“‘LIBTARD’ is a portmanteau of liberal and retard, conflating liberal ideology with developmental disability. Of course, dropping ad homs while advocating rational discourse is fairly obvious contradiction. PressF injects emotion into his claim in a clause dependent on the very sentence where he pretends to reason. I call PressF's claim self-disproved.”
Ad Homs refer to attacks made on a particular person’s character. As I have not made attacks against any particular person, by definition, it does not qualify as an ad hom.
Make sure to define all your terms from now on. Your opponent used the lack of definition for "better" as a way to steer the argument. Just something to take note of.
“So... let me get this straight:
10-0.999...=0.000...1, correct? Ok.
The "..." in 0.000...1 represents an infinite number of 0s before the one. Since there is an infinite number of zeros, there will always be a 0 after the previous one. Thus, the 1 at the end will never be reached. Since the 1 will never be reached, 0.000...1 is the same as 0.000..., which is simply 0.
Going back to our equation, we now have 10-0.999...=0
Adding 0.999... on both sides, we get 10 = 0.999...
QED”
Where you see a 10, replace it with a 1. I had big brain fart lol
Since 0.999… is a repeating decimal, it can be expressed as an infinite sum:
0.999… = 0.9 + 0.09 + 0.009 + 0.0009 + …
S∞ = The entire infinite sum → S∞ = 0.999...
A = The first term → A = 0.9
R = The constant multiple (this can be derived from dividing the second term from the first)
R = (9/100)/(9/10) ← Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second.
R = (9/100)*(10/9)
R = 90/900 ← Simplify
R = 1/10
Since |1/10| < 1, we will use the second case of the equation. We will now plug in our numbers into the equation.
0.999… = 0.9/(1 - 1/10)
0.999… = (9/10)/(10/10 - 1/10)
0.999… = (9/10)/(9/10) ← Any number divided by itself is 1
0.999… = 1
For the second case, remember that x^n will increase (gets closer to infinity) as n increases (gets closer to infinity). In this case, N = ∞. So, R^N = R^∞ = ∞. With that, we get this:
S∞ = A(1 - R^∞)/(1 - R), |R| > 1
S∞ = A(1 - ∞)/(some negative number)
S∞ = A(-∞ + 1)/(some negative number)
At this point, it is important to note that infinity plus/minus any finite number is still infinity. Similarly, negative infinity plus/minus any finite number is still negative infinity. Thus:
S∞ = A(-∞ + 1)/(some negative number)
S∞ = A(-∞)/(some negative number)
S∞ = -A(∞)/-(some positive number) ← remove the negative from the negative number
S∞ = A(∞)/(some positive number)
S∞ = ∞/(some positive number) ← Infinity/(Any finite number) = Infinity
S∞ = ∞
For this case, we can see that the infinite geometric series equation will equal to infinity. This leaves us with the third case:
S∞ = A(1 - R^∞)/(1 - R), |R| < 1
For the third case, remember that x^n will decrease (gets closer to zero) as n increases (gets closer to infinity). In this case, N = ∞. So, R^N = R^∞ = 0. With that, we get this:
S∞ = A(1 - R^∞)/(1 - R), |R| < 1
S∞ = A(1 - 0)/(1 - R)
S∞ = A(1)/(1 - R) ← Remember: A(1) = A
S∞ = A/(1 - R)
With that said, there are two possible cases for this equation:
S∞ = A(1 - R^∞)/(1 - R), |R| > 1 → S∞ = ∞
S∞ = A/(1 - R), |R| < 1
I will now use this equation to prove that 0.999… = 1.
Now that we have the equation for the geometric series, we need to establish what an infinite geometric series is.
An infinite geometric series is one which has an infinite amount of terms in its sequence. Since N is the number of terms in the sequence, and an infinite geometric series has an infinite amount of terms in its sequence (by definition), N = ∞. Let the entire infinite geometric series be S∞.
Retrieving the equation for the geometric series, and plugging ∞ into N, we get this:
S∞ = A(1 - R^∞)/(1 - R)
At this point, it is important to note the nature of exponents. Any number (x) to the power of something (n) will get larger or smaller when n gets larger, depending on what x is. The following will be the case for positive and negative numbers, so we can put an absolute value sign (| |) around x. This will make x an absolute value number. What an absolute value sign does is that it turns a negative number into a positive one. Positive numbers will not be affected.
If |x| > 1, then x^n will increase as n increases. (eg, 2^2 = 4, 2^3 = 8, etc.)
If |x| < 1, then x^n will decrease as n increases. (eg, (½)^2 = ¼, (½)^3 = ⅛)
If |x| = 1, then x^n will remain the same.
With this in mind, we can address the equation at hand. An absolute value number increasing can be thought of as getting closer to infinity, while an absolute value number decreasing can be thought of as getting closer to zero. There are three possible cases for the equation:
S∞ = A(1 - R^∞)/(1 - R), |R| = 1
S∞ = A(1 - R^∞)/(1 - R), |R| > 1
S∞ = A(1 - R^∞)/(1 - R), |R| < 1
For the first case, we get this:
S∞ = A(1 - R^∞)/(1 - R), |R| = 1 → S∞ = A(1 - 1^∞)/(1 - 1) → S∞ = A(1 - 1^∞)/0
There are two problems with this. Firstly, we cannot divide by zero. Secondly, 1^∞ is indeterminate. Thus, we can outright eliminate this case.
This goes on for the amount of terms that are in the sequence. Let's call the number of terms in the sequence "N", and the sequence as a whole as "s".(Again, what we call these by is completely arbitrary)
So, a geometric sequence can be shown as:
s = [A, AR, AR^2, ..., AR^(N-3), AR^(N-2), AR^(N-1)]
The reason that the final number in the sequence is AR^(N-1) and not AR^N is because on the 1st term, N = 0 (R^0 = 1, and A*1 is simply A).
Since a series is simply the sum of a sequence, a geometric series (let's call the variable for the entire series "S") can be shown as:
S = A + AR + AR^2 + ... + AR^(N-3) + AR^(N-2) + AR^(N-1)
This is our first equation for the proof.
If we multiply R by both sides, we get:
SR = R * [A + AR + AR^2 + ... + AR^(N-3) + AR^(N-2) + AR^(N-1)]
Remember that the exponent simply shows the number of times that something is being multiplied by. So, if we multiply something raised to the power of n (n is simply the exponent of that something) by that same something, then it is the same as adding 1 to n. With that in mind, we get:
SR = AR + AR^2 + AR^3 + ... + AR^(N-3+1) + AR^(N-2+1) + AR^(N-1+1)
= AR + AR^2 + AR^3 + ... + AR^(N-2) + AR^(N-1) + AR^N
This is our second equation.
Now, we subtract the second equation from the first (First equation is S, second equation is SR):
S - SR = [A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1)] - [AR + AR^2 + AR^3 + ... + AR^(N-2) + AR^(N-1) + AR^N]
The minus sign in front of the first series can be distributed into each number within the series:
S - SR = [A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1)] + [-AR - AR^2 - AR^3 - ... - AR^(N-2) - AR^(N-1) - AR^N]
We can get rid of the brackets:
S - SR = A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1) - AR - AR^2 - AR^3 - ... - AR^(N-2) - AR^(N-1) - AR^N
Then we can rearrange the equation:
S - SR = A - AR^N + AR - AR + AR^2 - AR^2 + AR^3 - AR^3 + ... - ... + AR^(N-2) - AR^(N-2) + AR^(N-1) - AR^(N-1)
I'll show you the infinite geometric series proof that I forgot to explain in full last time.
Before I go on to proving that 0.999...=1, I need to establish what an infinite geometric series is. Before I can do that, we need to establish what a geometric series is. Before that, I need to define what a series is.
A series is simply the sum of a sequence. A sequence is just a set of numbers. For example, 15 is the sum of the series [1,2,3,4,5] (the square brackets are just the formal way of showing a set). If you add all the numbers of the sequence together, you will get 15.
A geometric series is a series in which each subsequent number in the series' sequence is multiplied by some constant. There will be the first number in the sequence, and then that number will be multiplied by something to get the next number, which will be multiplied again by that something to get the next, and so on and so forth. Let's define the first number as "A" and the constant which "A" is being multiplied by as "R". These variables are completely arbitrary, and we can call them whatever we want. Given these two variables, we can now define our geometric sequence in a mathematical sense.
The first number of the sequence is just A, by definition.
The second is A*R, since a geometric sequence is, by definition, in which each subsequent number in it is the previous one multiplied by a constant. This can be rewritten as AR.
The third is AR*R, since we are multiplying the last term by the same constant (R). Since R*R can be rewritten as R^2, this term can be rewritten as AR^2. One important thing to remember is that the exponent of something simply denotes how many times that variable is multiplied by.
The fourth term is AR^3, since the previous term is AR^2 and we are multiplying it by R again.
So... let me get this straight:
10-0.999...=0.000...1, correct? Ok.
The "..." in 0.000...1 represents an infinite number of 0s before the one. Since there is an infinite number of zeros, there will always be a 0 after the previous one. Thus, the 1 at the end will never be reached. Since the 1 will never be reached, 0.000...1 is the same as 0.000..., which is simply 0.
Going back to our equation, we now have 10-0.999...=0
Adding 0.999... on both sides, we get 10 = 0.999...
As I stated before, if ...9999.0 is infinite (as seems to be the case with the ... to the left of the 9s), then you’re left with ♾-10♾. 10♾ is still infinity, so you’re left with ♾- ♾. This is indeterminate, meaning that there is no definite or definable variable. You can’t calculate an answer from an indeterminate equation. Since your entire argument hinges on the basis that ...9999.0-...9990.0=9 (essentially, ♾-♾=9), and this is shown not to be the case, the argument falls apart.
In that case, I guess I’m gonna have to look at the specific arguments presented. However, this debate suffers the same problem as the last (it’s about the same topic), that several “holy” texts are automatically eliminated by the presumption that the Judeo-Christian god is the true god.
Are you arguing that the Bible is the only true Holy Book because God created it, or that the Bible is the only true Holy Book because it is the only one that is truly about God?
I asked you a simple question, to which you still haven't answered. Unless and until you do so, by providing sources to your claim, my point (any claim presented without evidence can be dismissed without evidence [Hitchens' Razor]) still stands.
“I've posted links”
Where did you post the link to the US Department of Justice statement labelling White Americans as domestic terrorists? I have yet to see it.
“& we've all seen it on the news.”
Can you show me a news article stating that the US DOJ labelled White Americans as domestic terrorists?
So far everything you said is fluff, and you’re still dodging my original question:
Do you have sources? Yes or no
Basically yes
He's saying that PE should not be made optional
How so?
Sparknotes version plz
https://images7.memedroid.com/images/UPLOADED775/5bf3735373712.jpeg
I'll do this debate in full, if you're interested
Cuz I like a challenge
GG EZ Dub
Meme appeal IS how you win presidency
Yes he’s already won California
Comrade
SLAP MY SALAMI, THIS SQUID'S A COMMIE
Thanks! And yes, it was morse code. What you CP'd, I think.
Oops
Dw though I flagged my vote for removal and will revote
ORO's Sandcastle Status:
STOMPED ON
“‘LIBTARD’ is a portmanteau of liberal and retard, conflating liberal ideology with developmental disability. Of course, dropping ad homs while advocating rational discourse is fairly obvious contradiction. PressF injects emotion into his claim in a clause dependent on the very sentence where he pretends to reason. I call PressF's claim self-disproved.”
Ad Homs refer to attacks made on a particular person’s character. As I have not made attacks against any particular person, by definition, it does not qualify as an ad hom.
Yeet Yeet
Take a seat
Listen to the beat and
Taste defeat
👌 fanks
Make it one week and I’ll be down
Well I’M not MOST DEBATERS.
I make my arguments based on IMPERATIVE FACTS and not LIBTARDED FEELINGS.
#facts
Two days for argument?
Bruh
Yeet Yeet
Taste defeat
-mah bootiful poem
YEEESSSSS!!!
This is the question that no one knew they wanted but that everyone needs.
No prob!
#11 is also something to always watch out for
In that case, the entire premise is opinionated.
Make sure to define all your terms from now on. Your opponent used the lack of definition for "better" as a way to steer the argument. Just something to take note of.
“So... let me get this straight:
10-0.999...=0.000...1, correct? Ok.
The "..." in 0.000...1 represents an infinite number of 0s before the one. Since there is an infinite number of zeros, there will always be a 0 after the previous one. Thus, the 1 at the end will never be reached. Since the 1 will never be reached, 0.000...1 is the same as 0.000..., which is simply 0.
Going back to our equation, we now have 10-0.999...=0
Adding 0.999... on both sides, we get 10 = 0.999...
QED”
Where you see a 10, replace it with a 1. I had big brain fart lol
No prob!
If you want me to show you the problem with the equation in this debate using this math, let me know ;)
Dang I feel like a Math Professor now lol
Since 0.999… is a repeating decimal, it can be expressed as an infinite sum:
0.999… = 0.9 + 0.09 + 0.009 + 0.0009 + …
S∞ = The entire infinite sum → S∞ = 0.999...
A = The first term → A = 0.9
R = The constant multiple (this can be derived from dividing the second term from the first)
R = (9/100)/(9/10) ← Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second.
R = (9/100)*(10/9)
R = 90/900 ← Simplify
R = 1/10
Since |1/10| < 1, we will use the second case of the equation. We will now plug in our numbers into the equation.
0.999… = 0.9/(1 - 1/10)
0.999… = (9/10)/(10/10 - 1/10)
0.999… = (9/10)/(9/10) ← Any number divided by itself is 1
0.999… = 1
QED
For the second case, remember that x^n will increase (gets closer to infinity) as n increases (gets closer to infinity). In this case, N = ∞. So, R^N = R^∞ = ∞. With that, we get this:
S∞ = A(1 - R^∞)/(1 - R), |R| > 1
S∞ = A(1 - ∞)/(some negative number)
S∞ = A(-∞ + 1)/(some negative number)
At this point, it is important to note that infinity plus/minus any finite number is still infinity. Similarly, negative infinity plus/minus any finite number is still negative infinity. Thus:
S∞ = A(-∞ + 1)/(some negative number)
S∞ = A(-∞)/(some negative number)
S∞ = -A(∞)/-(some positive number) ← remove the negative from the negative number
S∞ = A(∞)/(some positive number)
S∞ = ∞/(some positive number) ← Infinity/(Any finite number) = Infinity
S∞ = ∞
For this case, we can see that the infinite geometric series equation will equal to infinity. This leaves us with the third case:
S∞ = A(1 - R^∞)/(1 - R), |R| < 1
For the third case, remember that x^n will decrease (gets closer to zero) as n increases (gets closer to infinity). In this case, N = ∞. So, R^N = R^∞ = 0. With that, we get this:
S∞ = A(1 - R^∞)/(1 - R), |R| < 1
S∞ = A(1 - 0)/(1 - R)
S∞ = A(1)/(1 - R) ← Remember: A(1) = A
S∞ = A/(1 - R)
With that said, there are two possible cases for this equation:
S∞ = A(1 - R^∞)/(1 - R), |R| > 1 → S∞ = ∞
S∞ = A/(1 - R), |R| < 1
I will now use this equation to prove that 0.999… = 1.
Now that we have the equation for the geometric series, we need to establish what an infinite geometric series is.
An infinite geometric series is one which has an infinite amount of terms in its sequence. Since N is the number of terms in the sequence, and an infinite geometric series has an infinite amount of terms in its sequence (by definition), N = ∞. Let the entire infinite geometric series be S∞.
Retrieving the equation for the geometric series, and plugging ∞ into N, we get this:
S∞ = A(1 - R^∞)/(1 - R)
At this point, it is important to note the nature of exponents. Any number (x) to the power of something (n) will get larger or smaller when n gets larger, depending on what x is. The following will be the case for positive and negative numbers, so we can put an absolute value sign (| |) around x. This will make x an absolute value number. What an absolute value sign does is that it turns a negative number into a positive one. Positive numbers will not be affected.
If |x| > 1, then x^n will increase as n increases. (eg, 2^2 = 4, 2^3 = 8, etc.)
If |x| < 1, then x^n will decrease as n increases. (eg, (½)^2 = ¼, (½)^3 = ⅛)
If |x| = 1, then x^n will remain the same.
With this in mind, we can address the equation at hand. An absolute value number increasing can be thought of as getting closer to infinity, while an absolute value number decreasing can be thought of as getting closer to zero. There are three possible cases for the equation:
S∞ = A(1 - R^∞)/(1 - R), |R| = 1
S∞ = A(1 - R^∞)/(1 - R), |R| > 1
S∞ = A(1 - R^∞)/(1 - R), |R| < 1
For the first case, we get this:
S∞ = A(1 - R^∞)/(1 - R), |R| = 1 → S∞ = A(1 - 1^∞)/(1 - 1) → S∞ = A(1 - 1^∞)/0
There are two problems with this. Firstly, we cannot divide by zero. Secondly, 1^∞ is indeterminate. Thus, we can outright eliminate this case.
[continued in next comment]
Since any number subtracted by itself is 0, we are left with:
S - SR = A - AR^N + 0 + 0 + 0 + ...(all of the terms in here are 0)... + 0 + 0
Simplified, we get this:
S(1) - SR = A(1) - AR^N
Since any number times 1 is itself, the number can be restated as itself times 1.
We can factor out the S and the A to get this:
S(1 - R) = A(1 - R^N)
Since we are trying to find S, we divide both sides by 1-R. As a result, we get:
S = A(1 - R^N)/(1 - R)
And this is the equation for the geometric series.
[Continued in next comment]
This goes on for the amount of terms that are in the sequence. Let's call the number of terms in the sequence "N", and the sequence as a whole as "s".(Again, what we call these by is completely arbitrary)
So, a geometric sequence can be shown as:
s = [A, AR, AR^2, ..., AR^(N-3), AR^(N-2), AR^(N-1)]
The reason that the final number in the sequence is AR^(N-1) and not AR^N is because on the 1st term, N = 0 (R^0 = 1, and A*1 is simply A).
Since a series is simply the sum of a sequence, a geometric series (let's call the variable for the entire series "S") can be shown as:
S = A + AR + AR^2 + ... + AR^(N-3) + AR^(N-2) + AR^(N-1)
This is our first equation for the proof.
If we multiply R by both sides, we get:
SR = R * [A + AR + AR^2 + ... + AR^(N-3) + AR^(N-2) + AR^(N-1)]
Remember that the exponent simply shows the number of times that something is being multiplied by. So, if we multiply something raised to the power of n (n is simply the exponent of that something) by that same something, then it is the same as adding 1 to n. With that in mind, we get:
SR = AR + AR^2 + AR^3 + ... + AR^(N-3+1) + AR^(N-2+1) + AR^(N-1+1)
= AR + AR^2 + AR^3 + ... + AR^(N-2) + AR^(N-1) + AR^N
This is our second equation.
Now, we subtract the second equation from the first (First equation is S, second equation is SR):
S - SR = [A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1)] - [AR + AR^2 + AR^3 + ... + AR^(N-2) + AR^(N-1) + AR^N]
The minus sign in front of the first series can be distributed into each number within the series:
S - SR = [A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1)] + [-AR - AR^2 - AR^3 - ... - AR^(N-2) - AR^(N-1) - AR^N]
We can get rid of the brackets:
S - SR = A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1) - AR - AR^2 - AR^3 - ... - AR^(N-2) - AR^(N-1) - AR^N
Then we can rearrange the equation:
S - SR = A - AR^N + AR - AR + AR^2 - AR^2 + AR^3 - AR^3 + ... - ... + AR^(N-2) - AR^(N-2) + AR^(N-1) - AR^(N-1)
[continued in next comment]
I'll show you the infinite geometric series proof that I forgot to explain in full last time.
Before I go on to proving that 0.999...=1, I need to establish what an infinite geometric series is. Before I can do that, we need to establish what a geometric series is. Before that, I need to define what a series is.
A series is simply the sum of a sequence. A sequence is just a set of numbers. For example, 15 is the sum of the series [1,2,3,4,5] (the square brackets are just the formal way of showing a set). If you add all the numbers of the sequence together, you will get 15.
A geometric series is a series in which each subsequent number in the series' sequence is multiplied by some constant. There will be the first number in the sequence, and then that number will be multiplied by something to get the next number, which will be multiplied again by that something to get the next, and so on and so forth. Let's define the first number as "A" and the constant which "A" is being multiplied by as "R". These variables are completely arbitrary, and we can call them whatever we want. Given these two variables, we can now define our geometric sequence in a mathematical sense.
The first number of the sequence is just A, by definition.
The second is A*R, since a geometric sequence is, by definition, in which each subsequent number in it is the previous one multiplied by a constant. This can be rewritten as AR.
The third is AR*R, since we are multiplying the last term by the same constant (R). Since R*R can be rewritten as R^2, this term can be rewritten as AR^2. One important thing to remember is that the exponent of something simply denotes how many times that variable is multiplied by.
The fourth term is AR^3, since the previous term is AR^2 and we are multiplying it by R again.
[continued in next comment]
So... let me get this straight:
10-0.999...=0.000...1, correct? Ok.
The "..." in 0.000...1 represents an infinite number of 0s before the one. Since there is an infinite number of zeros, there will always be a 0 after the previous one. Thus, the 1 at the end will never be reached. Since the 1 will never be reached, 0.000...1 is the same as 0.000..., which is simply 0.
Going back to our equation, we now have 10-0.999...=0
Adding 0.999... on both sides, we get 10 = 0.999...
QED
YOU ARE NOW A WHITE DOMESTIC TERRORIST!!!
-mairj23
As I stated before, if ...9999.0 is infinite (as seems to be the case with the ... to the left of the 9s), then you’re left with ♾-10♾. 10♾ is still infinity, so you’re left with ♾- ♾. This is indeterminate, meaning that there is no definite or definable variable. You can’t calculate an answer from an indeterminate equation. Since your entire argument hinges on the basis that ...9999.0-...9990.0=9 (essentially, ♾-♾=9), and this is shown not to be the case, the argument falls apart.
Good on you for debating from multiple perspectives, but the error I pointed out (if ...9999.0 is infinite) pretty much defeats the entire argument.
And by that, I mean explaining what ...9999.0 means
In that case, I guess I’m gonna have to look at the specific arguments presented. However, this debate suffers the same problem as the last (it’s about the same topic), that several “holy” texts are automatically eliminated by the presumption that the Judeo-Christian god is the true god.
I’m just saying, you might wanna clarify your terms before proceeding
If ...9999.0 is suppose to be infinity (as the “...” to the left of the 9s implies), then you end up with ♾-♾, which is indeterminate.
Question:
What number is ...9999.0?
Are you arguing that the Bible is the only true Holy Book because God created it, or that the Bible is the only true Holy Book because it is the only one that is truly about God?
I asked you a simple question, to which you still haven't answered. Unless and until you do so, by providing sources to your claim, my point (any claim presented without evidence can be dismissed without evidence [Hitchens' Razor]) still stands.
“I've posted links”
Where did you post the link to the US Department of Justice statement labelling White Americans as domestic terrorists? I have yet to see it.
“& we've all seen it on the news.”
Can you show me a news article stating that the US DOJ labelled White Americans as domestic terrorists?
So far everything you said is fluff, and you’re still dodging my original question:
Do you have sources? Yes or no
Do you have sources? Yes or no