Proof and Problem Solving - Quantifiers Example 03

This example works with the universal quantifier (i.e. the "for all" symbol) and the existential quantifier (i.e. the "there exists" symbol). A variety of logical expressions are provided and the validity (i.e. true or false) of each statement is determined.

[Atlastheyote]

I've been trying to wrap my head around discreet mathematics for my computational science unit at university and your videos are finally helping it all click in my head <3

[dillonberger]

At around 5:32 you said if you change it to a weak inequality then the statement would be true, but I think you omitted the fact that y can still be equal to 6, since y just has to be greater than 5. 

[avgerageCSStudent]

For the first question, would it still be true if we switched the quantifiers to "There exists a y for all x such that x-y=0"?

[krishna]

Thankyou! Your video is straightforward

[feras]

Thank you u really helped me with my exam 😘

[juricjr.gequillana]

You mentioned in letter c that it only needs one reason for a statement to be false, so in letter a and b I could not choose y=x or y=3x i could choose a lot of different numbers that will make the statement false. Based on your reasoning that it only needs one spot for a statement to be called false then why statements a and b are true with the fact there are possibilities that it could be false?

[SaiKobe]

Wouldn't part f still be false if u consider the number 6 after u changed it greater than or equal to?

[DatCrossGuy]

I'm trying to determine the truth value of the following: 
There exists x such that, all y and all z, x+y=z

There exists x such that all y and all z, z>y implies z>x+y

The domain for both is set to all real numbers

The way I understand it is that I need to find 1 value for x where no matter what values y and z have, the statements are true. This seems impossible to me though. Any suggestions?

[GC-Haendlach]

Yes but a bit fancy for beginning alg leve, so it may lack the same “shock and awe value as this:

.X=X for any real number X. Introduce a new variable name say Y, Such that
Y=X. Then squaring both sides, we obtain X^2 = Y^2. Now X^2-Y^2=0 because we are always allowed to apply the same operation to both sides. Factoring the diff of squares, gives (X+Y)(X-Y) = 0, that is to be clear, I’m not changing the VALUE of
The Left side-just rewriting its form.
So now, divide both sides by the quantity ( X-Y), which gives:

(X+Y) = 0, bc div by both sides is peachy so long as we are
Dividing BOTH sides by the same number…Hence X= -Y, for all values
Of the variables. Now, invoking the same real number properties again,
Dividing both sides by X gives:
1=-1…Yikes, untrue!!
Where did I go wrong??

[liamhoward]

How can you tell if it is an “existence proof” or a “for all” proof???

[souha]

Why C is false
We can assign 9 to x and 3 to y and it is equal to 0

[c.danielpremkumar]

2:54 x-3y = 0 if x=0 then y=0 therefore the answer should be TRUE.

[saifchannel]

Consider a statement :

For every integer x and every integer y there is an integer n such that.

if x>0 then nx>y
Translate the statement into symbol.

Can anyone help me

[jazz]

(∀x)(∃y)(x < y2). Is this true or false?