- 10 Jul '08 00:41When doing mental math for the sake of amusment, I lke to square numbers. When doing larger numbers starting with 3 digit squares I like to use the technique illustrated below

436^2 = (472)(400) + 36^2

It vastly simplifies the problem for mental calculation., but what is the proof for this working for all natural, or whole numbers?

somthing like this?

2 digit square

(10a + b)^2 = 100a^2 + 20ab + b^2

100a^2 + 20ab + b^2 - 10a ( 10a + 2b ) = b^2

and it does, but is this a proof? And would there need to be some other type of generalization? I'm interested in how specific situations are recognized that give notion to extend the proof to cover more obscure information.

thanks - 10 Jul '08 07:22You wish to find x^2, x is a positive integer

]Let k be the smallest positive integer such that x^2-k=a*10^b, where a is an element of {1, 2... 9} and b is a positive integer.

Your method calculates (x-k)(x+k)+k^2. Expanding this we get x^2-k^2+k^2 which is equal to x^2.

It seemed that you already had everything you needed to prove this. Instead of proving it for 2 digit numbers, then 3 digits, etc. just prove it for all integers in general! - 17 Jul '08 10:38

(10a+b)^2 =?= (10a+2b)(10a)+b^2*Originally posted by joe shmo***When doing mental math for the sake of amusment, I lke to square numbers. When doing larger numbers starting with 3 digit squares I like to use the technique illustrated below**

436^2 = (472)(400) + 36^2

It vastly simplifies the problem for mental calculation., but what is the proof for this working for all natural, or whole numbers?

somthing like t ...[text shortened]... recognized that give notion to extend the proof to cover more obscure information.

thanks

100a^2+20ab+b^2

(10a)(10a+2b)+b^2

Equality proven. What you need to do is change one and only one side until it is identical with the other. You're not supposed to manipulate the equality by doing stuff to both sides. - 17 Jul '08 12:52

I don't know much of anything about proofs, just toying with the concept over summer...It should be a few semesters until I get involed in proofs.....*Originally posted by AThousandYoung***(10a+b)^2 =?= (10a+2b)(10a)+b^2**

100a^2+20ab+b^2

(10a)(10a+2b)+b^2

Equality proven. What you need to do is change one and only one side until it is identical with the other. You're not supposed to manipulate the equality by doing stuff to both sides.

Anyhow, I was just trying to show that what is left after simplifying the problem in that particular manner that all that remains is the conjugate's square ( I dont think that conjugate is the right terminology, I mean the number added and subtracted for purpose of simplification) - 17 Jul '08 18:08

There's no "supposed to". The proof is either solid or not. Doing stuff to both sides is fine, as long as you're careful about it (e.g. everything has to be reversible - you can't prove 1 = -1 by squaring both side).*Originally posted by AThousandYoung***You're not supposed to manipulate the equality by doing stuff to both sides.** - 17 Jul '08 20:36

(x-k)(x+k) + k^2 = x^2 - k^2 +k^2 = x^2*Originally posted by Dejection***You wish to find x^2, x is a positive integer**

]Let k be the smallest positive integer such that x^2-k=a*10^b, where a is an element of {1, 2... 9} and b is a positive integer.

Your method calculates (x-k)(x+k)+k^2. Expanding this we get x^2-k^2+k^2 which is equal to x^2.

It seemed that you already had everything you needed to prove this. Instead of proving it for 2 digit numbers, then 3 digits, etc. just prove it for all integers in general!

Since all we are doing are expanding terms, and no squaring or square rooting is taking place, all steps are reversible.