Question

i need someone to walk me step by step on a Pythagorean theorem i am having trouble with it and just a little bit of extra help

Answer 1

Zed

Answer 2

hey tinkerbell :)

Answer 3

help

Answer 4

sure, just reading the file

Answer 5

ok

Answer 6

hello satellite :)

Answer 7

brb zed im going to get a drink

Answer 8

man that is confusing. pick two numbers, m and n, one even one odd. lets say i pick m = 4, n 1
then take
\[m^2-n^2=4^2-1^2=15\] also
\[2mn=2\times 4\times 1=8\] and finally
\[m^2+n^2=4^2+1^2=17\] those three number,
\[15,8,17\] will be a pythagorean triple. that is
\[15^2+8^2=17^2\]

Answer 9

watchmath!! long time no see. i just mentioned you the other day

Answer 10

saying i missed your interesting questions/puzzles

Answer 11

yes, you are a legend now here :). Good job!

Answer 12

legend smegend. when i get to 10,000 medals i am going to have a whiskey and call it a day

Answer 13

hope to see you more regularly.

Answer 14

:). I will think some interesting problem for you

Answer 15

ok im here sorry had to grab a bottle of water i have been sick but im back now

Answer 16

ok, but right now i am puzzling over how to make truth tables interesting because they are boring me to death. if you have any good puzzles or exercises involving them, let me know.

Answer 17

@tinkerbell, i wrote out a method for finding triples for you above. the worksheet you sent is rather vague (method on line) etc

Answer 18

i have been down ill with the flu really bad

Answer 19

i can walk you through another one if you like

Answer 20

tinkerbell do you need to use different methods each time?

Answer 21

It can be proved as the 2 dimensional case of the Parseval equality I think! Functional Analysis is goood. If not, then just remember this: The square on the hypotenuse is equal to the sum of the squares on the other two sides!

Answer 22

this is the only one i see , a^2+b^2 = c^2.

Answer 23

maybe it is not clear what you are looking for. not any three numbers a, b and c, but three whole numbers a, b and c with
\[a^2+b^2=c^2\]

Answer 24

thats what i see

Answer 25

did you open the link on the worksheet?

Answer 26

my assignment

Answer 27

this link
http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples

Answer 28

that confused me

Answer 29

i see. it looks like your worksheet is asking for 5 different triples. i wrote one above. would you like to do another one?

Answer 30

can you rewrite it and explain as you go

Answer 31

ok we can try the first one
If a is odd, then b = a2/2 − 1/2 and c = b + 1
If a is even, then b = a2/4 − 1 and c = b + 2

pick an odd number

Answer 32

7

Answer 33

ok then to find "b" we compute
\[b=\frac{7^2}{2}-\frac{1}{2}\] what do you get?

Answer 34

there any other way we can do this with out fractions

Answer 35

will 7^2 =49 so it would be 49/2-1/2

Answer 36

this is a fraction but don't fret
\[\frac{7^2}{2}-\frac{1}{2}=\frac{49-1}{2}=\frac{48}{2}=24\]

Answer 37

a nice whole number
so
\[a=7,b=24\] and now
\[c=b+1=24+1=25\] and that is your "triple"
\[7,24,25\]

Answer 38

and you are supposed to check that
\[7^2+24^2=25^2\] which you can do with a calculator

Answer 39

slow down

Answer 40

let me know if you have any questions about what i wrote

Answer 41

no i have no questions so far let me type it into the template real qick

Answer 42

ok that gives me 625

Answer 43

good so we have one. now we can try another one

Answer 44

yes please

Answer 45

Ready when you are

Answer 46

ok now we try a different method

Answer 47

pick two numbers m and n, where one is even and one is odd. make them not too big

Answer 48

88 and 13

Answer 49

whoa nice and small!

Answer 50

you did not say that

Answer 51

we want to make this easy, not hard. pick small numbers

Answer 52

11 and 17

Answer 53

ok i will pick them, one even and the other odd. i pick 5 and 2
now we compute
\[5^2-2^2=25-4=21\]

Answer 54

then i take
\[2\times 5\times 2=20\]
and finally
\[5^2+2^2=25+4=29\] and the three numbers
\[21,20,29\] are also a triple and you can check that
\[21^2+20^2=29^2\]

Answer 55

ok

Answer 56

now we have two. do you have to use a different method for each one?

Answer 57

yes i would like if it is possible

Answer 58

i will le let you know when im done entering it in the template

Answer 59

well maybe we can do one more, but if i were you i would use the last method again with two different numbers

Answer 60

ok hang on im 3^2-8^2=9-64=55

Answer 61

am i right so far

Answer 62

you are on the right track, but you should make it
\[8^2-3^2=64-9=55\] so you don't get negative numbers

Answer 63

so you picked 8 and 3,and now you have 55
next two numbers will be
\[2\times 8\times 3\] and
\[8^2+3^2\]

Answer 64

2x 8x 3=48

Answer 65

8^2 + 3^2 is to check right

Answer 66

no that is the third number, the long side

Answer 67

73

Answer 68

the three numbers are
\[55,48,73\] and the check is
\[55^2+48^2=73^2\]

Answer 69

ok be right with you

Answer 70

unfortunately i have to run. but you have 3 so far, and it looks like you know what you are doing, so i think you should be in good shape

Answer 71

i just have to rememer to put the bigger number in front