Question

Whoa, I didn't know this.
So I know that 2 x 2 = 2 + 2, and I thought that the number 2 was the only number that had such property where multiplying itself is the same as adding by itself, but there are other numbers with this property!

Answer 1

\[\large \sqrt{3}\times \sqrt{3}\times \sqrt{3}=\sqrt{3}+\sqrt{3}+\sqrt{3}\]

Answer 2

That equation is false, Steve. Multiplication and addition are VERY different operations.

Answer 3

@mathmale Check again, it is true.

Answer 4

\[\sqrt{3} \cdot \sqrt{3}=3 \\ \text{ so } \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{3}=3 \sqrt{3} \\ \text{ and } \sqrt{3}+\sqrt{3}+\sqrt{3}=3 \sqrt{3}\]

Answer 5

The burden of proof is on your shoulders.

\[\sqrt{3}\sqrt{3}=3;~but ~\sqrt{3}+\sqrt{3}=2\sqrt{3}\]

Answer 6

I think he is just noticing something neat

Answer 7

Better to keep an open mind, Steve, instead of refuting what a retired professor tells you about a math problem.

Answer 8

@mathmale That wasn't what I was saying.

Answer 9

and yes being that it is neat is an opinion :p

Answer 10

Steve, if you are convinced that you are right, then defend your statement.

Answer 11

@freckles just proved my statement and proved you wrong!

Answer 12

OK, Steve, have it your way. I'm here to help students who want to learn, not those who want primarily to argue. Take care, over and out.

Answer 13

I don't know what you are not getting mathmale, I am trying to say that root 3 multiplied by itself three times is the same as adding root 3 three times.

Answer 14

I never wanted to start an argument in the first place mathmale.

Answer 15

AND, I do want to learn.

Answer 16

Perhaps Freckles will have more patience with you than I do at the moment.

Answer 17

we should look at finding more values a with the fact that
\[(a)^n=\sum_{i=1}^{n} a\]

Answer 18

where n is any integer that we choose

Answer 19

http://www.wolframalpha.com/input/?i=sqrt(3)*sqrt(3)*sqrt(3)+%3D+sqrt(3)+%2B+sqrt(3)+%2B+sqrt(3)

Answer 20

guess i could have just wrote a^n=n*a

Answer 21

freckles: can you prove that your statement is true for all n?

2*2 = 4; 2^2=4; YES

3*3 = 9; 3^3 = 27; NO

Answer 22

To me, it is NOT a property!! because it is true at P(2) ,
Like 2*2 = 2+2
from P(3) 2*2*2 \(\neq 2+2+2\)
It has nothing to be called "property"

Answer 23

I didn't say a^n=n*a is true for all n

Answer 24

I said let's try to find values of a such that a^n=n*a is true for some value of n we choose

Answer 25

I agree with Loser66. He just won (in spite of his name).

Answer 26

Freckles: I'd go along with that latest statement of yours.
How would you go about finding n for this situation?

Answer 27

for example steve stated another one which was 2*2=2+2

Answer 28

That was my original statement if you go back up :p

Answer 29

I agree with you and Steven in this particular example. One example doesn't make a rule, does it?

Answer 30

no one claimed a^n=a*n was a rule

Answer 31

for all a and n

Answer 32

OK. How would you wrap up this discussion? What have we discovered?

Answer 33

To me the whole point of this discussion in which you are trying to derail is trying to find values of a such that a^n=n*a is true for some value of n we choose

Answer 34

right now that is what I'm trying to do is find a solution to that equation

Answer 35

besides the ones given which were
a=2 , n=2

a=sqrt(3) , n=3

Answer 36

You may need to choose a particular value for "a," such as "a=2" and then work at solving the resulting equation for n.

Answer 37

So using frecle's equation, I found another solution when n = 4
\[\large (2^\frac{ 3 }{ 4 })^4=2^\frac{ 3 }{ 4 }*4\]

Answer 38

that one doesn't work
that one side is 8 while the other isn't

Answer 39

Steve: can you prove that using a calculator?

Answer 40

Waiit, hold on I wrote it wrong. The exponent is supposed to be 2/3

Answer 41

Freckles essentially did that (prove the "equation" false by doing actual calculations).

Answer 42

\[\large (2^\frac{ 2 }{ 3 })^4=2^\frac{ 2 }{ 3 }*4\]

Answer 43

oh nice job steve

Answer 44

and I think I found one for n=5

Answer 45

oops n=6
\[a=6^\frac{1}{5} \text{ when } n=6\]

Answer 46

yep that seems an easy way
just choosing values of n which results in a polynomial to solve in terms of a
I guess we could also use non-real values

Answer 47

ohhh, so does this work for any n value?

Answer 48

Steve: That's a good question, and one that I hope you can answer yourself.
My initial response to y our post here was NO! I didn't stop to check whether there might be exceptions to that. 2*2 = 2^2 is true and is thus an exception.

Answer 49

Make up your own examples and then look for a pattern.

Answer 50

yep
\[a^n=an \\ a^n-an=0 \\ a(a^{n-1}-n)=0 \\ a=0 \text{ or } a^{n-1}-n=0 \\ a=0 \text{ or } a^{n-1}=n \\ a=0 \text{ or } a=n^{\frac{1}{n-1}}\]

Answer 51

This is soooo beautiful.

Answer 52

Hmm this is really cool post.

So is this what we figured out?\[\large\rm \sum_{i=1}^n n^{\frac{1}{n-1}}\quad=\quad\prod_{i=1}^n n^{\frac{1}{n-1}}\]Does that look correct maybe?

Answer 53

One thing: how did you go from \[a^{n-1}=n\]to\[a=n ^{\frac{ 1 }{ n-1 }}\]

Answer 54

To undo (n-1) exponent, we take the (n-1)th root of each side, ya?

Answer 55

ohhh, ok I see!

Answer 56

Is \(k^{\frac{1}{k-1}}\) an integer for more than k=2?

Answer 57

Steve, to your initial question \(\color{black}{0\times 0=0+0}\).
(\(\color{black}{2}\) is not the only number like this)
--------------------------------------

I guess maybe what you noticed is ... \(\tiny \\[0.5em]\)
\(\color{black}{2(2^{1/1})=(2^{1/1})^2}\)
\(\color{black}{3(3^{1/2})=(3^{1/2})^3}\)
\(\color{black}{4(4^{1/3})=(4^{1/3})^4}\)
\(\color{black}{5(5^{1/4})=(5^{1/4})^5}\)

and so .... \(\color{black}{k\left[k^{\frac{1}{k-1} }\right]=\left[k^{\frac{1}{k-1} }\right]^k}\) for all \(\color{black}{ k \in \mathbb{N}}\).

Answer 58

and actually I like that conclusion, Steve:)

Answer 59

\(\color{red}{\bf Correction:}\) This is true for all integers n\(\ge\)2.
(1/0 is undefined)

Answer 60

\(\color{black}{k\left[k^{\frac{1}{k-1} }\right]=\left[k^{\frac{1}{k-1} }\right]^k }\)

\(\color{black}{k=\left[k^{\frac{1}{k-1} }\right]^{k-1} }\)

\(\color{black}{k=k }\)

Answer 61

this is a brief proof of what you noticed:)

Answer 62

So, we can also assert, by the same token,
\(\large \color{black}{k\left[k^{\frac{1}{k-\alpha}}\right]=\left[k^{\frac{1}{k-\alpha}}\right]^{k+1-\alpha } }\)

Answer 63

but it all comes out from k=k.

Answer 64

Solomon Actually Proved it O_O