Question

Theorem: Let a and be be real numbers. If (a+b)^2 <= 6ab then a =b.
Proof: suppose that a=b. Then (a+b)^2 =4a^2 and 6ab +6a^2, and since 4a^2 <= 6a^2 is true, the theorem is true.
Is the proof correct and is the theorem correct? giving clear reasons for each

Answer 1

Welcome to Open Study.

Answer 2

I think you have to use induction to prove

Answer 3

Basic step:
Suppose it is true for the case a =b =1
\(P(1) := (1+1)^2 \leq 6*1*1\\~~~~~~~~~~~~~~~~~~~~~~~~~~~2\leq 6\)
Hypothesis step:
Assume it is true for a = b =n
That is
\[P(_n):= (n+n)^2\leq 6n*n\\~~~~~~~~~~~~~~~~~~~~~~~~4n^2\leq 6n^2\]

Answer 4

Now, prove \(P_{n+1}\) which is
\[P_{n+1}:= ((n+1)+(n+1))^2\leq 6*(n+1)*(n+1)\]

Answer 5

I don't think you can't handle it. :)

Answer 6

you know you only suppose the if part...
if you wanted to try to prove the contrapositive though:
you could assume a doesn't equal b and show (a+b)^2>6ab.

If this "theorem" is false, then we should be able to find a counterexample.
(a+b)^2 will be positive or zero for all real a and b
6ab will be negative for some real values a and b
so I think you could actually find a counterexample for (a+b)^2<=6ab

Answer 7

@OOOPS a proof by induction doesn't work here. We're working with real numbers here. There is no basis case.

Answer 8

its if then statement ,
\(P\rightarrow Q \)
there is 3 ways to prove :-
1- direct proof , assume P is true to reach Q is true
2- contrapositive assume not Q in true to reach not P is true
3- contradiction , assume what ever u want to be true but at the end ur result would contradict ur assumption

what do you wanna use ? ( im assuming cu i dint find a ounter example xD)

Answer 9

I haven't heard anything from @jbn123
are you there?

Answer 10

This is what i did:

Suppose that (a+b)^2 <= 6ab, then a^2+2ab+b^2 <= 6ab
and a^2-4ab+b^2<=0

then i wanted to let a=b, so that you eend up with b^2-4b^2+b^2<=0
so -2b^2<=0

but it seems incorrect

Answer 11

I'm not sure if the theorem is correct or not, i was looking for a counter example. But surely by letting a=b, then the theorem would then become 4a^2 <= 6a^2 which meas for any value of a, the theorem will be true?

Answer 12

yeah its true

Answer 13

because what we considered initially has led to the right result or proved us right by satisfying the condition ...so our assumption and theorm both are true

Answer 14

The statement is false. Take \(a=-2\) and \(b=2\). Then you have: $$
(-2+2)^2\le 6(-2)(2)\Longrightarrow 0\le -24,$$ which is false.

Answer 15

lolll a=b we have to take

Answer 16

2 is not equal to -2
is 2=-2???

Answer 17

The statement is, "If \((a+b)^2\le 6ab \), then \(a=b\). You can't assume \(a=b\) to prove the statement.

Answer 18

You are trying to prove that \((a+b)^2\le 6ab\) forces \(a=b\), which is not the case since I provided a counter example.

Answer 19

they said if (a+b)^2<=6ab

Answer 20

than a=b

Answer 21

If the problem had been, "Prove if \(a=b\), then \((a+b)^2\le 6ab\).", then you start by assume \(a=b\).

Answer 22

you get 0<=-24

Answer 23

which is not true

Answer 24

means a not equal to be

Answer 25

so the theorm is true

Answer 26

@nerdguy2535

Answer 27

Let's make sure we agree what the problem wants us to prove. Using what you posed earlier, this is a "\(P\longrightarrow Q\)" problem, with $$P=(a+b)^2\le 6ab$$and $$Q=(a=b).$$ Do we agree on this?

Answer 28

no no no

Answer 29

if P is true than Q

Answer 30

if P comes true than a=b

Answer 31

Yes. Thats what I typed means. If P is true then Q is also true. Thats the exactly what \(P\longrightarrow Q\) means.

Answer 32

okay....further

Answer 33

Cool

Answer 34

Ok, so I claim this statement is false. The statement (as claimed by the original poster) says it should work for all real numbers a and b (The second sentance, "Let a and b be real numbers...). My counter example is letting \(a=-2\) and \(b=2\). For these real numbers, the statement doesn't hold.

Answer 35

check again

Answer 36

there is a word IF before this

Answer 37

which means if P is true

Answer 38

a=-2 and b=2

Answer 39

0<=-24

Answer 40

AH!

Answer 41

I understand now. You are definitely right.

Answer 42

thanks @nerdguy2535

Answer 43

Ok, what about this?
\(a=4 ,b=5\)?
Then \((5+4)^2=81\) and \(6(4)(5)=120\), so \((a+b)^2\le 6ab\) is true, but, \(4\ne 5\).

Answer 44

applause to u buddy ..u r right now

Answer 45

u proved it wrong

Answer 46

so rather than try to write a proof for the statement being true or false, we can simply prove it to be false by a counter example?

Answer 47

yeah we can prove it wrong

Answer 48

but by counter example its hard to prove wrong