 3 at 15:39
The ideas are generally fine, but there is one factual inaccuracy (and the proof can be shortened significantly by focusing only on Case 2). It is not always true that $alpha x$ can be written as $a+b$ where $ain mathbb Z$ and $b in (0,1)$: the exception is when $x=0$. But you don't need $b>0$, so you can relax this assertion (and then greatly simplify the proof).
– Erick Wong
Aug 3 at 13:35
The ideas are generally fine, but there is one factual inaccuracy (and the proof can be shortened significantly by focusing only on Case 2). It is not always true that $alpha x$ can be written as $a+b$ where $ain mathbb Z$ and $b in (0,1)$: the exception is when $x=0$. But you don't need $b>0$, so you can relax this assertion (and then greatly simplify the proof).
– Erick Wong
Aug 3 at 13:35
Your proof is fine, but if $yin mathbbR$ is any real number, then $0leq y-E(y)<1$. Why? Well, you've proved this kinda multiple times: If $y=n+r$ with $nin mathbbZ$ and $rin mathbbIcap [0,1)$, then $y-E(y)=r$.
– Hamed
Aug 3 at 13:40
Your proof is fine, but if $yin mathbbR$ is any real number, then $0leq y-E(y)<1$. Why? Well, you've proved this kinda multiple times: If $y=n+r$ with $nin mathbbZ$ and $rin mathbbIcap [0,1)$, then $y-E(y)=r$.
– Hamed
Aug 3 at 13:40
@ErickWong Thank you, that's a good point, I've included $0$ in the interval for $b$ which makes $ain mathbb Z,; b in [0, 1)$ a valid statement for $x = 0$
– roman
Aug 3 at 13:47
@ErickWong Thank you, that's a good point, I've included $0$ in the interval for $b$ which makes $ain mathbb Z,; b in [0, 1)$ a valid statement for $x = 0$
– roman
Aug 3 at 13:47
@roman Notice now that Step 2 works equally well for $alpha x$ as it does for $rx$, making Step 1 redundant.
– Erick Wong
Aug 3 at 15:39
@roman Notice now that Step 2 works equally well for $alpha x$ as it does for $rx$, making Step 1 redundant.
– Erick Wong
Aug 3 at 15:39
add a comment |Â
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The ideas are generally fine, but there is one factual inaccuracy (and the proof can be shortened significantly by focusing only on Case 2). It is not always true that $alpha x$ can be written as $a+b$ where $ain mathbb Z$ and $b in (0,1)$: the exception is when $x=0$. But you don't need $b>0$, so you can relax this assertion (and then greatly simplify the proof).
– Erick Wong
Aug 3 at 13:35
Your proof is fine, but if $yin mathbbR$ is any real number, then $0leq y-E(y)<1$. Why? Well, you've proved this kinda multiple times: If $y=n+r$ with $nin mathbbZ$ and $rin mathbbIcap [0,1)$, then $y-E(y)=r$.
– Hamed
Aug 3 at 13:40
@ErickWong Thank you, that's a good point, I've included $0$ in the interval for $b$ which makes $ain mathbb Z,; b in [0, 1)$ a valid statement for $x = 0$
– roman
Aug 3 at 13:47
@roman Notice now that Step 2 works equally well for $alpha x$ as it does for $rx$, making Step 1 redundant.
– Erick Wong
Aug 3 at 15:39