# Convert 653 494 985.947 893 58 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

## How to convert the decimal number 653 494 985.947 893 58(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

### 1. First, convert to the binary (base 2) the integer part: 653 494 985. Divide the number repeatedly by 2.

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

• division = quotient + remainder;
• 653 494 985 ÷ 2 = 326 747 492 + 1;
• 326 747 492 ÷ 2 = 163 373 746 + 0;
• 163 373 746 ÷ 2 = 81 686 873 + 0;
• 81 686 873 ÷ 2 = 40 843 436 + 1;
• 40 843 436 ÷ 2 = 20 421 718 + 0;
• 20 421 718 ÷ 2 = 10 210 859 + 0;
• 10 210 859 ÷ 2 = 5 105 429 + 1;
• 5 105 429 ÷ 2 = 2 552 714 + 1;
• 2 552 714 ÷ 2 = 1 276 357 + 0;
• 1 276 357 ÷ 2 = 638 178 + 1;
• 638 178 ÷ 2 = 319 089 + 0;
• 319 089 ÷ 2 = 159 544 + 1;
• 159 544 ÷ 2 = 79 772 + 0;
• 79 772 ÷ 2 = 39 886 + 0;
• 39 886 ÷ 2 = 19 943 + 0;
• 19 943 ÷ 2 = 9 971 + 1;
• 9 971 ÷ 2 = 4 985 + 1;
• 4 985 ÷ 2 = 2 492 + 1;
• 2 492 ÷ 2 = 1 246 + 0;
• 1 246 ÷ 2 = 623 + 0;
• 623 ÷ 2 = 311 + 1;
• 311 ÷ 2 = 155 + 1;
• 155 ÷ 2 = 77 + 1;
• 77 ÷ 2 = 38 + 1;
• 38 ÷ 2 = 19 + 0;
• 19 ÷ 2 = 9 + 1;
• 9 ÷ 2 = 4 + 1;
• 4 ÷ 2 = 2 + 0;
• 2 ÷ 2 = 1 + 0;
• 1 ÷ 2 = 0 + 1;

### 3. Convert to the binary (base 2) the fractional part: 0.947 893 58.

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

• #) multiplying = integer + fractional part;
• 1) 0.947 893 58 × 2 = 1 + 0.895 787 16;
• 2) 0.895 787 16 × 2 = 1 + 0.791 574 32;
• 3) 0.791 574 32 × 2 = 1 + 0.583 148 64;
• 4) 0.583 148 64 × 2 = 1 + 0.166 297 28;
• 5) 0.166 297 28 × 2 = 0 + 0.332 594 56;
• 6) 0.332 594 56 × 2 = 0 + 0.665 189 12;
• 7) 0.665 189 12 × 2 = 1 + 0.330 378 24;
• 8) 0.330 378 24 × 2 = 0 + 0.660 756 48;
• 9) 0.660 756 48 × 2 = 1 + 0.321 512 96;
• 10) 0.321 512 96 × 2 = 0 + 0.643 025 92;
• 11) 0.643 025 92 × 2 = 1 + 0.286 051 84;
• 12) 0.286 051 84 × 2 = 0 + 0.572 103 68;
• 13) 0.572 103 68 × 2 = 1 + 0.144 207 36;
• 14) 0.144 207 36 × 2 = 0 + 0.288 414 72;
• 15) 0.288 414 72 × 2 = 0 + 0.576 829 44;
• 16) 0.576 829 44 × 2 = 1 + 0.153 658 88;
• 17) 0.153 658 88 × 2 = 0 + 0.307 317 76;
• 18) 0.307 317 76 × 2 = 0 + 0.614 635 52;
• 19) 0.614 635 52 × 2 = 1 + 0.229 271 04;
• 20) 0.229 271 04 × 2 = 0 + 0.458 542 08;
• 21) 0.458 542 08 × 2 = 0 + 0.917 084 16;
• 22) 0.917 084 16 × 2 = 1 + 0.834 168 32;
• 23) 0.834 168 32 × 2 = 1 + 0.668 336 64;
• 24) 0.668 336 64 × 2 = 1 + 0.336 673 28;
• 25) 0.336 673 28 × 2 = 0 + 0.673 346 56;
• 26) 0.673 346 56 × 2 = 1 + 0.346 693 12;
• 27) 0.346 693 12 × 2 = 0 + 0.693 386 24;
• 28) 0.693 386 24 × 2 = 1 + 0.386 772 48;
• 29) 0.386 772 48 × 2 = 0 + 0.773 544 96;
• 30) 0.773 544 96 × 2 = 1 + 0.547 089 92;
• 31) 0.547 089 92 × 2 = 1 + 0.094 179 84;
• 32) 0.094 179 84 × 2 = 0 + 0.188 359 68;
• 33) 0.188 359 68 × 2 = 0 + 0.376 719 36;
• 34) 0.376 719 36 × 2 = 0 + 0.753 438 72;
• 35) 0.753 438 72 × 2 = 1 + 0.506 877 44;
• 36) 0.506 877 44 × 2 = 1 + 0.013 754 88;
• 37) 0.013 754 88 × 2 = 0 + 0.027 509 76;
• 38) 0.027 509 76 × 2 = 0 + 0.055 019 52;
• 39) 0.055 019 52 × 2 = 0 + 0.110 039 04;
• 40) 0.110 039 04 × 2 = 0 + 0.220 078 08;
• 41) 0.220 078 08 × 2 = 0 + 0.440 156 16;
• 42) 0.440 156 16 × 2 = 0 + 0.880 312 32;
• 43) 0.880 312 32 × 2 = 1 + 0.760 624 64;
• 44) 0.760 624 64 × 2 = 1 + 0.521 249 28;
• 45) 0.521 249 28 × 2 = 1 + 0.042 498 56;
• 46) 0.042 498 56 × 2 = 0 + 0.084 997 12;
• 47) 0.084 997 12 × 2 = 0 + 0.169 994 24;
• 48) 0.169 994 24 × 2 = 0 + 0.339 988 48;
• 49) 0.339 988 48 × 2 = 0 + 0.679 976 96;
• 50) 0.679 976 96 × 2 = 1 + 0.359 953 92;
• 51) 0.359 953 92 × 2 = 0 + 0.719 907 84;
• 52) 0.719 907 84 × 2 = 1 + 0.439 815 68;
• 53) 0.439 815 68 × 2 = 0 + 0.879 631 36;

### 9. 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;
• 1 052 ÷ 2 = 526 + 0;
• 526 ÷ 2 = 263 + 0;
• 263 ÷ 2 = 131 + 1;
• 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;

## Number 653 494 985.947 893 58 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point: 0 - 100 0001 1100 - 0011 0111 1001 1100 0101 0110 0100 1111 1001 0101 0100 1001 0011

(64 bits IEEE 754)

• 0

63

• 1

62
• 0

61
• 0

60
• 0

59
• 0

58
• 0

57
• 1

56
• 1

55
• 1

54
• 0

53
• 0

52

• 0

51
• 0

50
• 1

49
• 1

48
• 0

47
• 1

46
• 1

45
• 1

44
• 1

43
• 0

42
• 0

41
• 1

40
• 1

39
• 1

38
• 0

37
• 0

36
• 0

35
• 1

34
• 0

33
• 1

32
• 0

31
• 1

30
• 1

29
• 0

28
• 0

27
• 1

26
• 0

25
• 0

24
• 1

23
• 1

22
• 1

21
• 1

20
• 1

19
• 0

18
• 0

17
• 1

16
• 0

15
• 1

14
• 0

13
• 1

12
• 0

11
• 1

10
• 0

9
• 0

8
• 1

7
• 0

6
• 0

5
• 1

4
• 0

3
• 0

2
• 1

1
• 1

0

## Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

 653 494 985.947 893 58 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:24 UTC (GMT) 456.21 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:23 UTC (GMT) 11.428 571 3 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:23 UTC (GMT) 89.4 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:23 UTC (GMT) 172.625 6 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:23 UTC (GMT) -274.374 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:23 UTC (GMT) 2.333 333 333 36 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:22 UTC (GMT) -6.3 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:22 UTC (GMT) -2 329.032 983 7 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:22 UTC (GMT) -1.61 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:21 UTC (GMT) 0.000 000 002 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:20 UTC (GMT) 0.367 26 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:20 UTC (GMT) 12 347 to 64 bit double precision IEEE 754 binary floating point = ? Dec 03 00:20 UTC (GMT) All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

## 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