This guy must be an ethical man, a man of pure vision in the societal, and personal spheres of his life .His love for all learners gleams in his style of teaching Mathematics, reaching the edges of chosen topics in real time. Stay Blessed.
I don’t know how to describe it but most teachers I had taught math in a very cold and sterile way. You explain things in such a warm compassionate manner. Even if I don’t grasp the concept right away, I don’t feel like an imbecile after watching. Really glad I found your channel.
The proof is easy without induction.if a is odd then a-1, and a+1 are even numbers and since they are consecutive even numbers then one of them is divisible by 4. So that their multiplication is divosible by 8. And their multiplication is (a-1)(a+1) =a²-1
10:33 remark that 4(k^2 + k) = 4*k(k+1) ---> k and k+1 are 2 consecutive number ----> one of them is even ---> a^2 - 1 is divisible by 8. My demo is : Let a = 2k+ 1 an odd number ---> a^2-1 = (a+1)(a-1) = (2k+1 +1)(2k+1 -1) = (2k+2)(2k) = 4*(k+1)*k --->divisible by 8
Oh, fascinating dive into the intricacies of odd numbers and their mystical relationship with the number 8! Reminds me of the profound lyrics from "We Need Cash" - "I bought an electric guitar, I rented a shiny car." Just like odd numbers, life's equations seem to dance in mysterious patterns. By the way, did anyone notice that elusive secret cat in the background? A subtle nod to the feline chorus singing "nya ;3" in the song. Life's oddities and hidden references, what a peculiar symphony! Keep on calculating the enigma of existence!
A quick way to prove that a^2 - 1 is divisible by 8, for every odd number a, is :- Let a = 2k + 1, where k is an integer. Let x = a^2 - 1 => x = (a+1)(a-1) = (2k + 1 + 1)(2k + 1 - 1) = 2(k+1) * 2k = 4 * k(k+1) => x = 8 * k(k+1)/2 Now, we know that n(n+1)/2 is just the sum of all natural numbers from 1 to n (the sum being another natural number), and since all natural numbers are also integers, we can say that the sum is also an integer (note : it is of no importance whether this integer is positive, negative, or even zero). So, x = 8 * [integer] => x / 8 = [integer] => (a^2 - 1)/8 = [integer] => (a^2 - 1) mod 8 = 0 Hence, a^2 - 1 is divisible by 8, for any and all odd numbers (here, 'a'). P.S. :- Food for thought : How would we prove that n(n+1)/2 is necessarily an integer for an integral n, given that n(n+1)/2 is a natural number for a natural n? Or, is this assumption wrong in itself?
If a²-1 is divisible by 8, Then, (a²-1)/8=n =>a²-1=8n -(i) (n belongs to set of integers) {(a+2)²-1}/8=k =>(a+2)²-1=8k -(ii) (k belongs to set of integers) From (i) & (ii), It is clear that (a²-1) and {(a+2)²-1} are even numbers as they both equate to a certain integer being multiplied by 8. So, =>(a+2)²-1-(a²-1)=2x =>a²+4a+4-1-a²+1=2x =>4a+4=2x =>a=(x/2)+1 Here,For every value of x for which a is an integer is where a is odd. As,a is odd when a²-1 is divisible by 8 then conversely a²-1 is divisible by 8 when a is odd.
I noticed a sequence I'm familiar within this proof. If A=2K+1 then (A^2-1)/8 = K*(K+1)/2 It's the triangle numbers. It even holds up if K is a negative number.
Start off with an integer a Double it Add one (this is k) Square it Minus 1 And divide it by 8 Following this you get a 2a 2a + 1 4a^2 + 4a + 1 4a(a + 1) and now it should be clear that when we divide it by 8 to get a(a + 1)/2 either a or a + 1 is even because a is an integer showing that we can divide one of those by 2 and multiply the remaining numbers to form a new integer.
Once we have noted that a² - 1 = 4k² + 4 k, we can, taking 4k as common, write a² - 1 = 4k (k+1) = 4 . k(k+1). Notice that k(k+1) is the product of two consecutive numbers, the product of which is always an even number, say 2t , where t is any even or odd number. Then our a² - 1 becomes 8t, that is divisible by 8.
If a = 1, then a^2 - 1 = 0, which is divisible by 8. Assume that k is an odd number such that, if a = k, then a^2 - 1 is divisible by 8. Consider a = k+2. So, a^2 - 1 = (k + 2)^2 - 1 = k^2 + 4k + 4 - 1 = k^2 - 1 + 4(k+1). Since k^2 - 1 is divisible by 8, we only need to show that 4(k+1) is divisible by 8. Note that k is odd, so k+1 is even. 4 times an even number is a multiple of 8, so 4(k+1) is a multiple of 8. Since it's true for a = 1, then by what we just argued, it's true for a = 3. Therefore, it's true for a = 5. etc...
Newton, I choose a² -1 = (a + 1)(a - 1). So, a + 1 = 2k + 2, a - 1 = 2k, and a² -1 = (a + 1)(a - 1) = (2k + 2)*2k = 4k² + 4k. After, putting 4k in evidence, we have 4k(k + 1). So, if k is odd, k + 1 is even, and if k is even, k + 1 is odd. Any "Even times Odd" resullts in a Even. Hence, we have 4k(k + 1) = 4.Even = 4.2t = 8t, divisible by 8
Or just note that (a^2-1) = (a-1)(a+1), so if a=2n+1 (ie an odd number), then (a^2-1)=2n.2.(n+1) which always contains 2x2x2 whether n is odd or even => divisible by 8
Yes, that was also my own though. And you even only have to check 1,3,5,7 which I would call trivial. The "problem" is you have to give an argument that modulo arithmetic works (i.e. a*b (mod n) = a (mod n) * b(mod n) and a+b (mod n) = a (mod n) + b (mod n)). Not hard, but not self-evident.
a = 2k + 1. a² - 1 = (a + 1)(a - 1) = (2k + 1 + 1)(2k + 1 - 1) = (2k + 2)(2k) = 4k(k + 1). If k is even, then 4k(k + 1) = 4(2q)(k + 1) = 8q(k + 1), which is divisible by 8. If k is odd, then k + 1 is even, so 4k(k + 1) = 4k(2q) = 8kq, which is divisible by 8.
The second proof you did in the video is similar to a proof I did in a university project. I was deriving the formula for something in Knot Theory. Essentially, the formula needed to output an integer, and I actually had two parts of it. One of the parts hypothetically produced double of what I was needing (so I multiplied that by 1/2) but I didn't have a guarantee that it was even. The only guarantee I found was if I could show (x^2 - x) was always even, so I did something similar to what you did. I later realised that you could factor (x^2 - x) into x(x-1). If the input to that function is an integer, then x(x-1), or in your case x(x+1) is the product of two consecutive integers. This would prove its even, as I think there's a theorem out there where for any n consecutive integers, you are guaranteed that one of them will be divisible by n. When n = 2, one of the 2 consecutive integers is divisible by 2, meaning whatever product of the consecutive integers would have 2 as a factor, making it even. The way we both did it is fine, but that was another way it could be proven.
My solution was simply: a² - 1 = (a + 1)(a - 1) Since a is odd, (a + 1) and (a - 1) are even. In any two consecutive even numbers, one of them will be a multiple of 4. Therefore, their product is divisible by 2 x 4 = 8.
and it seems like if you were to do it by induction would it be (2(k+1)+1)^2 - 1 that you need to check. so, (2k+3)^2 - 1. It's pretty nice since it follows basically the same process in factoring out an 8 as this one. (2k+3)^2 - 1 = 4k^2+12k+8 = 4(k^2+3k+2) if k is even, m is some other interger: = 4(even + even + 2); (even numbers squared are even; 2k1(2k2+1) = 4k1k2+2k1 = 2(2k1k2+k1), so even times odd is even) = 4(2m1+2m2+2) = 8(m1+m2+1) if odd, q for some other interger: = 4(odd+odd+2); (odd nums squared are odd, odd*odd = odd) = 4(2q1+1+2q2+1+2) = 8(q1+q2+2)
clarification, the error lies in the assumption that P(K+1) = P(K) + P(1). You can assume that you survive day K and if you can show you survive day K+1 then it would have to be the case that you can live forever. The reason your joke proof doesn't work is because the function cannot be broken the way you claim, not because you can't assume K.
Grouping 4k(k+1) in 4(k^2+k), it easily turns out that k(k+1) is always even, either k is odd or even (if k is even, k+1 is odd and vice versa, thus k*(k+1) is always even)
Let a = 8k + r, where k and r are integers and 0 ≤ r ≤ 7. But since a is odd, then r = 1, 3, 5, or 7. So, a² = (8k + r)² = 8(8k² + 2kr) + r² So a² - 1 is divisible by 8 if r² - 1 is divisble by 8. We note that the only possible values of r² - 1 are 0, 8, 24, 48 which are all divisible by 8. So a² - 1 is divisible by 8 if a is odd.
Hi good day Prime Newtons. I love your videos and i have a question. How does this proof implies that a^2-1 is divisible by 8, for values of a which are odd only? Is it that an odd number squared is always odd?
if a is odd, then a^2 is odd. but the -1 makes it even. the proof in the video is to write a = 2k + 1 where k is an integer. a = 7 can be written as 2*3+1. a = 13 can be written as 2*6+1. a = -5 can be written as 2*(-3)+1. hope that helps clarify a few things
Well: (5:20) Just by adjusting from an undisputed "survive every day" to a probability of surviving a day (maybe even in dependence from age and other factors), we make a valid scientific base for survivability analysis. Nonetheless, IF you INDEED survive every daily unit, you INDEED live forever. That is absolutely true - even if that "if" is not at all undisputed in real life. But IF it would hold, the "life forever" holds as well. You are allowed to make assumptions in your models. You just have to track what you do for what purpose. I don't count that as an "illogical" construct. It just is an example for the responsibility of a scientist for judgement of his data source and the judgement of the conclusions he draws from his data. You just showed in your model that infinite life is possible by infinitely surviving every day - completely analogue to counting to infinity or to dividing into infinitely small differentials - which you may instantiate as a model for certain examinations, DESPITE never being able to factually do so in real life.
since a is odd, a = 2n + 1 where n is an integer so a^2 - 1 = (a - 1)(a + 1) = 2n(2n + 2) = 4n(n + 1) but n(n + 1) is divisible by 2 because either n or n + 1 is even. therefore 4n(n + 1) is divisible by 8
Also kinda cool, btw: for all n >= 0: 2^(2(2n+1)) + 1 = 0 mod 5 Easy proof by induction too. In binary, they are these numbers: 101, 1000001, 10000000001, 100000000000001, etc.
it is guaranteed of course it is that you can live forever John 11:25 Jesus said unto her, I AM the resurrection and the life; he that believes in me, though he is dead, yet shall he live; John 11:26 and whosoever lives and believes in me shall never die. Believest thou this?
I have a better way to (maybe not) live forever. I. You have the ability to live for 1 day. II. Let's assume you have the ability to live for k days. What about k + 1 days ? well + is commutative, so k+1 = 1 +k We know you have the ability to live for 1 day because of I. it remains k days, and thanks to II. we knoe youhave the ability to live for k days. therefor you can live for 1+k days Conclusion you will live forever.
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This guy must be an ethical man, a man of pure vision in the societal, and personal spheres of his life .His love for all learners gleams in his style of teaching Mathematics, reaching the edges of chosen topics in real time. Stay Blessed.
Thank you. You sound very wise.
You should also hear him sing with the Bob Marley tribute band, the Junior Whalers.
(2k+1)^2 -1 = 2k * 2(k+1) = 4k*(k+1)
It doesn't matter if k is even or odd, one of k or k+1 is divisible by 2 (and so is their product)
I knew I should that but somehow, I went the way of the wasteful 😂
I arrived at that logic too.
That's what I came up with in less than a minute.
@@PrimeNewtons
It's never a waste of effort to solve a problem a different way.
Using mathematics to light up smiles is a such a blessing. Thank you Coach! 🙏
I don’t know how to describe it but most teachers I had taught math in a very cold and sterile way. You explain things in such a warm compassionate manner. Even if I don’t grasp the concept right away, I don’t feel like an imbecile after watching. Really glad I found your channel.
I found it funny that he said "let's get into the video ☺️" almost halfway through the video
The proof is easy without induction.if a is odd then a-1, and a+1 are even numbers and since they are consecutive even numbers then one of them is divisible by 4. So that their multiplication is divosible by 8. And their multiplication is (a-1)(a+1) =a²-1
Smart!
That was my proof.
That's an awesome way of doing it!
Loveable and lovely. Enjoyable every second of the presentation
10:33 remark that 4(k^2 + k) = 4*k(k+1) ---> k and k+1 are 2 consecutive number ----> one of them is even --->
a^2 - 1 is divisible by 8.
My demo is :
Let a = 2k+ 1 an odd number ---> a^2-1 = (a+1)(a-1) = (2k+1 +1)(2k+1 -1) = (2k+2)(2k) = 4*(k+1)*k --->divisible by 8
Oh, fascinating dive into the intricacies of odd numbers and their mystical relationship with the number 8! Reminds me of the profound lyrics from "We Need Cash" - "I bought an electric guitar, I rented a shiny car." Just like odd numbers, life's equations seem to dance in mysterious patterns. By the way, did anyone notice that elusive secret cat in the background? A subtle nod to the feline chorus singing "nya ;3" in the song. Life's oddities and hidden references, what a peculiar symphony! Keep on calculating the enigma of existence!
I never expected a Tally Hall reference in a math teacher's comments.
A quick way to prove that a^2 - 1 is divisible by 8, for every odd number a, is :-
Let a = 2k + 1, where k is an integer.
Let x = a^2 - 1
=> x = (a+1)(a-1) = (2k + 1 + 1)(2k + 1 - 1) = 2(k+1) * 2k = 4 * k(k+1)
=> x = 8 * k(k+1)/2
Now, we know that n(n+1)/2 is just the sum of all natural numbers from 1 to n (the sum being another natural number), and since all natural numbers are also integers, we can say that the sum is also an integer (note : it is of no importance whether this integer is positive, negative, or even zero).
So, x = 8 * [integer]
=> x / 8 = [integer] => (a^2 - 1)/8 = [integer]
=> (a^2 - 1) mod 8 = 0
Hence, a^2 - 1 is divisible by 8, for any and all odd numbers (here, 'a').
P.S. :-
Food for thought : How would we prove that n(n+1)/2 is necessarily an integer for an integral n, given that n(n+1)/2 is a natural number for a natural n? Or, is this assumption wrong in itself?
Presentation is awesome
That was a fun problem thanks mate 😊 I like these simpler ones I can understand and can get engaged in. ✅
If a²-1 is divisible by 8,
Then,
(a²-1)/8=n
=>a²-1=8n -(i)
(n belongs to set of integers)
{(a+2)²-1}/8=k
=>(a+2)²-1=8k -(ii)
(k belongs to set of integers)
From (i) & (ii),
It is clear that (a²-1) and {(a+2)²-1} are even numbers as they both equate to a certain integer being multiplied by 8.
So,
=>(a+2)²-1-(a²-1)=2x
=>a²+4a+4-1-a²+1=2x
=>4a+4=2x
=>a=(x/2)+1
Here,For every value of x for which a is an integer is where a is odd.
As,a is odd when a²-1 is divisible by 8 then conversely a²-1 is divisible by 8 when a is odd.
I noticed a sequence I'm familiar within this proof. If A=2K+1 then (A^2-1)/8 = K*(K+1)/2
It's the triangle numbers. It even holds up if K is a negative number.
Start off with an integer a
Double it
Add one (this is k)
Square it
Minus 1
And divide it by 8
Following this you get
a
2a
2a + 1
4a^2 + 4a + 1
4a(a + 1) and now it should be clear that when we divide it by 8 to get a(a + 1)/2 either a or a + 1 is even because a is an integer showing that we can divide one of those by 2 and multiply the remaining numbers to form a new integer.
You can do the second proof with induction, which seems useful in the context of the first proof, showing how it is done properly.
…we conclude that life is non-linear 😊
Once we have noted that a² - 1 = 4k² + 4 k, we can, taking 4k as common, write a² - 1 = 4k (k+1) = 4 . k(k+1). Notice that k(k+1) is the product of two consecutive numbers, the product of which is always an even number, say 2t , where t is any even or odd number. Then our a² - 1 becomes 8t, that is divisible by 8.
If a = 1, then a^2 - 1 = 0, which is divisible by 8.
Assume that k is an odd number such that, if a = k, then a^2 - 1 is divisible by 8. Consider a = k+2. So, a^2 - 1 = (k + 2)^2 - 1 = k^2 + 4k + 4 - 1 = k^2 - 1 + 4(k+1). Since k^2 - 1 is divisible by 8, we only need to show that 4(k+1) is divisible by 8. Note that k is odd, so k+1 is even. 4 times an even number is a multiple of 8, so 4(k+1) is a multiple of 8.
Since it's true for a = 1, then by what we just argued, it's true for a = 3. Therefore, it's true for a = 5. etc...
Newton, I choose a² -1 = (a + 1)(a - 1). So, a + 1 = 2k + 2, a - 1 = 2k, and a² -1 = (a + 1)(a - 1) = (2k + 2)*2k = 4k² + 4k. After, putting 4k in evidence, we have 4k(k + 1). So, if k is odd, k + 1 is even, and if k is even, k + 1 is odd. Any "Even times Odd" resullts in a Even. Hence, we have 4k(k + 1) = 4.Even = 4.2t = 8t, divisible by 8
Or just note that (a^2-1) = (a-1)(a+1), so if a=2n+1 (ie an odd number), then (a^2-1)=2n.2.(n+1) which always contains 2x2x2 whether n is odd or even => divisible by 8
With mathematical induction I might have the chance to live forever.
It is sufficient to study the remainders of a modulo 8 which are 0,1,2,3,4,5,6,7
Yes, that was also my own though. And you even only have to check 1,3,5,7 which I would call trivial.
The "problem" is you have to give an argument that modulo arithmetic works (i.e. a*b (mod n) = a (mod n) * b(mod n) and a+b (mod n) = a (mod n) + b (mod n)). Not hard, but not self-evident.
a = 2k + 1.
a² - 1 = (a + 1)(a - 1) = (2k + 1 + 1)(2k + 1 - 1) = (2k + 2)(2k) = 4k(k + 1).
If k is even, then 4k(k + 1) = 4(2q)(k + 1) = 8q(k + 1), which is divisible by 8.
If k is odd, then k + 1 is even, so 4k(k + 1) = 4k(2q) = 8kq, which is divisible by 8.
The second proof you did in the video is similar to a proof I did in a university project. I was deriving the formula for something in Knot Theory. Essentially, the formula needed to output an integer, and I actually had two parts of it. One of the parts hypothetically produced double of what I was needing (so I multiplied that by 1/2) but I didn't have a guarantee that it was even. The only guarantee I found was if I could show (x^2 - x) was always even, so I did something similar to what you did. I later realised that you could factor (x^2 - x) into x(x-1). If the input to that function is an integer, then x(x-1), or in your case x(x+1) is the product of two consecutive integers. This would prove its even, as I think there's a theorem out there where for any n consecutive integers, you are guaranteed that one of them will be divisible by n. When n = 2, one of the 2 consecutive integers is divisible by 2, meaning whatever product of the consecutive integers would have 2 as a factor, making it even. The way we both did it is fine, but that was another way it could be proven.
My solution was simply:
a² - 1 = (a + 1)(a - 1)
Since a is odd, (a + 1) and (a - 1) are even.
In any two consecutive even numbers, one of them will be a multiple of 4.
Therefore, their product is divisible by 2 x 4 = 8.
and it seems like if you were to do it by induction would it be (2(k+1)+1)^2 - 1 that you need to check.
so, (2k+3)^2 - 1.
It's pretty nice since it follows basically the same process in factoring out an 8 as this one.
(2k+3)^2 - 1 = 4k^2+12k+8
= 4(k^2+3k+2)
if k is even, m is some other interger:
= 4(even + even + 2); (even numbers squared are even; 2k1(2k2+1) = 4k1k2+2k1 = 2(2k1k2+k1), so even times odd is even)
= 4(2m1+2m2+2)
= 8(m1+m2+1)
if odd, q for some other interger:
= 4(odd+odd+2); (odd nums squared are odd, odd*odd = odd)
= 4(2q1+1+2q2+1+2)
= 8(q1+q2+2)
a is odd --> a=2k±1
a²-1=(2k±1)²-1
=4k²±4k
=4k(k±1)
Note that 2 | (k±1)
Hence 8 | a²-1
clarification, the error lies in the assumption that P(K+1) = P(K) + P(1). You can assume that you survive day K and if you can show you survive day K+1 then it would have to be the case that you can live forever. The reason your joke proof doesn't work is because the function cannot be broken the way you claim, not because you can't assume K.
You could also just factor k^2+k into k(k+1) one of them will be even number.
Grouping 4k(k+1) in 4(k^2+k), it easily turns out that k(k+1) is always even, either k is odd or even (if k is even, k+1 is odd and vice versa, thus k*(k+1) is always even)
U got me addicted to maths
Can you consider doing some basic Topology - I never grasped that subject!
Ha, that's a big ask. I'll see if I can get some friends to fill in for me
It obvious that for any number n, n²= 1,4,0 mod 8, 1 if n is odd and 4,0 when n is evenm
Since a is odd a²= 1 mod 8
Thus a²-1 is divisible by 8.
Let a = 8k + r, where k and r are integers and
0 ≤ r ≤ 7.
But since a is odd, then r = 1, 3, 5, or 7.
So, a² = (8k + r)² = 8(8k² + 2kr) + r²
So a² - 1 is divisible by 8 if r² - 1 is divisble by 8.
We note that the only possible values of
r² - 1 are 0, 8, 24, 48 which are all divisible by 8.
So a² - 1 is divisible by 8 if a is odd.
Beautiful
Can you prove that: n^3 - n is divisible by 6, where n is a natural number?
Hi good day Prime Newtons. I love your videos and i have a question. How does this proof implies that a^2-1 is divisible by 8, for values of a which are odd only? Is it that an odd number squared is always odd?
if a is odd, then a^2 is odd. but the -1 makes it even.
the proof in the video is to write a = 2k + 1 where k is an integer.
a = 7 can be written as 2*3+1.
a = 13 can be written as 2*6+1.
a = -5 can be written as 2*(-3)+1.
hope that helps clarify a few things
@@nanamacapagal8342 Hey there , thank you
Well: (5:20) Just by adjusting from an undisputed "survive every day" to a probability of surviving a day (maybe even in dependence from age and other factors), we make a valid scientific base for survivability analysis. Nonetheless, IF you INDEED survive every daily unit, you INDEED live forever. That is absolutely true - even if that "if" is not at all undisputed in real life. But IF it would hold, the "life forever" holds as well.
You are allowed to make assumptions in your models. You just have to track what you do for what purpose. I don't count that as an "illogical" construct. It just is an example for the responsibility of a scientist for judgement of his data source and the judgement of the conclusions he draws from his data. You just showed in your model that infinite life is possible by infinitely surviving every day - completely analogue to counting to infinity or to dividing into infinitely small differentials - which you may instantiate as a model for certain examinations, DESPITE never being able to factually do so in real life.
Can you please make a video about log and natural log
That's a wide topic. Is this algebra or calculus?
Both
Maybe calculus
since a is odd,
a = 2n + 1 where n is an integer
so a^2 - 1 = (a - 1)(a + 1) = 2n(2n + 2)
= 4n(n + 1)
but n(n + 1) is divisible by 2 because either n or n + 1 is even.
therefore 4n(n + 1) is divisible by 8
Sir I need your help please
2k+1 , k>1. As the first + odd is 3
I think I found a counter-example
Can you do the Cbrt of i
we're gonna need some euler's formula e^(iT) = cos(T) + isin(T)
get ready
Assume my head is frozen on day k.
what if you were a stillborn
But Souls are eternal as fire-fly.
Simply, P(1) would not be true.
Then you wouldn't have been watching this video.
Also kinda cool, btw:
for all n >= 0:
2^(2(2n+1)) + 1 = 0 mod 5
Easy proof by induction too.
In binary, they are these numbers:
101,
1000001,
10000000001,
100000000000001, etc.
The answer is: [Life = Yeshua]
I meant k> 0
it is guaranteed of course it is that you can live forever John 11:25 Jesus said unto her, I AM the resurrection and the life; he that believes in me, though he is dead, yet shall he live;
John 11:26 and whosoever lives and believes in me shall never die. Believest thou this?
Hello
I have a better way to (maybe not) live forever.
I. You have the ability to live for 1 day.
II. Let's assume you have the ability to live for k days.
What about k + 1 days ?
well + is commutative, so k+1 = 1 +k
We know you have the ability to live for 1 day because of I.
it remains k days,
and thanks to II. we knoe youhave the ability to live for k days.
therefor you can live for 1+k days
Conclusion you will live forever.