Post mathematic related questions here
OP
rst
Youngling
Re: mathematic related questions
Absolutely right
But bolded above step is complicated
instead of it you can use:-
PQ² = PR² + QR² (here R is used where dotted line meet PA)
PQ² = (4-2x)² + 3²
PQ² = (4-7/4)² + 9
PQ² = (81/16) + 9
pQ=15/4
Absolutely right
But bolded above step is complicated
instead of it you can use:-
PQ² = PR² + QR² (here R is used where dotted line meet PA)
PQ² = (4-2x)² + 3²
PQ² = (4-7/4)² + 9
PQ² = (81/16) + 9
pQ=15/4
Niilesh
Padawan
OP
rst
Youngling
Re: mathematic related questions
when you divide above mentioned numbers by 4 then remainder will be either 0 or 1
thats why, The only squares mod 4 are 0 and 1.
Here square means numbers like 1,4,9,16,25,36,49,................
when you divide above mentioned numbers by 4 then remainder will be either 0 or 1
thats why, The only squares mod 4 are 0 and 1.
OP
rst
Youngling
Re: mathematic related questions
lets see this proof in simple language
every square number will give remainder 0 and 1 if it is divisible by 4
this means If a number gives remainder other than 0 and 1 then it is not square number
we will this result
but it doesnot means that if remainder is o or 1 then it is perfect square
for example 40 gives remainder 0 when divisible by 4 (but it is not perfect square)
Now in a^2 + 6b^2
a ,b can be even or odd
case 1
a is even and b is odd
lets take a=2 and b=3
then a^2 + 6b^2 =58
58 will give remainder 2 when divided by 4
so a^2 + 6b^2 is not perfect square (as remainder is other than 0 and 1)
case 2
a is odd and b is even
similar to case one
case 3
both a and b are odd
let a=1 ,b=3
then a^2 + 6b^2 =1+18=19
19 will give remainder 3 when divided by 4
so a^2 + 6b^2 is not perfect square (as remainder is other than 0 and 1)
case 4
both a and b are even
let a=2,b=4
then a^2 + 6b^2 =100
so a^2 + 6b^2 is perfect square (as 100 is perfect square)
let a=2,b=4
then (b)^2 + 6(a)^2 =16+24=40
(b)^2 + 6(a)^2 is not perfect square
so a^2 + 6b^2 and (b)^2 + 6(a)^2 is not perfect square for a and b both even [ (0,0) is exception]
so answer is (0,0)
this proof is quite good
lets see this proof in simple language
every square number will give remainder 0 and 1 if it is divisible by 4
this means If a number gives remainder other than 0 and 1 then it is not square number
we will this result
but it doesnot means that if remainder is o or 1 then it is perfect square
for example 40 gives remainder 0 when divisible by 4 (but it is not perfect square)
Now in a^2 + 6b^2
a ,b can be even or odd
case 1
a is even and b is odd
lets take a=2 and b=3
then a^2 + 6b^2 =58
58 will give remainder 2 when divided by 4
so a^2 + 6b^2 is not perfect square (as remainder is other than 0 and 1)
case 2
a is odd and b is even
similar to case one
case 3
both a and b are odd
let a=1 ,b=3
then a^2 + 6b^2 =1+18=19
19 will give remainder 3 when divided by 4
so a^2 + 6b^2 is not perfect square (as remainder is other than 0 and 1)
case 4
both a and b are even
let a=2,b=4
then a^2 + 6b^2 =100
so a^2 + 6b^2 is perfect square (as 100 is perfect square)
let a=2,b=4
then (b)^2 + 6(a)^2 =16+24=40
(b)^2 + 6(a)^2 is not perfect square
so a^2 + 6b^2 and (b)^2 + 6(a)^2 is not perfect square for a and b both even [ (0,0) is exception]
so answer is (0,0)
OP
rst
Youngling
Re: mathematic related questions
If we want to proof that a theorem or result is not correct
Then in such situation even examples are enough
But If we want to proof a theorem or result is correct
Then we have to prove it with general terms
---------------------------------------------------------
Anyway I didn't try to prove any thing
I just tried to explain the proof with examples
If we want to proof that a theorem or result is not correct
Then in such situation even examples are enough
But If we want to proof a theorem or result is correct
Then we have to prove it with general terms
---------------------------------------------------------
Anyway I didn't try to prove any thing
I just tried to explain the proof with examples
Last edited:
OP
rst
Youngling
Re: mathematic related questions
I didn't try to proof the problem
proof is already given by Vignesh B
I just tried to explain the proof with examples
----------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION
Also plz explain your answer
(Also you can ask your mathematics related questions or queries
we will try to solve them)
I didn't try to proof the problem
proof is already given by Vignesh B
I just tried to explain the proof with examples
----------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION
Also plz explain your answer
(Also you can ask your mathematics related questions or queries
we will try to solve them)
Last edited:
Niilesh
Padawan
Re: mathematic related questions
It seems too easy :/
My ans - 50Kg
Solution-
Note: Assuming 99/98% is given weight by weight
Original condition - 1 Kg real matter + 99 Kg water
New condition - 1 Kg real matter + 'x' Kg water
Now it is given that
x/(x+1) = 98/100
2x=98
x=49
So ans = x+1= 50Kg
It seems too easy :/
My ans - 50Kg
Solution-
Note: Assuming 99/98% is given weight by weight
Original condition - 1 Kg real matter + 99 Kg water
New condition - 1 Kg real matter + 'x' Kg water
Now it is given that
x/(x+1) = 98/100
2x=98
x=49
So ans = x+1= 50Kg
OP
rst
Youngling
Re: mathematic related questions
great !!
Thanks a lot
-------------------------------------------------------
NEW OBJECTIVE QUESTION
Also plz explain your answer
(Also you can ask your mathematics related questions or queries
we will try to solve them)
great !!
Thanks a lot
Its better than mineI know my writing is not that good
-------------------------------------------------------
NEW OBJECTIVE QUESTION
Also plz explain your answer
(Also you can ask your mathematics related questions or queries
we will try to solve them)
Last edited: