# Convert the Number -45 474 668.446 669 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number. Detailed Explanations

## Number -45 474 668.446 669(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

### 2. First, convert to binary (in base 2) the integer part: 45 474 668. Divide the number repeatedly by 2.

#### We stop when we get a quotient that is equal to zero.

• division = quotient + remainder;
• 45 474 668 ÷ 2 = 22 737 334 + 0;
• 22 737 334 ÷ 2 = 11 368 667 + 0;
• 11 368 667 ÷ 2 = 5 684 333 + 1;
• 5 684 333 ÷ 2 = 2 842 166 + 1;
• 2 842 166 ÷ 2 = 1 421 083 + 0;
• 1 421 083 ÷ 2 = 710 541 + 1;
• 710 541 ÷ 2 = 355 270 + 1;
• 355 270 ÷ 2 = 177 635 + 0;
• 177 635 ÷ 2 = 88 817 + 1;
• 88 817 ÷ 2 = 44 408 + 1;
• 44 408 ÷ 2 = 22 204 + 0;
• 22 204 ÷ 2 = 11 102 + 0;
• 11 102 ÷ 2 = 5 551 + 0;
• 5 551 ÷ 2 = 2 775 + 1;
• 2 775 ÷ 2 = 1 387 + 1;
• 1 387 ÷ 2 = 693 + 1;
• 693 ÷ 2 = 346 + 1;
• 346 ÷ 2 = 173 + 0;
• 173 ÷ 2 = 86 + 1;
• 86 ÷ 2 = 43 + 0;
• 43 ÷ 2 = 21 + 1;
• 21 ÷ 2 = 10 + 1;
• 10 ÷ 2 = 5 + 0;
• 5 ÷ 2 = 2 + 1;
• 2 ÷ 2 = 1 + 0;
• 1 ÷ 2 = 0 + 1;

### 4. Convert to binary (base 2) the fractional part: 0.446 669.

#### Stop when we get a fractional part that is equal to zero.

• #) multiplying = integer + fractional part;
• 1) 0.446 669 × 2 = 0 + 0.893 338;
• 2) 0.893 338 × 2 = 1 + 0.786 676;
• 3) 0.786 676 × 2 = 1 + 0.573 352;
• 4) 0.573 352 × 2 = 1 + 0.146 704;
• 5) 0.146 704 × 2 = 0 + 0.293 408;
• 6) 0.293 408 × 2 = 0 + 0.586 816;
• 7) 0.586 816 × 2 = 1 + 0.173 632;
• 8) 0.173 632 × 2 = 0 + 0.347 264;
• 9) 0.347 264 × 2 = 0 + 0.694 528;
• 10) 0.694 528 × 2 = 1 + 0.389 056;
• 11) 0.389 056 × 2 = 0 + 0.778 112;
• 12) 0.778 112 × 2 = 1 + 0.556 224;
• 13) 0.556 224 × 2 = 1 + 0.112 448;
• 14) 0.112 448 × 2 = 0 + 0.224 896;
• 15) 0.224 896 × 2 = 0 + 0.449 792;
• 16) 0.449 792 × 2 = 0 + 0.899 584;
• 17) 0.899 584 × 2 = 1 + 0.799 168;
• 18) 0.799 168 × 2 = 1 + 0.598 336;
• 19) 0.598 336 × 2 = 1 + 0.196 672;
• 20) 0.196 672 × 2 = 0 + 0.393 344;
• 21) 0.393 344 × 2 = 0 + 0.786 688;
• 22) 0.786 688 × 2 = 1 + 0.573 376;
• 23) 0.573 376 × 2 = 1 + 0.146 752;
• 24) 0.146 752 × 2 = 0 + 0.293 504;
• 25) 0.293 504 × 2 = 0 + 0.587 008;
• 26) 0.587 008 × 2 = 1 + 0.174 016;
• 27) 0.174 016 × 2 = 0 + 0.348 032;
• 28) 0.348 032 × 2 = 0 + 0.696 064;
• 29) 0.696 064 × 2 = 1 + 0.392 128;
• 30) 0.392 128 × 2 = 0 + 0.784 256;
• 31) 0.784 256 × 2 = 1 + 0.568 512;
• 32) 0.568 512 × 2 = 1 + 0.137 024;
• 33) 0.137 024 × 2 = 0 + 0.274 048;
• 34) 0.274 048 × 2 = 0 + 0.548 096;
• 35) 0.548 096 × 2 = 1 + 0.096 192;
• 36) 0.096 192 × 2 = 0 + 0.192 384;
• 37) 0.192 384 × 2 = 0 + 0.384 768;
• 38) 0.384 768 × 2 = 0 + 0.769 536;
• 39) 0.769 536 × 2 = 1 + 0.539 072;
• 40) 0.539 072 × 2 = 1 + 0.078 144;
• 41) 0.078 144 × 2 = 0 + 0.156 288;
• 42) 0.156 288 × 2 = 0 + 0.312 576;
• 43) 0.312 576 × 2 = 0 + 0.625 152;
• 44) 0.625 152 × 2 = 1 + 0.250 304;
• 45) 0.250 304 × 2 = 0 + 0.500 608;
• 46) 0.500 608 × 2 = 1 + 0.001 216;
• 47) 0.001 216 × 2 = 0 + 0.002 432;
• 48) 0.002 432 × 2 = 0 + 0.004 864;
• 49) 0.004 864 × 2 = 0 + 0.009 728;
• 50) 0.009 728 × 2 = 0 + 0.019 456;
• 51) 0.019 456 × 2 = 0 + 0.038 912;
• 52) 0.038 912 × 2 = 0 + 0.077 824;
• 53) 0.077 824 × 2 = 0 + 0.155 648;

### 10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

#### Use the same technique of repeatedly dividing by 2:

• division = quotient + remainder;
• 1 048 ÷ 2 = 524 + 0;
• 524 ÷ 2 = 262 + 0;
• 262 ÷ 2 = 131 + 0;
• 131 ÷ 2 = 65 + 1;
• 65 ÷ 2 = 32 + 1;
• 32 ÷ 2 = 16 + 0;
• 16 ÷ 2 = 8 + 0;
• 8 ÷ 2 = 4 + 0;
• 4 ÷ 2 = 2 + 0;
• 2 ÷ 2 = 1 + 0;
• 1 ÷ 2 = 0 + 1;

## The base ten decimal number -45 474 668.446 669 converted and written in 64 bit double precision IEEE 754 binary floating point representation: 1 - 100 0001 1000 - 0101 1010 1111 0001 1011 0110 0011 1001 0010 1100 0111 0011 0010

(64 bits IEEE 754)

• 1

63

• 1

62
• 0

61
• 0

60
• 0

59
• 0

58
• 0

57
• 1

56
• 1

55
• 0

54
• 0

53
• 0

52

• 0

51
• 1

50
• 0

49
• 1

48
• 1

47
• 0

46
• 1

45
• 0

44
• 1

43
• 1

42
• 1

41
• 1

40
• 0

39
• 0

38
• 0

37
• 1

36
• 1

35
• 0

34
• 1

33
• 1

32
• 0

31
• 1

30
• 1

29
• 0

28
• 0

27
• 0

26
• 1

25
• 1

24
• 1

23
• 0

22
• 0

21
• 1

20
• 0

19
• 0

18
• 1

17
• 0

16
• 1

15
• 1

14
• 0

13
• 0

12
• 0

11
• 1

10
• 1

9
• 1

8
• 0

7
• 0

6
• 1

5
• 1

4
• 0

3
• 0

2
• 1

1
• 0

0

## How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

### Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

• 1. If the number to be converted is negative, start with its the positive version.
• 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
• 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
• 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
• 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
• 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
• 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
• 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
• 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

### Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

|-31.640 215| = 31.640 215

• 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
• division = quotient + remainder;
• 31 ÷ 2 = 15 + 1;
• 15 ÷ 2 = 7 + 1;
• 7 ÷ 2 = 3 + 1;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;
• We have encountered a quotient that is ZERO => FULL STOP
• 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

31(10) = 1 1111(2)

• 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
• #) multiplying = integer + fractional part;
• 1) 0.640 215 × 2 = 1 + 0.280 43;
• 2) 0.280 43 × 2 = 0 + 0.560 86;
• 3) 0.560 86 × 2 = 1 + 0.121 72;
• 4) 0.121 72 × 2 = 0 + 0.243 44;
• 5) 0.243 44 × 2 = 0 + 0.486 88;
• 6) 0.486 88 × 2 = 0 + 0.973 76;
• 7) 0.973 76 × 2 = 1 + 0.947 52;
• 8) 0.947 52 × 2 = 1 + 0.895 04;
• 9) 0.895 04 × 2 = 1 + 0.790 08;
• 10) 0.790 08 × 2 = 1 + 0.580 16;
• 11) 0.580 16 × 2 = 1 + 0.160 32;
• 12) 0.160 32 × 2 = 0 + 0.320 64;
• 13) 0.320 64 × 2 = 0 + 0.641 28;
• 14) 0.641 28 × 2 = 1 + 0.282 56;
• 15) 0.282 56 × 2 = 0 + 0.565 12;
• 16) 0.565 12 × 2 = 1 + 0.130 24;
• 17) 0.130 24 × 2 = 0 + 0.260 48;
• 18) 0.260 48 × 2 = 0 + 0.520 96;
• 19) 0.520 96 × 2 = 1 + 0.041 92;
• 20) 0.041 92 × 2 = 0 + 0.083 84;
• 21) 0.083 84 × 2 = 0 + 0.167 68;
• 22) 0.167 68 × 2 = 0 + 0.335 36;
• 23) 0.335 36 × 2 = 0 + 0.670 72;
• 24) 0.670 72 × 2 = 1 + 0.341 44;
• 25) 0.341 44 × 2 = 0 + 0.682 88;
• 26) 0.682 88 × 2 = 1 + 0.365 76;
• 27) 0.365 76 × 2 = 0 + 0.731 52;
• 28) 0.731 52 × 2 = 1 + 0.463 04;
• 29) 0.463 04 × 2 = 0 + 0.926 08;
• 30) 0.926 08 × 2 = 1 + 0.852 16;
• 31) 0.852 16 × 2 = 1 + 0.704 32;
• 32) 0.704 32 × 2 = 1 + 0.408 64;
• 33) 0.408 64 × 2 = 0 + 0.817 28;
• 34) 0.817 28 × 2 = 1 + 0.634 56;
• 35) 0.634 56 × 2 = 1 + 0.269 12;
• 36) 0.269 12 × 2 = 0 + 0.538 24;
• 37) 0.538 24 × 2 = 1 + 0.076 48;
• 38) 0.076 48 × 2 = 0 + 0.152 96;
• 39) 0.152 96 × 2 = 0 + 0.305 92;
• 40) 0.305 92 × 2 = 0 + 0.611 84;
• 41) 0.611 84 × 2 = 1 + 0.223 68;
• 42) 0.223 68 × 2 = 0 + 0.447 36;
• 43) 0.447 36 × 2 = 0 + 0.894 72;
• 44) 0.894 72 × 2 = 1 + 0.789 44;
• 45) 0.789 44 × 2 = 1 + 0.578 88;
• 46) 0.578 88 × 2 = 1 + 0.157 76;
• 47) 0.157 76 × 2 = 0 + 0.315 52;
• 48) 0.315 52 × 2 = 0 + 0.631 04;
• 49) 0.631 04 × 2 = 1 + 0.262 08;
• 50) 0.262 08 × 2 = 0 + 0.524 16;
• 51) 0.524 16 × 2 = 1 + 0.048 32;
• 52) 0.048 32 × 2 = 0 + 0.096 64;
• 53) 0.096 64 × 2 = 0 + 0.193 28;
• We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
• 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

• 6. Summarizing - the positive number before normalization:

31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

• 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

31.640 215(10) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

• 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign: 1 (a negative number)

Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

• 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
100 0000 0011(2)

• 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

• Conclusion:

Sign (1 bit) = 1 (a negative number)

Exponent (8 bits) = 100 0000 0011

Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100