How can I be this dumb? (1 Viewer)

dark_tranquility

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It happens, I feel the same way with certain things/games.
 
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athenaaa

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How do you think I failed pre-calculus 5 times? I certainly didn't do it by getting the correct answer.
Damn 💀 and I thought me getting a pitty pass for gr9 math was bad
 
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Myth

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So I was playing Yahtzee and I rolled two twos on the dice. And somehow I write 10 in that column. In what world does 2+2=10???​
In your own world and i want to live in it :sanapray:
 

litc

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So I was playing Yahtzee and I rolled two twos on the dice. And somehow I write 10 in that column. In what world does 2+2=10???​
well not 2 + 2

but 2 in binary is written as 10

and binary is base 2

so if instead of 2 + 2

it was (2) in base (2) it would be = 10

2 as you got it is also in the decimal representation, which is base 10, so that too

for decimal representation 2 digits:
first digit is 10 (base)^1 so tens place
second digit is 10 (base) ^ 0 (1) so ones place

so 2 is 0 * 10^1 + 2 * 10^0 (1) = 2
and 10 is 1 * 10^1 + 0 * 10^0 (1) = 10

binary representation 2 digits:
first digit is 2(base)^1 so twos place
second digit is 2 (base)^0 (1) so ones place

so 10 in binary is 1 * 2^1(2) + 0 * 2^0(1) = 1 * 2 + 0 * 0 = 2 in decimal representation

for any base representation system, each "place" starting from the very right (before any decimal point if any) is the base raised to (how many places it is from the rightmost/ones place)

so like for 10 (decimal) in binary
4 digits
1010
1 0 1 0
2 ^ 3 place (8s place), 2 ^ 2 place (4s place), 2 ^ 1 place (2s place), 2 ^ 0 place (1s place)

so 1010 in binary is 10 in decimal bc
1 * 2 ^ 3 + 0 * 2 ^ 2 + 1 * 2 ^ 1 + 0 * 2 ^ 0 = 1 * 8 + 0 * 4 + 1 * 2 + 0 * 1 = 8 + 2 = 10
one 8, one 2

like for 4 digits in decimal we have
10 ^ 3 place (thousands place), 10 ^ 2 (hundreds place), 10 ^ 1 (tens place), 10 ^ 0 (ones place)

this place value is the same for any base
like if we had base 3, 4 digits would be
3 ^ 3 place (27s place), 3 ^ 2 place (9s place), 3 ^ 1 place (3s place), 3 ^ 0 place (1s place)

converting from decimal to any base is first dividing the number by the base and keeping track of the remainder

step 1, 10 to binary: 10 / 2 = 5 R0

step 1, 10 to base 3: 10 / 3 = 3 R1

then you divide the quotient by the base, and keep track of the remainder, and keep doing this until the quotient becomes 0

10 to binary:
- 5 / 2 = 2 R1
- 2 / 2 = 1 R0
- 1 / 2 = 0 R1

10 to base 3:
- 3 / 3 = 1 R0
- 1 / 3 = 0 R1

then the converted decimal number in the new base representation, is the remainders written in from most recent (reverse order of you getting them)

so 10 in binary = 1010 (showed already)

and 10 in base 3 = 101
Check?
1 * 3 ^ 2 (9s place) + 0 * 3 ^ 1 (3s place) + 1 * 3 ^ 0 (1s place) = 1 * 9 + 0 * 3 + 1 * 1 = 9 + 1 = 10


but for 2 + 2 = 10 exactly and specifically
we know 2 + 2 = 4

so then 10 must be equal to 4

how would that work? what base?
the 1 is in the (base) ^ 1 place (which is just the base) and the 0 is in the (base) ^ 0 place (which is just the ones place)

so 1 * base = 4,

so the base is 4, 10 in base 4 is 4 in decimal representation

the decimal representation 2 is still 2 in base 4, because it is one digit and in the ones place

so is 2 (base 4) + 2 (base 4) = 10 (base 4)?

yes because we end up carrying.

The ones place in any base representation goes from 0 to the (base - 1) before, requiring another digit and being carried to the next place

For example, for decimal representation, base 10, the ones place goes from 0 - 9, and 10 requires 2 digits. In fact, for any place, the digits we can use are 0 - 9, and for any base representation, the digits we can have in any place, not just ones place, are 0 - (base - 1)

9 is 9 * 10^(0) (1)= 9 * 1 = 9
10 is 1 * 10^1 (10) + 0 * 10^0 (1) = 1 * 10 + 0 * 1 = 10


so adding 2 and 2 in base 4 gets 10 in base 4, 1 * 4 + 0 * 1 = 4 + 0 = 4
for base 4, the digits we have are 0 - 3

so the world in which 2 + 2 = 10 is base 4

I forgot about that being true in base 4 at first hence the binary explanation bc I thought 10 in binary is 2 right away

another conversion, if it helps make it sense?

10 (in decimal) to base 4

10 / 4 = 2 R2
2 / 4 = 0 R2

10 (in decimal) = 22 in base 4

Check/Proof:
2 * 4^1 + 2 * 4^0
2 * 4 + 2 * 1
8 + 2
10
10 = 10

lemme know if anything didn't make sense
 
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litc

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well not 2 + 2

but 2 in binary is written as 10

and binary is base 2

so if instead of 2 + 2

it was (2) in base (2) it would be = 10

2 as you got it is also in the decimal representation, which is base 10, so that too

for decimal representation 2 digits:
first digit is 10 (base)^1 so tens place
second digit is 10 (base) ^ 0 (1) so ones place

so 2 is 0 * 10^1 + 2 * 10^0 (1) = 2
and 10 is 1 * 10^1 + 0 * 10^0 (1) = 10

binary representation 2 digits:
first digit is 2(base)^1 so twos place
second digit is 2 (base)^0 (1) so ones place

so 10 in binary is 1 * 2^1(2) + 0 * 2^0(1) = 1 * 2 + 0 * 0 = 2 in decimal representation

for any base representation system, each "place" starting from the very right (before any decimal point if any) is the base raised to (how many places it is from the rightmost/ones place)

so like for 10 (decimal) in binary
4 digits
1010
1 0 1 0
2 ^ 3 place (8s place), 2 ^ 2 place (4s place), 2 ^ 1 place (2s place), 2 ^ 0 place (1s place)

so 1010 in binary is 10 in decimal bc
1 * 2 ^ 3 + 0 * 2 ^ 2 + 1 * 2 ^ 1 + 0 * 2 ^ 0 = 1 * 8 + 0 * 4 + 1 * 2 + 0 * 1 = 8 + 2 = 10
one 8, one 2

like for 4 digits in decimal we have
10 ^ 3 place (thousands place), 10 ^ 2 (hundreds place), 10 ^ 1 (tens place), 10 ^ 0 (ones place)

this place value is the same for any base
like if we had base 3, 4 digits would be
3 ^ 3 place (27s place), 3 ^ 2 place (9s place), 3 ^ 1 place (3s place), 3 ^ 0 place (1s place)

converting from decimal to any base is first dividing the number by the base and keeping track of the remainder

step 1, 10 to binary: 10 / 2 = 5 R0

step 1, 10 to base 3: 10 / 3 = 3 R1

then you divide the quotient by the base, and keep track of the remainder, and keep doing this until the quotient becomes 0

10 to binary:
- 5 / 2 = 2 R1
- 2 / 2 = 1 R0
- 1 / 2 = 0 R1

10 to base 3:
- 3 / 3 = 1 R0
- 1 / 3 = 0 R1

then the converted decimal number in the new base representation, is the remainders written in from most recent (reverse order of you getting them)

so 10 in binary = 1010 (showed already)

and 10 in base 3 = 101
Check?
1 * 3 ^ 2 (9s place) + 0 * 3 ^ 1 (3s place) + 1 * 3 ^ 0 (1s place) = 1 * 9 + 0 * 3 + 1 * 1 = 9 + 1 = 10


but for 2 + 2 = 10 exactly and specifically
we know 2 + 2 = 4

so then 10 must be equal to 4

how would that work? what base?
the 1 is in the (base) ^ 1 place (which is just the base) and the 0 is in the (base) ^ 0 place (which is just the ones place)

so 1 * base = 4,

so the base is 4, 10 in base 4 is 4 in decimal representation

the decimal representation 2 is still 2 in base 4, because it is one digit and in the ones place

so is 2 (base 4) + 2 (base 4) = 10 (base 4)?

yes because we end up carrying.

The ones place in any base representation goes from 0 to the (base - 1) before, requiring another digit and being carried to the next place

For example, for decimal representation, base 10, the ones place goes from 0 - 9, and 10 requires 2 digits. In fact, for any place, the digits we can use are 0 - 9, and for any base representation, the digits we can have in any place, not just ones place, are 0 - (base - 1)

9 is 9 * 10^(0) (1)= 9 * 1 = 9
10 is 1 * 10^1 (10) + 0 * 10^0 (1) = 1 * 10 + 0 * 1 = 10


so adding 2 and 2 in base 4 gets 10 in base 4, 1 * 4 + 0 * 1 = 4 + 0 = 4
for base 4, the digits we have are 0 - 3

so the world in which 2 + 2 = 10 is base 4

I forgot about that being true in base 4 at first hence the binary explanation bc I thought 10 in binary is 2 right away

another conversion, if it helps make it sense?

10 (in decimal) to base 4

10 / 4 = 2 R2
2 / 4 = 0 R2

10 (in decimal) = 22 in base 4

Check/Proof:
2 * 4^1 + 2 * 4^0
2 * 4 + 2 * 1
8 + 2
10
10 = 10

lemme know if anything didn't make sense
so you were right, but in base 4:nekosweat:
 

athenaaa

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well not 2 + 2

but 2 in binary is written as 10

and binary is base 2

so if instead of 2 + 2

it was (2) in base (2) it would be = 10

2 as you got it is also in the decimal representation, which is base 10, so that too

for decimal representation 2 digits:
first digit is 10 (base)^1 so tens place
second digit is 10 (base) ^ 0 (1) so ones place

so 2 is 0 * 10^1 + 2 * 10^0 (1) = 2
and 10 is 1 * 10^1 + 0 * 10^0 (1) = 10

binary representation 2 digits:
first digit is 2(base)^1 so twos place
second digit is 2 (base)^0 (1) so ones place

so 10 in binary is 1 * 2^1(2) + 0 * 2^0(1) = 1 * 2 + 0 * 0 = 2 in decimal representation

for any base representation system, each "place" starting from the very right (before any decimal point if any) is the base raised to (how many places it is from the rightmost/ones place)

so like for 10 (decimal) in binary
4 digits
1010
1 0 1 0
2 ^ 3 place (8s place), 2 ^ 2 place (4s place), 2 ^ 1 place (2s place), 2 ^ 0 place (1s place)

so 1010 in binary is 10 in decimal bc
1 * 2 ^ 3 + 0 * 2 ^ 2 + 1 * 2 ^ 1 + 0 * 2 ^ 0 = 1 * 8 + 0 * 4 + 1 * 2 + 0 * 1 = 8 + 2 = 10
one 8, one 2

like for 4 digits in decimal we have
10 ^ 3 place (thousands place), 10 ^ 2 (hundreds place), 10 ^ 1 (tens place), 10 ^ 0 (ones place)

this place value is the same for any base
like if we had base 3, 4 digits would be
3 ^ 3 place (27s place), 3 ^ 2 place (9s place), 3 ^ 1 place (3s place), 3 ^ 0 place (1s place)

converting from decimal to any base is first dividing the number by the base and keeping track of the remainder

step 1, 10 to binary: 10 / 2 = 5 R0

step 1, 10 to base 3: 10 / 3 = 3 R1

then you divide the quotient by the base, and keep track of the remainder, and keep doing this until the quotient becomes 0

10 to binary:
- 5 / 2 = 2 R1
- 2 / 2 = 1 R0
- 1 / 2 = 0 R1

10 to base 3:
- 3 / 3 = 1 R0
- 1 / 3 = 0 R1

then the converted decimal number in the new base representation, is the remainders written in from most recent (reverse order of you getting them)

so 10 in binary = 1010 (showed already)

and 10 in base 3 = 101
Check?
1 * 3 ^ 2 (9s place) + 0 * 3 ^ 1 (3s place) + 1 * 3 ^ 0 (1s place) = 1 * 9 + 0 * 3 + 1 * 1 = 9 + 1 = 10


but for 2 + 2 = 10 exactly and specifically
we know 2 + 2 = 4

so then 10 must be equal to 4

how would that work? what base?
the 1 is in the (base) ^ 1 place (which is just the base) and the 0 is in the (base) ^ 0 place (which is just the ones place)

so 1 * base = 4,

so the base is 4, 10 in base 4 is 4 in decimal representation

the decimal representation 2 is still 2 in base 4, because it is one digit and in the ones place

so is 2 (base 4) + 2 (base 4) = 10 (base 4)?

yes because we end up carrying.

The ones place in any base representation goes from 0 to the (base - 1) before, requiring another digit and being carried to the next place

For example, for decimal representation, base 10, the ones place goes from 0 - 9, and 10 requires 2 digits. In fact, for any place, the digits we can use are 0 - 9, and for any base representation, the digits we can have in any place, not just ones place, are 0 - (base - 1)

9 is 9 * 10^(0) (1)= 9 * 1 = 9
10 is 1 * 10^1 (10) + 0 * 10^0 (1) = 1 * 10 + 0 * 1 = 10


so adding 2 and 2 in base 4 gets 10 in base 4, 1 * 4 + 0 * 1 = 4 + 0 = 4
for base 4, the digits we have are 0 - 3

so the world in which 2 + 2 = 10 is base 4

I forgot about that being true in base 4 at first hence the binary explanation bc I thought 10 in binary is 2 right away

another conversion, if it helps make it sense?

10 (in decimal) to base 4

10 / 4 = 2 R2
2 / 4 = 0 R2

10 (in decimal) = 22 in base 4

Check/Proof:
2 * 4^1 + 2 * 4^0
2 * 4 + 2 * 1
8 + 2
10
10 = 10

lemme know if anything didn't make sense
SO I WAS CORRECT? :pepecry1:
 
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well not 2 + 2

but 2 in binary is written as 10

and binary is base 2

so if instead of 2 + 2

it was (2) in base (2) it would be = 10

2 as you got it is also in the decimal representation, which is base 10, so that too

for decimal representation 2 digits:
first digit is 10 (base)^1 so tens place
second digit is 10 (base) ^ 0 (1) so ones place

so 2 is 0 * 10^1 + 2 * 10^0 (1) = 2
and 10 is 1 * 10^1 + 0 * 10^0 (1) = 10

binary representation 2 digits:
first digit is 2(base)^1 so twos place
second digit is 2 (base)^0 (1) so ones place

so 10 in binary is 1 * 2^1(2) + 0 * 2^0(1) = 1 * 2 + 0 * 0 = 2 in decimal representation

for any base representation system, each "place" starting from the very right (before any decimal point if any) is the base raised to (how many places it is from the rightmost/ones place)

so like for 10 (decimal) in binary
4 digits
1010
1 0 1 0
2 ^ 3 place (8s place), 2 ^ 2 place (4s place), 2 ^ 1 place (2s place), 2 ^ 0 place (1s place)

so 1010 in binary is 10 in decimal bc
1 * 2 ^ 3 + 0 * 2 ^ 2 + 1 * 2 ^ 1 + 0 * 2 ^ 0 = 1 * 8 + 0 * 4 + 1 * 2 + 0 * 1 = 8 + 2 = 10
one 8, one 2

like for 4 digits in decimal we have
10 ^ 3 place (thousands place), 10 ^ 2 (hundreds place), 10 ^ 1 (tens place), 10 ^ 0 (ones place)

this place value is the same for any base
like if we had base 3, 4 digits would be
3 ^ 3 place (27s place), 3 ^ 2 place (9s place), 3 ^ 1 place (3s place), 3 ^ 0 place (1s place)

converting from decimal to any base is first dividing the number by the base and keeping track of the remainder

step 1, 10 to binary: 10 / 2 = 5 R0

step 1, 10 to base 3: 10 / 3 = 3 R1

then you divide the quotient by the base, and keep track of the remainder, and keep doing this until the quotient becomes 0

10 to binary:
- 5 / 2 = 2 R1
- 2 / 2 = 1 R0
- 1 / 2 = 0 R1

10 to base 3:
- 3 / 3 = 1 R0
- 1 / 3 = 0 R1

then the converted decimal number in the new base representation, is the remainders written in from most recent (reverse order of you getting them)

so 10 in binary = 1010 (showed already)

and 10 in base 3 = 101
Check?
1 * 3 ^ 2 (9s place) + 0 * 3 ^ 1 (3s place) + 1 * 3 ^ 0 (1s place) = 1 * 9 + 0 * 3 + 1 * 1 = 9 + 1 = 10


but for 2 + 2 = 10 exactly and specifically
we know 2 + 2 = 4

so then 10 must be equal to 4

how would that work? what base?
the 1 is in the (base) ^ 1 place (which is just the base) and the 0 is in the (base) ^ 0 place (which is just the ones place)

so 1 * base = 4,

so the base is 4, 10 in base 4 is 4 in decimal representation

the decimal representation 2 is still 2 in base 4, because it is one digit and in the ones place

so is 2 (base 4) + 2 (base 4) = 10 (base 4)?

yes because we end up carrying.

The ones place in any base representation goes from 0 to the (base - 1) before, requiring another digit and being carried to the next place

For example, for decimal representation, base 10, the ones place goes from 0 - 9, and 10 requires 2 digits. In fact, for any place, the digits we can use are 0 - 9, and for any base representation, the digits we can have in any place, not just ones place, are 0 - (base - 1)

9 is 9 * 10^(0) (1)= 9 * 1 = 9
10 is 1 * 10^1 (10) + 0 * 10^0 (1) = 1 * 10 + 0 * 1 = 10


so adding 2 and 2 in base 4 gets 10 in base 4, 1 * 4 + 0 * 1 = 4 + 0 = 4
for base 4, the digits we have are 0 - 3

so the world in which 2 + 2 = 10 is base 4

I forgot about that being true in base 4 at first hence the binary explanation bc I thought 10 in binary is 2 right away

another conversion, if it helps make it sense?

10 (in decimal) to base 4

10 / 4 = 2 R2
2 / 4 = 0 R2

10 (in decimal) = 22 in base 4

Check/Proof:
2 * 4^1 + 2 * 4^0
2 * 4 + 2 * 1
8 + 2
10
10 = 10

lemme know if anything didn't make sense
Omg, I love you for this response. When I take pre calculusfor the 6th time what will you charge for doing my homework?
 
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litc

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something also cool

to convert from binary to base 4

since digits in base 4 go from 0 - 3, and two digits of binary represent 0 - 3 (00 = 0, 01 = 1, 10 = 2, 11 = 3)

you can convert from binary to base 4 by splitting the binary representation into groups of 2 digits, then converting those into decimal, and those become the digits for base 4

example with 10 in decimal being 1010 in binary
1010
10 | 10
2 | 2
22 in base 4 (as I showed before)

this works for any power of 2, grouping the binary representation by as many digits as the power of 2 the other base is

so for 8, 2^3

10 in decimal, 1010 in binary
1010
001010
001 | 010
1 | 2
12 in base 8

check
1 * 8^1 + 2 * 8^0
1 * 8 + 2 * 1
8 + 2
10 in decimal

this works for converting from any base to another base that's a power of the original base

converting 10 in decimal, 101 in base 3 to base 9
101
0101
01 | 01
1 | 1
11 in base 9

check
1 * 9^1 + 1 * 9^0
1 * 9 + 1 * 1
9 + 1
10 in decimal
 

litc

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Omg, I love you for this response. When I take pre calculusfor the 6th time what will you charge for doing my homework?
I am literally reactivating again bc I remembered smth from pre Calc honors @StayCsPinkPanda

:jisoosmh:

So in our class we had participation credit, out of 10 points

part of that was, we would go up and explain problems from the previous nights homework on the whiteboard. So first students would say what problems they had trouble with on the hw and the teacher wrote the numbers on the board. Then people would volunteer to demonstrate them on the board, they would write their steps, show their work, and the solution, and how they got to it, and would verbally go through the problem and explain how to solve it

so since the participation was very vague, like how much do we do to get a 10, I was scared and so I volunteered almost every single class (all classes except the ones I couldn’t finish the hw in time) to demonstrate a problem on the board. It was also embarrassing cuz I was 15 but my voice sounded 5, but I explained as step by step and thoroughly and basically as I could

but somehow I got a 9/10 for participation despite all my volunteering. I was raging in the group chat rip. Ppl who only volunteered like 4 times got 10/10. Note there are 100+ days of school!

she hated my guts I guess. Math teachers typically don’t like me for wtv reason.

maybe she was salty that kids were telling me I explained better than the teacher and helped them understand, and would also just ask me for help besides the showing on the board:susPepe:

the kids I helped always had better grades than me so I always take it as I’m better at teaching than learning, and I’m good at teaching bc I’m bad at learning, cuz it was hard for me so I know how to break it down to who else is struggling:pandashrug:
 
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I am literally reactivating again bc I remembered smth from pre Calc honors @StayCsPinkPanda

:jisoosmh:

So in our class we had participation credit, out of 10 points

part of that was, we would go up and explain problems from the previous nights homework on the whiteboard. So first students would say what problems they had trouble with on the hw and the teacher wrote the numbers on the board. Then people would volunteer to demonstrate them on the board, they would write their steps, show their work, and the solution, and how they got to it, and would verbally go through the problem and explain how to solve it

so since the participation was very vague, like how much do we do to get a 10, I was scared and so I volunteered almost every single class (all classes except the ones I couldn’t finish the hw in time) to demonstrate a problem on the board. It was also embarrassing cuz I was 15 but my voice sounded 5, but I explained as step by step and thoroughly and basically as I could

but somehow I got a 9/10 for participation despite all my volunteering. I was raging in the group chat rip. Ppl who only volunteered like 4 times got 10/10. Note there are 100+ days of school!

she hated my guts I guess. Math teachers typically don’t like me for wtv reason.

maybe she was salty that kids were telling me I explained better than the teacher and helped them understand, and would also just ask me for help besides the showing on the board:susPepe:

the kids I helped always had better grades than me so I always take it as I’m better at teaching than learning, and I’m good at teaching bc I’m bad at learning, cuz it was hard for me so I know how to break it down to who else is struggling:pandashrug:
So, Yes! You'll do my homework! (I'm joking, if I'm going to suffer through more math classes I want to actually learn something)
 

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