Maraming salamat po ma'am Isa Kang tunay na bayani, napakalaking tulong po ito Lalo na ngayong nasa pandemic at nag seself study Ang mga studyante. God bless po mam aabangan ko po mga susunod na video nyo. From siargao💗💕
Mairon pang mga pinapahanapan ng counter examples pero walang mahanap ng counter examples? So sasabihin mo nalang na "therefore (x + 9)(x-13)/(x-13) = x-9 is true and don't have any counter examples to make this statement false"
When n = 10 n^2+n+11 10^2 + 10 + 11 100 + 10 + 11= 121 121 is not a prime number because it has more than 1 factor which are 1×121=121 and 11×11=121. When x is 10 disproves the statement.
According to number 1 problem; (x+y )/2 = whole number But when x= 1 and y= 2 ; (x+y)/2 (1+2)/2 = 3/2 or 1.5 The answer is not divisible by 2. Therefore, the statement is false when the two numbers are 1 and 2. We can also say that "The sum of two whole numbers wherein one number is an even number and the other is an odd number is not divisible by 2."
New subscriber here po.. naguguluhan po ako. how about this one po if n is a whole number, then n2+n+11 is prime number. the question po is give counterexamples on the following false statements. so is the answer is like this.. is n is even numbers then n2+n+11 is prime number..tama po ba
If n= 10 (whole number) n^2+n+11 (10)^2 + 10 + 11 = 121 121 is not a prime number because it has two factors, 121×1 =121 and 11×11= 121. So when n is equals to 10 is the counterexample of the statement.
Kung ganun nga ang equation katulad ng sinabi ko, you can use x=1 as a counterexample to the statement. When x=1 (x+3)/x = x+1 (x+3)/x = (1+3)/1 = 4/1= 4 x+1= 1+1= 2 4 is not equal to 2, which means that when x is 1 the statement will be false.
When x=1 (x+4)^2 = x^2 + 16 For left equation; (x+4)^2 (1+4)^2 = (5)^2 (5)^2 = 25 For right equation; x^2 + 16 (1)^2 + 16 = 1+16 1+16 = 17 25 is not equal to 17 so when x is 1 the statement is invalid. Let's try when x= 2 For left equation; (x+4)^2 (2+4)^2 = (6)^2 (6)^2 = 36 For right equation; x^2 + 16 (2)^2 + 16 = 4+16 4+16 = 20 36 is not equal to 20 so the statement is invalid when x is 2. What if x = 0 For left equation; (0+4)^2 (0+4)^2 = (4)^2 (4)^2 = 16 For right equation; x^2 + 16 (0)^2 + 16 = 0+16 0+16 = 16 The statement is valid when x is equal to zero (0). Therefore, for all numbers except x = 0, the equation (x+4)^2 = x^2 + 16 is invalid.
When n = 6 6 is a multiple of 3 since 3×2 =6 6 is a multiple of 6 since 6×1= 6 or you can do this; n/3 = whole number (no remainder) n/6 = whole number (no remainder) when n is equals to 6 ( n=6) n/3= 6/3= 2 n/6= 6/6 = 1 No remainder, meaning n=6 is a multiple of 3 and 6.
Dapat para maging multiple of 3 and 6 ang "any integer" , kailangan; n is greater than or equal to 6 ang mga number na pwede sa multiple 3 and 6 : 6,12,18,24,30 kapag ang integer ay multiple of 6 meaning multiple of 3 na din yun
2.) when x = -5 |x+3| = |x|+3 |x+3| = |-5+3| = |-2| = 2 |x|+3 = |-5| + 3 = 5+3 = 8 2 is not equal to 8.. Therefore, when x is -5 disproves the statement.
x^2 > x when x = 1/2 x^2 = (1/2)^2 = 1/4 1/4 is lesser than 1/2; 1/4 < 1/2 So, the statement is invalid when x is equal to 1/2. How about, x = 0 x^2 > x (0)^2 =0 0 = 0 Hence, the statement x^2 > x is invalid if x is 0. Let's try x=1 x^2 = (1)^2 =1 x^2 > x 1>1 is wrong it shoud be; 1=1 So, the statement is invalid when x is 1. But if x=2 x^2 = (2)^2= 4 x^2 > x (2)^2 > 2 4 > 2 is true Hence, the statement is valid when x is 2. Therefore, for all numbers except when the value of x is 2 and above, the statement x^2 > x is invalid.
@@nherinadarr2696 FIND THE NUMBER THAT PROVIDES A COUNTER-EXAMPLE TO SHOW THAT THE GIVEN STATEMENT IS FALSE. 1.∀ real numbers x,x³≥x 2. ∀ real numbers x,|x+3|=|x|+3
2.) when x = -5 |x+3| = |x|+3 |x+3| = |-5+3| = |-2| = 2 |x|+3 = |-5| + 3 = 5+3 = 8 2 is not equal to 8.. Therefore, when x is -5 disproves the statement.
Try to use number which is negative, for example: when x is -2 then the statement will be like this; √x^2 = x √(-2)^2 = 2 x= -2 Therefore when substituting -2 as the value of x the statement will be disprove.
Actually, you can choose any number for x para ma disprove ang statement, for example: when x=1 : √x^2+4 = x+2 √(1)^2 + 4 = √5 = 2.24 .. then for the right side equation : x+2 1+2 = 3 which means that 2.24 is not equals to 3. Meaning when x is 1 the statement will be false.
Maraming salamat po ma'am Isa Kang tunay na bayani, napakalaking tulong po ito Lalo na ngayong nasa pandemic at nag seself study Ang mga studyante. God bless po mam aabangan ko po mga susunod na video nyo. From siargao💗💕
Napaka laking tulong po ito sa tulad ko na na may Discrete math na sub ❤
the best video on this topic. very simple to understand. i subscribed, thxs for the helpful vid. keep up the good work!!
Thank u so much po talaga ng siksik liglig garabe laking tulong sa akin to😍😍😍😍
Wow galing po thank you.
ateeeeee moree, iba talaga pag tagalog inexplain
thank you po!😊❤️
Pud po magpaturo sa counterexample nag message po ako sa fb page po ninyo salamat po😃😃😃
Thank you po☺️☺️
Thank u po talagaaaaaaa mam
Mairon pang mga pinapahanapan ng counter examples pero walang mahanap ng counter examples? So sasabihin mo nalang na "therefore (x + 9)(x-13)/(x-13) = x-9 is true and don't have any counter examples to make this statement false"
thankkyouuuuu poooo!!!
How to disprove
All two digit even numbers are divisible by both 2 and 4.
All numbers divisible by 3 is divisible by 9 also.
thank u po
Thank you po
hi po may i ask? pano po ang "If n is a whole number, then n^2+ n + 11 is a prime number." thank you po.
Idisprove po ba yung statement?
When n = 10
n^2+n+11
10^2 + 10 + 11
100 + 10 + 11= 121
121 is not a prime number because it has more than 1 factor which are 1×121=121 and 11×11=121.
When x is 10 disproves the statement.
Opo e dis approve po, thank u po ng marami. Love lots po
May I ask po, how many counterexample s are needed to prove that a statement is false???
1 counterexample is enough
Ano po counter example nito po😊...
1. The sum of any two whole numbers is divisible by 2.?
According to number 1 problem;
(x+y )/2 = whole number
But when x= 1 and y= 2 ;
(x+y)/2
(1+2)/2 = 3/2 or 1.5
The answer is not divisible by 2.
Therefore, the statement is false when the two numbers are 1 and 2.
We can also say that "The sum of two whole numbers wherein one number is an even number and the other is an odd number is not divisible by 2."
HI maam diba po may square root yung 9 sa the square root of y squared +9 .
oo merong square root ang dalawa actually, √y^2+9
New subscriber here po.. naguguluhan po ako. how about this one po if n is a whole number, then n2+n+11 is prime number. the question po is give counterexamples on the following false statements. so is the answer is like this.. is n is even numbers then n2+n+11 is prime number..tama po ba
If n= 10 (whole number)
n^2+n+11
(10)^2 + 10 + 11 = 121
121 is not a prime number because it has two factors, 121×1 =121 and 11×11= 121.
So when n is equals to 10 is the counterexample of the statement.
thank you soo much 🙏🙏🙏💞💞💞
hi good afternoon patulong naman po sa counterexampple ng x+1>x
UP HAHAHAHA
Is this still part of inductive reasoning?
is it a theorem? prove. if it is not, give counter example. how po... n^2≥0 for some counting number n
may i ask po is x+3 over x =x+1 .how can i disprove this ?
ganito po ba yan?
(x+3)/x = x+1 ??
Kung ganun nga ang equation katulad ng sinabi ko, you can use x=1 as a counterexample to the statement.
When x=1
(x+3)/x = x+1
(x+3)/x = (1+3)/1 = 4/1= 4
x+1= 1+1= 2
4 is not equal to 2, which means that when x is 1 the statement will be false.
How to disapprove this statement (𝑥 + 4)2 = 𝑥2 + 16
When x=1
(x+4)^2 = x^2 + 16
For left equation;
(x+4)^2
(1+4)^2 = (5)^2
(5)^2 = 25
For right equation;
x^2 + 16
(1)^2 + 16 = 1+16
1+16 = 17
25 is not equal to 17 so when x is 1 the statement is invalid.
Let's try when x= 2
For left equation;
(x+4)^2
(2+4)^2 = (6)^2
(6)^2 = 36
For right equation;
x^2 + 16
(2)^2 + 16 = 4+16
4+16 = 20
36 is not equal to 20 so the statement is invalid when x is 2.
What if x = 0
For left equation;
(0+4)^2
(0+4)^2 = (4)^2
(4)^2 = 16
For right equation;
x^2 + 16
(0)^2 + 16 = 0+16
0+16 = 16
The statement is valid when x is equal to zero (0).
Therefore, for all numbers except
x = 0, the equation
(x+4)^2 = x^2 + 16 is invalid.
@@nherinadarr2696 OMGGG THANK YOU ATEEE LIFE SAVER🥰
How about po dito pano po ito I solve.
Prove all integer n which is a multiple of 3 are multiple 6
Na confuse po ako sa 3 and 6 , sana po matulungan ninyo po ako
When n = 6
6 is a multiple of 3 since 3×2 =6
6 is a multiple of 6 since 6×1= 6
or you can do this;
n/3 = whole number (no remainder)
n/6 = whole number (no remainder)
when n is equals to 6 ( n=6)
n/3= 6/3= 2
n/6= 6/6 = 1
No remainder, meaning n=6 is a multiple of 3 and 6.
Dapat para maging multiple of 3 and 6 ang "any integer" , kailangan;
n is greater than or equal to 6
ang mga number na pwede sa multiple 3 and 6 :
6,12,18,24,30
kapag ang integer ay multiple of 6 meaning multiple of 3 na din yun
Thank you po bali po ang sagot ay 6 sa counter example
yes when n=6 ang answer kung i-prove yung statement
How to disapprove this statement ate (a+b)(a-b)/(a-b)=a+b?
Naguguluhan ako pwedi po patunlong neto
XX=1
x+33=x+1
How to disprove (A-B)' =A'-B' po?
Maam pano po pag ganto?
|x + 3| = |x| + 3
yan po kasi mostly tanong hehe
2.) when x = -5
|x+3| = |x|+3
|x+3| = |-5+3| = |-2| = 2
|x|+3 = |-5| + 3 = 5+3 = 8
2 is not equal to 8..
Therefore, when x is -5 disproves the statement.
@@nherinadarr2696 Yun! maraming salama ma'am ❤
𝑥2 > x how to disprove this statement??
x^2 > x
when x = 1/2
x^2 = (1/2)^2 = 1/4
1/4 is lesser than 1/2;
1/4 < 1/2
So, the statement is invalid when x is equal to 1/2.
How about, x = 0
x^2 > x
(0)^2 =0
0 = 0
Hence, the statement x^2 > x is invalid if x is 0.
Let's try x=1
x^2 = (1)^2 =1
x^2 > x
1>1 is wrong it shoud be;
1=1
So, the statement is invalid when x is 1.
But if x=2
x^2 = (2)^2= 4
x^2 > x
(2)^2 > 2
4 > 2 is true
Hence, the statement is valid when x is 2.
Therefore, for all numbers except when the value of x is 2 and above, the statement x^2 > x is invalid.
@@nherinadarr2696 Maraming salamat po sa inyo☺️ malaking tulong po talaga yung yt channel niyo lalo na may MMW po kamii
🤗🤗
Paano po ba i-disprove ito x+3/3=x+1?
Paano po kaya ito idisprove? √x^2=x
Paano po I disprove ito x+3/3=x+3? Pa help po pleasee
•∀ real numbers x,x³≥x.
•∀ real numbers x,|x+3|=|x|+3
Pasagot po thank you,medj naguguluhan pa rin po ako eh
i-disproved po ba ang statement?
@@nherinadarr2696 FIND THE NUMBER THAT PROVIDES A COUNTER-EXAMPLE TO SHOW THAT THE GIVEN STATEMENT IS FALSE.
1.∀ real numbers x,x³≥x
2. ∀ real numbers x,|x+3|=|x|+3
1.) When x=1/2
x^3 = (1/2)^3 = 1/8
x=1/2
1/8 is not greater than or equal to 1/2
Therefore, when x is 1/2 disproves the statement.
2.) when x = -5
|x+3| = |x|+3
|x+3| = |-5+3| = |-2| = 2
|x|+3 = |-5| + 3 = 5+3 = 8
2 is not equal to 8..
Therefore, when x is -5 disproves the statement.
@@nherinadarr2696 thank you pooo
im stuck here...how to disprove if problem is like this √𝑥^2=𝑥
Hi idol 👋 new subscribed pa notice naman oh
How to disprove x
Is the statement goes like this:
x is less than or equal to x raise to two?
@@nherinadarr2696 yes
You can use, when x is 1/2 or x=1/2 to disprove the statement.
Solution:
x1/4
Thank you last how about √x^2 = x?
Try to use number which is negative, for example: when x is -2 then the statement will be like this;
√x^2 = x
√(-2)^2 = 2
x= -2
Therefore when substituting -2 as the value of x the statement will be disprove.
I chatted you po on Fb. Hope you can reply
|x|>0 disprove statement using counterexample ano po sagot dito?
When x=0
|0| > 0 is false, it is suppose to be like this:
0=0
when x is zero is the counter example of that statement
How about √x^2+4 = x+2
ang square root ba hanggang x^2 lang o hanggang +4?
@@nherinadarr2696 hanggang +4
Actually, you can choose any number for x para ma disprove ang statement, for example:
when x=1 :
√x^2+4 = x+2
√(1)^2 + 4 = √5 = 2.24
.. then for the right side equation :
x+2
1+2 = 3
which means that 2.24 is not equals to 3.
Meaning when x is 1 the statement will be false.
I messaged you on fb po hoping you would respond 🙏
Ma'am paano po ididisaprove tong statement na "if u>-5 then,u is a positive ....