Maths trivia

[ZVI]

Heared that before; is there proof for that statement? It seems to me quite "forced" (don't know a better english word for this), especially when looking to 7 and 9.

Hi Erik,
Agree with you that it seems more amusing than historically proved
Vladimir

 

[pgc01]

One I always found very interesting

We know that a rational to the power of another rational can have a rational result, like

2<SUP>2</SUP> = 4

We also know that an irrational to the power of a rational can have a rational result, like

sqrt(2)<SUP>2</SUP> = 2

But can an irrational to the power of an irrational have a rational result?

Suprisingly the answer is yes, it has been proven that such pairs of irrational numbers exist!

 

[Joe4]

But can an irrational to the power of an irrational have a rational result?

Suprisingly the answer is yes, it has been proven that such pairs of irrational numbers exist!

Pedro,

That is very interesting! Do you have any examples, or links to an example? I would love to see one.

 

[pgc01]

Joe

I don't remember any example but I remember the proof, and it's very simple. It's also funny because it proves that such a pair exists but we don't know which.

Using the known property of power to power (a^b)^c = a^(b*c)

(sqrt(2)^sqrt(2))^sqrt(2) = sqrt(2)^(sqrt(2)*sqrt(2)) = sqrt(2)^2 = 2

Using just the 2 extremes of the expression

(sqrt(2)^sqrt(2))^sqrt(2) = 2

One of 2 things must be true:

- either sqrt(2)^sqrt(2) is rational and then our pair is (sqrt(2),sqrt(2))

- or sqrt(2)^sqrt(2) is irrational and then our pair is (sqrt(2)^sqrt(2),sqrt(2))

either way there's one pair (irrational,irrational) that satisfies irrational^irrational = rational

I'm sure there are works done on this but I never looked for any, I just remember this because the first time I saw it I thought it was an interesting result.

 

[baitmaster]

amaze your friends with your ability to divide by 7 - I used this trick at a restaurant and everyone was shocked that I could split the bill to infinte decimal places...

1/7 = 0.142857 recurring
2/7 = 0.285714 " "
etc

series shifts and repeats consistently, and is easily remembered by the multiples of 7 within it, and the multiples of 14 that start it...

 

[baitmaster]

To multiply 2 numbers using only 2s:

1) Write both numbers.
2) Halve one number, and double the other. Ignore remainders
3) Write these under the originals, and repeat step until you reach 1
4) Add up the numbers on the "doubled" side, that correspond to any odd values on the "halved" side, including the original numbers if appropriate

The result is the multiple.

If you stick a 1 next to the numbers you included, and 0 against those you didn't, then read bottom to top, thats the answer in binary

 

[Andrew Fergus]

amaze your friends with your ability to divide by 7 - I used this trick at a restaurant and everyone was shocked that I could split the bill to infinte decimal places...

1/7 = 0.142857 recurring
2/7 = 0.285714 " "
etc

series shifts and repeats consistently, and is easily remembered by the multiples of 7 within it, and the multiples of 14 that start it...

Cool. This got me thinking......

Check out the sequence for 13ths - there are two different repeating series, and note the pattern as to which series is the relevant series.

And 17ths seems to be like the series for 7ths - just a little harder to remember.