30 responses so far ↓

• 1 Daniel // Oct 30, 2007 at 9:03 pm

  537
  x 23
  ———
  12351

• 2 Samira // Oct 30, 2007 at 11:40 pm

  Hi Daniel,
  Thanks for posting!
  One is not a prime number. The only numbers you can use are the prime digits 2, 3, 5, and 7. Keep trying, don’t give up!

• 3 ali(vancouver) // Oct 31, 2007 at 4:20 am

  so did you do it by “trial and error”??where’s the trick part?

• 4 bardia // Oct 31, 2007 at 11:38 pm

  buncha nerds!

• 5 ali(vancouver) // Nov 1, 2007 at 3:00 am

  haha n you always check the comments!!

• 6 Samira // Nov 1, 2007 at 10:28 am

  Hey guys…remember…ONE is NOT a prime number. You can only use the numbers 2, 3, 5, and 7.

• 7 Samira // Nov 1, 2007 at 10:31 am

  sigh…if only i had a nickel for every time i was called a nerd!

• 8 Ali - (York, UK) // Nov 1, 2007 at 12:49 pm

  Hi Samira,

  First of all, the answer to your question is:
  7 7 5
  X 3 3
  ————-
  2 3 2 5
  2 3 2 5
  ————-
  2 5 5 7 5

  then, unlike some of my friends who put comments here, I find it interesting. To me solving maths problems is a brain exercise and love to do it. To be honest, I thought I am so good at maths, but since watching your shows I changed my mind; I am not that good at maths I thought.
  Also, I have a question for you; why do you always have an apple in your hand at the end of the show?
  Finally, I wanted to thank you for forcing my brain to start doing some exercises again, and ask you to carry on with show with more and more energy.

  khaste nabashi

• 9 Rob W // Nov 2, 2007 at 8:27 am

  teachers love apples!

• 10 Hesham // Nov 2, 2007 at 12:27 pm

  Hi Samira,

  I wanted to ask you to have your own problems in the program. Everybody can find these question and their solutions on the net (like [1]).

  I hope soon you can come up with your own designed problems, to make the program interesting.

  [1] http://www.mscs.dal.ca/~rjn/probofweek/solution1.html

• 11 Samira // Nov 2, 2007 at 2:25 pm

  Hi Hesham,
  The problems I give have played a big role in Mathematics, so of course with enough “google-ing” you’d be able to find them on the internet. But…if you’re trying to solve the problem for yourself, why would you search for the solution? Don’t ruin the fun for yourself! If you’re interested in solving the problem, try it out yourself, then search for it to see if you are correct. Since these problems have played a big role, I think they’re more important and fun to give, rather than problems I make up. But I’ll try to squeeze some in for you

• 12 bardia // Nov 2, 2007 at 8:21 pm

  Baba Bikhial beshin torokhodaaaa

• 13 Sia // Nov 3, 2007 at 12:48 pm

  Great job, love the little math question, I have a question for you, can you show how
  1 + 1 = 5 or N
  N can be any number
  I remember this problem in Calculas 1 or 2
  this is fun.
  again you all doing a great job at bebin tv.

• 14 Ali - (York, UK) // Nov 3, 2007 at 5:53 pm

  ok, I think I get this one;

  0 = 0 + 0 + 0 + …
  0 = (1-1)+(1-1)+(1-1)+…
  0 = (1-1+1-1+1-1+…)
  0 = 1-(1-1+1-1+1-1+1-1+…)
  0 = 1-(0+0+0+0+0+…)
  0 = 1 - 0
  0 = 1 (1)

  from (1) => 0+1=1+1 => 1=2 (2)
  from (2) => 1+1=2+1 => 2=3 (3)
  from (3) => 2+1=3+1 => 3=4 (4)
  from (2),(3),(4) => 1=4 (5)
  and,
  from (5) => 1+1=4+1
  => 1+1=5

  obviously you can do this for any number N.
  furthermore, you are now able to destroy the hole world by the fact that 0=1 which destroys the logic foundation as it implies that
  “existence equals naught”.

• 15 bardia // Nov 4, 2007 at 11:47 am

  be ghoraan hamatoon ghati darin
  1+1=5 what the hell ?!?!?!? 1+1=4+1 are you all out of your mindsssss??!!?! plzzzzzz stop this madness…. all of you do u hear me?? no more take this show off it is killing me no more NO MORE

• 16 Ali - (York, UK) // Nov 4, 2007 at 4:43 pm

  If you don’t want to see this math MADNESS, what the hell are you doing here in the blog?!

• 17 Samira // Nov 4, 2007 at 7:37 pm

  oh bardia, the blog just wouldn’t be the same without you…

• 18 Samira // Nov 4, 2007 at 7:44 pm

  Thanks for the challenge, Sia, and nice proof Ali! I love that problem! Your proof reminds me of another false proof of 1 < 0. Have you seen that one?

• 19 Ali - (York, UK) // Nov 4, 2007 at 11:40 pm

  If it’s not the same, then no!
  Put it here please, I’d be glad to see that!
  and, Samira I didn’t really mean that bardia shouldn’t come to blog at all, I just wanted to make him calm down and think. I’m sure if he starts thinking about maths problems, then he’d be better than anybody else.

• 20 Samira // Nov 5, 2007 at 10:40 am

  I agree, Ali. I think Bardia secretly loves mathematics!

  Anyway, here’s the proof….

  Let x < 1

  take the log of both sides…

  ln(x) < ln(1)

  ln(x) < 0

  now divide both sides by ln(x) and you get…

  1 < 0

• 21 Kamran // Nov 5, 2007 at 11:52 am

  Samira man donbale yeh khanum e khoshgel keh math beduneh migardam, shoma soragh darin?

• 22 Ali - (York, UK) // Nov 5, 2007 at 4:18 pm

  wow! That was owesome!
  I didn’t know this one, but now I know. Thanks Samira.
  Give me more, if possible!
  again thanks

• 23 Vafa // Nov 25, 2007 at 2:34 am

  First of all, Ali your proof is not right. Why because you did not consider the fundemental laws of real number which is technically called ” the axioms of real numbers”. If you do anything by the axiom of real numbers, then your proof will be wrong. In fact the real numbers are made first by accepting the axioms of the real numbers. Lets see what are the axioms of the real numbers:

  1- $\forall x,\,y\in \mathbb{R},\,x+y\in\mathbb{R}$ and $xy\in\mathbb{R}$
  2- $\forall x,\,y\in\mathbb{R},\,\left(x+y\right)+z=x+\left(y+z\right)$ (Associative Law)
  3- $\forall x,\,y\in\mathbb{R},\,x+y=y+x$ (commutative law)
  4- $\exists 0\in\mathbb{R},\,x+0=0+x=x,\,\forall x\in\mathbb{R}$
  5- $\forall x\in\mathbb{R},\,\exists\left(-x\right)\in\mathbb{R},\,x+\left(-x\right)=\left(-x\right)+x=0

  6- $\forall x,\,y,\,z\in\mathbb{R},\, \left(xy\right)z=x\left(yz\right)$ (associative law)

  7- $\forall x,\,y\in\mathbb{R},\,xy=yx$ (commutative law)

  8- $\exists1\in\mathbb{R},\,1\times x=x\times1=x\,\forall x\in\mathbb{R}$

  9- $\forall x\in\mathbb{R},\,x\neq0,\,\exists x^{-1}\in\mathbb{R},\,x\left(x^{-1}\right)=\left(x^{-1}\right)x=1$

  10- $\forall x,\,y,\,z\in\mathbb{R},\, x\left(y+z\right)=xy+xz$ (distributive law)

  We also should be considering the order axioms for real numbers since they are ordered:

  1- $\forall x,\,y\in\mathbb{R},\,xy$

  2-$\begin{cases}x<y\\y<z\end{cases}\Rightarrow x<z$

  3- $x<y\Rightarrow x+z<y+z,\,\forall x,\,y,\,z\in\mathbb{R}$

  4- $\begin{cases}x0\end{cases}\Rightarrow xz>yz$

• 24 Vafa // Nov 25, 2007 at 2:59 am

  Samira, there is a problem with your proof as follow:

  ok, suppose we let $x<1$, assuming $x\in\mathbb{R}$, if we write a few $x$ values, then we have $\left\{0,-1,-2,…\right\}\in\mathbb{Z}$. Ok the problem is here

  If you take the logarithm of both of $x<1$, you should have $\log|x|<\log1$ and not $\log\,x<log1$. Why? because the logarithmic function is only defined for the values bigger than $0$. so weshoul be letting $|x|<1$ and not $x<1$.

  let’s continue
  $\log|x|<\log1$

  $\log|x|0$
  by dividing both sides of equation 1.1, I will get
  $10$
  That means $11$. but this contradicts our assumption because we assumed $|x|<1$, therefore we can not accept that $1<0$.

  2- $\log|x|0$ but here we have $01$, which is absolutely true.

  And this completes that $1>0$ not less than $0$.

• 25 Vafa // Nov 25, 2007 at 3:29 am

  Ok, I am sorry for the messy writing.

  Samira, please look at

  http://mathspassion.wordpress.com/2007/11/25/is-10/

  Ali, please look at

  http://mathspassion.wordpress.com/2007/11/25/the-real-numbers-axiom/

• 26 Vafa // Nov 25, 2007 at 9:37 am

  Samira, you said

  “Hi all! Hope you like this episode of Deriving School!! Hey…here’s a neat math fact. The number of decimal numbers between zero and one is infinity, right? And the number of counting numbers (that is, 1 2 3…) is infinity too, right? Well, did you know that the infinity of the decimal numbers is larger than the infinity of counting numbers?! That means there are more decimal numbers between zero and one than counting numbers from one to infinity! Who knew you can compare infinities!!”

  But infinity is not a number so it does not follow any axioms and that means you can not compare, add, subtract and erc. Infinity is just something that we define to be undefined. I ask one question and let’s see who can answer it:

  What is the difference between 0/0 and a/0, where a can be a complex number?

• 27 Samira // Nov 25, 2007 at 7:59 pm

  Hi Vafa, the proofs ARE in fact incorrect. They are false proofs, and the problem is to find where the fault is. Good job on finding them! There are more of those proofs too!

  As for the infinity comparison. Here is the proof. It’s pretty well-known.
  http://scidiv.bcc.ctc.edu/Math/diag.html
  You can find more papers on it too. It’s a beautiful proof. Enjoy!

  ~Samira

• 28 Vafa // Nov 26, 2007 at 8:40 am

  Hi, Samira. I looked at that link. That proof is correct .

  Basically, there is more than one infinity - and some are bigger than
  others. But we can not compare two infinity of the same type because infinity it is not a number and does not follow any axioms therefore it is not ordered. i.e. one infinity can not be bigger than the other one if they are all the same type.

  To put it simply, two sets have the same size if you can find a one-to-
  one and onto function between them. [That is, find a way of pairing
  off the elements, so each of the elements in one set is matched to
  precisely one element in the other set].

  This works just as well with infinite sets as with finite sets. But it
  does give surprising results. For instance, the number of fractions is
  exactly the same as the number of natural numbers.
  You can match them up as follows:

  1 -1
  2 0
  3 1
  4 -2
  5 -3/2
  6 -1/2
  7 1/2
  8 3/2
  9 2
  10 -3
  11 -8/3
  12 -7/3
  13 -5/3
  14 -4/3
  15 -2/3
  16 -1/3
  17 1/3
  etc.
  [Here, the fractions are listed by first listing those between -1 and
  1 with denominator <=1, then those between -2 and 2 with denominator
  <=2, then between -3 and 3 with denominator <=3, leaving out anything
  which has occurred earlier. Clearly, each fraction will occur exactly
  once in this list.]

  Notice this is surprising. There are an infinite number of fractions
  between 0 and 1, and infinite number between 1 and 2 etc, but here
  \inf * \inf is just the same as the number of natural numbers (which
  is the smallest possible infinity).

  However, there are bigger infinities. For instance, the number of real
  numbers is greater than the number of natural numbers.

  And for any set S, the number of subsets of S is always greater than
  the number of elements in S. [So if you start with an infinite set,
  repeating this over and over lets you build bigger and bigger
  inifinities].

  One of the unanswered questions in mathematics is whether there is an
  infinity BETWEEN the number of natural numbers and the number of real
  numbers. No-one knows yet - and some suspect that this is a question
  that actually can’t be answered. but recently Grisha Perelman who won Mathematics field medal in 2006 and refused it (the guy who had Proof of poincare conjecture as one of the result of his works) is currently researching on it. I did my honours research on number theory and thought alot about this last bit and I was progressing but at the end I faced a copntradiction which destroyed all of my researches.

• 29 Vafa // Nov 26, 2007 at 9:15 pm

  If you have two finite sets and want to know which one has more elements
  then you can pair off elements in one with elements in the other
  and if there are elements left over in one set when you’ve exhausted
  the other
  then the set with elements left over has more elements.

  For example, if A = {North, East, South, West}
  and B = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
  and you pair off
  North Sunday, East Monday, South Tuesday, West
  Wednesday,
  then you’ve exhausted A and still have things left over in B,
  so B has more elements.

  In the late 1800s a mathematician named Cantor showed that we can do
  pretty much the same thing with infinite sets. If there is a way
  to pair off two sets so that nothing is left over in either set,
  then we say the two sets have the same cardinality
  (a fancy word for “number of elements”).
  If no matter how we try to pair up the two sets
  the first one always has elements left over,
  we say the first set has greater cardinality than the second.

  So, for example, the odd numbers {1, 3, 5, 7, …}
  and the even numbers {2, 4, 6, 8, …} have the same cardinality
  because we can pair each odd number n with the even number n + 1
  (1 2, 3 4, 5 6, etc.)
  and nothing is left over in either set.

  But the even numbers also have the same cardinality
  as the set of all positive integers {1, 2, 3, 4, …}
  because we can pair each positive integer n with the even number 2n
  (1 2, 2 4, 3 6, etc)
  and nothing is left over in either set.

  If we write N for {1, 2, 3, 4, …}
  and N^2 for the set of all pairs of positive integers
  (that is, N^2 is the set whose elements are all the ordered pairs (a, b)
  where a and b are positive integers)
  then we can list the elements of N^2 as follows:
  (1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1), (1, 4), ….
  that is, in increasing order of a + b, and, among pairs that have
  the same value of a + b, list them in increasing order of a
  (so (3, 13) will come after (5, 10) because 3 + 13 > 5 + 10,
  but (3, 13) will come before (5, 11) because 3 + 13 = 5 + 11 and 3 < 5).
  Now we can pair each element in our list of N^2
  with the corresponding element in the list of N:
  (1, 1) 1, (1, 2) 2, (2, 1) 3, (1, 3) 4, (2, 2) 5,
  etc.,
  and nothing will be left out in either set,
  so N has the same cardinality as N^2.

  A similar argument can be used to show that if Q is the set of
  fractions,
  that is, Q is everything of the form a / b where a and b are positive
  integers
  and a / b is in lowest terms,
  then the cardinality of Q is the same as that of N.

  Now let I be the set of decimal numbers between 0 and 1.
  The argument on the website shows that no matter how you try to pair off
  the elements of I with the elements of N = {1, 2, 3, 4, …}
  there will always be elements of I left over.
  That proves that the cardinality of I is greater than the cardinality
  of N.

  Summary:

  When we say “infinity is not a number,” we mean it is not a decimal
  number
  like 3 or 2/7 or square root of 2 or pi, and it doesn’t satisfy the
  axioms
  the decimal numbers satisfy.

  Despite that, we can compare two infinite sets and say which, if
  either, is bigger.

• 30 Vafa // Nov 27, 2007 at 3:53 am

  I would like to apologise all people who are haters of maths and are its number one enemy and also think I am talking crap. In fact, Maths is my only motivation for being alive…! because I find nothing attractive more than maths.

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