Square root of 2 - infinitely long

[TonyMcC]

Many years ago I was shown a simple check for multiplication.

The check can indicate a definite mistake; but cannot indicate definite accuracy.

The check is performed by adding the individual digits of each part of the calculation. If this results in a number with more than one digit then add the digits of the new number and repeat as necessary until you have a single digit representing each part of the calculation.

Example:-

 

39 leads to 3+9=12, and 1+2=3

58 leads to 5+8=13, and 1+3=4

 

39 x 58=2262 leads to 2+2+6+2=12, and 1+2=3

 

If the single digit representing 39 (i.e.3) is multiplied by the single digit representing 58 (i.e.4) we get 12 which leads to 1+2=3. The fact that this agrees with the single digit representing the answer indicates that the calculation might be correct, but more importantly if the single digit representing the answer had been any other number the calculation would have been performed incorrectly.

 

Another example:-

 

58764 (5+8+7+6+4=30 ; 3+0=3)

3271 (3+2+7+1=13 ; 1+3=4)

------------

192217044 (1+9+2+2+1+7+0+4+4=30 ; 3+0=3)

 

Performing the check, 3 x 4=12 and 1+2=3

since this agrees with the answers single digit no error has been detected.

 

Although this description might seem complicated it is really very simple and can be done "in your head" with very little practice.

 

Examining the square root 0f 2

_________________________

 

Think of the calculation A x A= 2

So A must be the square root of 2

 

Someone suggests that the square root of 2 is 1.4

If they are correct then 14 x 14 =200

 

So using the test

14 (1+4=5)

14 ( 1+4=5) and 5x5 = 25, and 2+5=7

----

200 (2+0+0=2)

Since "7" and "2" do not have the same value this calculation must be incorrect.

 

Someone then suggests 1.41

 

So using the test

141 (1+4+1=6)

141 (1+4+1=6) and 6x6=36, and 3+6=9

---------

20000 (2+0+0+0+0 =2)

Since"9" is not "2" this calculation is incorrect.

 

My calculator gives the square root of 2 as 1.414213562

therefore 1414213562^2=2 x 10^18.

If you do the test the two relevant digits are "4" and "2" indicating an error.

 

The fact is you can add digits beyond 1.414213562 all the way to infinity without getting a "correct" indication!

 

This argument holds good for many (probably most) numbers.

 

Perhaps any whole number falls in to one of two categories. Perhaps it either has an easily found whole number square root or it is an infinitely long decimal number. Perhaps this is true of any number base you might use (octal, hexadecimal etc.)??

Edited February 22, 2010 by TonyMcC

[Bignose]

[math]\sqrt{2}[/math] is an irrational number. It will have an infinitely long non-repeating non-pattern forming decimal representation. All irrational numbers have this property. Like [math]\sqrt{3}[/math] or [math]\pi[/math] or [math]e[/math].

 

http://en.wikipedia.org/wiki/Irrational_number

[TonyMcC]

Thanks Maths Expert. Actually the fact that Root 2 is irrational was the reason I chose it to "test my test".

Not all infinitely long decimal representations are irrational and I suppose some have quite long repeating patterns. What I am wondering is :-

a) If a whole number does not have a whole number square root does it always result in an infinitely long square root?

b) If it does result in an infinitely long square root does that indicate probable irrationality for that square root?

 

Please note - I am wondering, not claiming. I would value your (or anybody's) opinion.

 

I don't want to end up with silly logic similar to:-

Animals have 4 legs, dogs have 4 legs so all animals are dogs!

Edited February 23, 2010 by TonyMcC

[khaled]

irrational numbers, with continues non-repeating decimal representation,

we round the value for use,

[ewmon]

Something interesting occurs with this check method when used on square roots of whole numbers. The “product” amounts to the whole number itself, and the product remains constant because any extra decimal places are filled with zeroes. The multipliers are the square root and, thus, are identical. Thus, there are nine possible check numbers (1², 2², 3², … 7², 8², and 9²) resulting in 1, 4, 9, … 49, 64, and 81. These, in turn, simplify into 1, 4, 9, 7, 7, 9, 4, 1, and 9. All together, the check numbers can only equal 1, 4, 7 or 9.

 

Thus, the square root of 2, regardless of the number of decimal places, will never check as “valid” because the check number can never equal 2. The same holds true for 3, 5, 6, 8, 11, 12, 14, 15, 17, 20, 21, 23, 24, 26, 29, etc, but not for 1, 4, 7, 9, 10, 13, 16, 18, 19, 22, 25, 27, 28, etc.

 

Notice that the set of numbers that can never check as valid don’t contain squares of whole numbers, so I’ll call them N numbers (for Never a square). The squares are all contained in the second set, so I’ll call them S numbers (for Sometimes a square).

 

Now it gets interesting! Analyzing the set of whole positive numbers, we find their check values. Some of these values equal 1, 4, 7 or 9, and others do not. Let’s look at the prime numbers.

 

Twin primes (P and P+2 are primes), except for 3 and 5, seem to follow the pattern of an N number followed by an S number. That is, the P prime is an N number, and the P+2 prime is an S number. This holds true for the 35 pairs of twin primes below 1,000 and for the somewhat large twins 3,557 and 3,559.

 

Cousin primes (P and P+4 are primes), except for 3 and 7, seem to follow the opposite pattern of S then N, which holds true for the 41 sets of cousin primes below 1,000.

 

Sexy prime pairs (P and P+6 are primes) seem to follow the pattern of both N or both S, which holds true for the 46 pairs of sexy primes below 500.

 

Sexy prime triplets (P and P+6 and P+12 are primes) seem to follow the pattern of all N or all S, which holds true for the 21 sets of triplets below 1,000.

 

Sexy prime quadruplets (P, P+6, P+12 and P+18 are primes) also seem to follow the same pattern of all N or all S, which holds true for the 7 sets of quadruplets below 1,000.

 

Cuban primes of either form seem to be all S numbers.

 

Eisenstein primes seem to be all N numbers.

 

I could go on, but it's getting late...

[TonyMcC]

Glad you found my check method interesting Ewmon. However I think you may have missed the main point; the number "2" is incidental - the METHOD works for all single digit numbers and usually works for multidigit numbers.

Example showing the square root of 7 is infinitely long:-

A X A =7 (making A the square root of 7)

my calculator says the square root of 7 is 2.645751311, thus it says 2645751311 X 2645751311 = 7000000000000000000.

(in fact my calculator says it is correct (7E18). However this must be incorrect as using the check:-

35 X 35 = 7

8 X 8 =7

64=7

1=7

Since adding digits can never make the LHS = (the square root of 7) X (the square root of 7), you will end up producing an infinitely long answer to the square root of 7.

By the way the method works for cube roots etc.. Have fun.

(Any bearing on Fermat's last theorem?!)

Edited April 25, 2010 by TonyMcC

[the tree]

the METHOD works for all single digit numbers and usually works for multidigit numbers.

Example showing the square root of 7 is infinitely long

The word really is irrational, "infinitely long" does kind of apply to the decimal expansions all real numbers (1=1.00000... etc). It can be easily proven that the square root of any prime is irrational and subsequently from there that an integer must either be a perfect square or have an irrational square root.

[TonyMcC]

Thank you "the tree". If you look at my earlier entries for this subject you will see that you seem to have answered my earlier supposition. However I still am not really happy with your explanation concerning irrationality. As I understand it the square root of 2 is accepted as irrational not because its decimal expansion extends to infinity but for a completely different reason. It is easily proved that when expressed as a fraction and in its most simplified form both parts of the fraction must be even numbers. That really is irrational! Is it a proven fact that all square roots of whole numbers are either whole numbers or irrational? How is the irrationality proved?

[sammy28]

Proof by contradiction.

 

any number that isnt rational must be irrational, therefore start by making the contradiction that its rational:

 

[math]\sqrt{2}=\frac{p}{q}[/math] where p, q are integers and q not equal to 0 with [math]\frac{p}{q}[/math] in lowest terms.

 

[math]\frac{p^2}{q^2}=2[/math]

 

so

 

[math]p^2=2q^2[/math]

 

so p is even. since p^2 is even, p must also be even (even x even = even) so you can write:

 

[math]p=2m[/math] where m is an integer

 

substituting this into the previous statement

 

[math]4m^2=2q^2[/math]

 

[math]2m^2=q^2[/math]

 

So q^2 is even and so q must be even.

 

But if both p and q are even then [math]\frac{p}{q}[/math] is not in lowest terms.

 

Since there is a contradiction, the original assumption must be false. That is [math]\sqrt{2}[/mat] cannot be rational.

 

therefore [math]\sqrt{2}[/math] is irrational

[the tree]

As I understand it the square root of 2 is accepted as irrational not because its decimal expansion extends to infinity but for a completely different reason. It is easily proved that when expressed as a fraction and in its most simplified form both parts of the fraction must be even numbers. That really is irrational!

To be clear with the definitions, a number is rational if it can be expressed as the ratio of two integers and irrational if it is not rational. This is equivalent to having a finite decimal expansion, no points for working out why.

 

Is it a proven fact that all square roots of whole numbers are either whole numbers or irrational? How is the irrationality proved?

Take a rational number that isn't an integer, call it [imath]\frac{p}{q}[/imath] so that [imath]p[/imath] and [imath]q[/imath] have no common factors.

 

[imath]p^2[/imath] and [imath]q^2[/imath] will clearly then have no common factors so [imath]\left( \frac{p}{q} \right)^2[/imath] will be a rational non integer.

 

So no rational non-integer has an integer square, as a corollary no integer can have a rational non-integer square root.