Question

Prove that if m and n are perfect squares then m + n +2 √ mn is also a perfect square. Using direct proof.

Answer 1

so what do you get to assume in a direct proof?

Answer 2

hint: p -> q

Answer 3

I assume that n is a perfect square

Answer 4

m and n are perfect squares

Answer 5

so if that's what we get to assume, it will be helpful if we can find a way to rewrite m and n

Answer 6

for example, an even number can be written as 2k, and an odd number can be written as 2k + 1 by definition. any ideas on how we can represent a perfect square?

Answer 7

Here's a hint:

A perfect square is a number that can be expressed as the product of two equal integers.

Answer 8

I'm actually excited to see this one. I have an idea of what I would use, but I don't want to hijack the OP and rob him of his eureka moment.

Answer 9

hehe

Answer 10

(n+2)^2

Answer 11

that is in the form of a perfect square, but i dont think it would be helpful in our proof. if a perfect square can be expressed as the product of two equal integers, then if we say x is some integer, then m = (x*x) = x^2 fits our definition pretty nicely

Answer 12

the same can be done for n now, so how would you write that

Answer 13

n = (x*x) = x^2

Answer 14

hm, i dont think we get to assume that m and n are the same perfect square, so instead of x, let's choose y

Answer 15

okay. n = (y*y) = y^2

Answer 16

okay, so now that we have definitions for m and n, we can apply them to what we are trying to prove: namely m + n + 2\[\sqrt{mn}\]

Answer 17

sqrt[mn] should be next to the 2, dont know why it moved down

Answer 18

go ahead and substitute for m and n

Answer 19

actually, now that i think about it, i think m and n are going to have to be equal.

Answer 20

but go ahead and continue

Answer 21

Are we sure about not being able to assume they can use the same bound variable? I'm really new to this also, but it looks to me like it works out if we just use x^2 for m and x^2 for n.

Answer 22

yeah, they will have to be, because in order for our proof to work in the end, m + n + 2sqrt[mn] will need to be a perfect square, and by our definition, a perfect square can be expressed as the product of two equal integers

Answer 23

do you know what the next step is @mtalampas21 ?

Answer 24

I'm working my way through Vellman's How to Prove It and I thought I remembered something about even with different free variables, it isn't a guarantee that different object will use them (without a y doens't equal x provision).

Answer 25

I'm not really sure

Answer 26

interesting @Ajk. I read like 1/3 of that of that book, then i got too busy. interesting read from what i gathered

Answer 27

okay, we are going to let m and n both = x^2, and ill show you why afterwards. since we have those definitions, we can use substitution, for what we are trying to prove.

so wherever you see m and n, put x^2 in there place

Answer 28

their*

Answer 29

I'm liking the next step being just filling in m and n with what @blurbendy suggested (x^2) and doing the math. I ended up with a perfect square once it was all added together in the end.

Answer 30

x^2 + x^2 + 2 sqrt(x^2)(x^2)

Answer 31

I don't know if I'm right, but that is what I did. Once you perform the root operation and add everything up . . .

Answer 32

okay so after some simplifying you should get 2x^2 + 2sqrt[x^4]

Answer 33

what is the square root of x^4?

Answer 34

dont give up, you're close!

Answer 35

x^2

Answer 36

good, so that gives us 2x^2 + 2x^2. looking good, we need it in the form of a perfect square, and a perfect square can be expressed as the product of two equal integers.

2x^2 + 2x^2 gives us 4x^2

Now, we can we rewrite 4x^2 so that it resembles the definition of a perfect square?

Answer 37

how can we **

Answer 38

2x * 2x

Answer 39

not quite

Answer 40

think of another way to write 4

Answer 41

actually, that might work

Answer 42

4(x*x)

Answer 43

that will work, i just did it differently

Answer 44

yeah, I think any. (2x)^2 2^2*x^2 2x*2x . This was a good question. Thanks guys.

Answer 45

yup, good job mtalampas. hope it makes sense. do we we see what we concluded though? 2x * 2x = the product of two equal integers = a perfect square which is what we were trying to show.

Answer 46

yup, it makes sense. Thanks a lot.

Answer 47

you bet.

Answer 48

may you give me another example of a direct proof? If it's okay

Answer 49

let me think.... okay

Show that the product of two rational numbers is rational.

Answer 50

We get to assume that we have two rational numbers, and we to show their product is rational

Answer 51

similar to what we did for a perfect square, we need some sort of definition to rewrite those numbers. Let's call them m and n again

Answer 52

any ideas on how to write a rational number?

Answer 53

m/n, where n is not equal to zero ?

Answer 54

perfect, and we also need to say a/b are in the set of integers (since we dont want fractions). also instead of m/n, let's let m = a/b, where b is not zero and a and b are in the set of integers
and
n = x /y where y is not zero and x and y are in the set of integers

Answer 55

so now that we have our definitions, we can substitute them in for m*n, and then we'll show that their product is rational

Answer 56

so, what will our product look like

Answer 57

a/b * x/y

Answer 58

yup, and that gives us (ax) / (by), and notice we have it in the form p / q where q isn't 0 and p and q are in the set of integers. this the definition of rational, so we have concluded that the product of two rational numbers is always rational

Answer 59

I get it, thanks.

Answer 60

Direct proofs are pretty straight forward. Label your assumptions. State a way to rewrite your assumptions according to some definition. apply your rewritten assumptions to what you are trying to prove, and the math should take care of itself.

Answer 61

How about Indirect proofs, by contrapositive and contradiction? Any examples.

Answer 62

for contrapositive and contradiction you have to be careful of what you get to assume.

for example if you wanted to prove using contraposition that:

if the product m*n is even, then m and n cannot both me odd

in contraposition, you get to assume the negation of q and you have to show the negation of p

so, in the above, you would get to assume m and n are both EVEN, and would have to show m*n is ODD

In contradiction, you get to assume p and the negation of q

so if you had:
There does not exist an integer x, such that x^2 + 1 is negative

You get to assume some integer x, and that x^2 + 1 is POSITIVE
In contradiction, you go until you break math

So if we are assuming there does not exist an integer x such that x^2 + 1 is POSITIVE, let's plug in x = 1, and show that that is true

(1)^2 + 1 = 2 which is positive
We reached a contradiction! Therefore, we can restate that our conjecture is true (There does not exist an integer x, such that x^2 + 1 is negative)

Answer 63

an indirect proof is the same as contradiction

Answer 64

well sort of

Answer 65

you basically show that it can't be false

Answer 66

i.e. the math never breaks

Answer 67

both contradiction and contraposition are indirect proofs if I'm not mistaken..

Answer 68

probably. you dont see many proofs using the indirect method solely. most rely on contraposition or contradiction instead.

Answer 69

I see. How about the counter proof?

Answer 70

these are the easiest in my opinion. they are mostly used as disproofs
so if you have a statement that said
There does not exist an integer x such that x^2 -1 is negative
we can do a proof by counter-example
since x is an integer, we can let x = 0
(0)^2 - 1 = -1 which is negative.
hence, the statement "There does not exist an integer x such that x^2 -1 is negative" is false.

Answer 71

disproof by counter-example *** didnt mean to say proof

Answer 72

I understand...Thank you =)

Answer 73

anytime. proofs are all about practice. do a bunch of em' and soon enough they'll become second nature.

Answer 74

Noted. I'm so glad that I found this site, you guyz are really helpful. May I know your age and profession?

Answer 75

I'm 21, and i'm going to school for computer science right now. i'm actually interning for OpenStudy this summer

Answer 76

Okay. I'm just curious. I'll go practice making proofs.

Answer 77

are you in highscool, college?

Answer 78

highschool*

Answer 79

2nd year college. I'm 17

Answer 80

I'm not good in math that's why.

Answer 81

cool, you'll be done early then

Answer 82

is this class part of your major?

Answer 83

yup, in Discrete math. I'm an Information technology student

Answer 84

oh, right on

Answer 85

in what country are you by the way?

Answer 86

United States, you?

Answer 87

Oh. Philippines

Answer 88

ah, I have a friend from there. it's getting late though, so i'm going to take off. hope to see you around here again!

Answer 89

Thanks again. This is not my last in this site for sure. =)

Answer 90

awesome. cya!