# Convert 0.785 398 163 397 448 309 615 660 89 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

## 0.785 398 163 397 448 309 615 660 89(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: 0. Divide the number repeatedly by 2.

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

• division = quotient + remainder;
• 0 ÷ 2 = 0 + 0;

### 3. Convert to the binary (base 2) the fractional part: 0.785 398 163 397 448 309 615 660 89.

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

• #) multiplying = integer + fractional part;
• 1) 0.785 398 163 397 448 309 615 660 89 × 2 = 1 + 0.570 796 326 794 896 619 231 321 78;
• 2) 0.570 796 326 794 896 619 231 321 78 × 2 = 1 + 0.141 592 653 589 793 238 462 643 56;
• 3) 0.141 592 653 589 793 238 462 643 56 × 2 = 0 + 0.283 185 307 179 586 476 925 287 12;
• 4) 0.283 185 307 179 586 476 925 287 12 × 2 = 0 + 0.566 370 614 359 172 953 850 574 24;
• 5) 0.566 370 614 359 172 953 850 574 24 × 2 = 1 + 0.132 741 228 718 345 907 701 148 48;
• 6) 0.132 741 228 718 345 907 701 148 48 × 2 = 0 + 0.265 482 457 436 691 815 402 296 96;
• 7) 0.265 482 457 436 691 815 402 296 96 × 2 = 0 + 0.530 964 914 873 383 630 804 593 92;
• 8) 0.530 964 914 873 383 630 804 593 92 × 2 = 1 + 0.061 929 829 746 767 261 609 187 84;
• 9) 0.061 929 829 746 767 261 609 187 84 × 2 = 0 + 0.123 859 659 493 534 523 218 375 68;
• 10) 0.123 859 659 493 534 523 218 375 68 × 2 = 0 + 0.247 719 318 987 069 046 436 751 36;
• 11) 0.247 719 318 987 069 046 436 751 36 × 2 = 0 + 0.495 438 637 974 138 092 873 502 72;
• 12) 0.495 438 637 974 138 092 873 502 72 × 2 = 0 + 0.990 877 275 948 276 185 747 005 44;
• 13) 0.990 877 275 948 276 185 747 005 44 × 2 = 1 + 0.981 754 551 896 552 371 494 010 88;
• 14) 0.981 754 551 896 552 371 494 010 88 × 2 = 1 + 0.963 509 103 793 104 742 988 021 76;
• 15) 0.963 509 103 793 104 742 988 021 76 × 2 = 1 + 0.927 018 207 586 209 485 976 043 52;
• 16) 0.927 018 207 586 209 485 976 043 52 × 2 = 1 + 0.854 036 415 172 418 971 952 087 04;
• 17) 0.854 036 415 172 418 971 952 087 04 × 2 = 1 + 0.708 072 830 344 837 943 904 174 08;
• 18) 0.708 072 830 344 837 943 904 174 08 × 2 = 1 + 0.416 145 660 689 675 887 808 348 16;
• 19) 0.416 145 660 689 675 887 808 348 16 × 2 = 0 + 0.832 291 321 379 351 775 616 696 32;
• 20) 0.832 291 321 379 351 775 616 696 32 × 2 = 1 + 0.664 582 642 758 703 551 233 392 64;
• 21) 0.664 582 642 758 703 551 233 392 64 × 2 = 1 + 0.329 165 285 517 407 102 466 785 28;
• 22) 0.329 165 285 517 407 102 466 785 28 × 2 = 0 + 0.658 330 571 034 814 204 933 570 56;
• 23) 0.658 330 571 034 814 204 933 570 56 × 2 = 1 + 0.316 661 142 069 628 409 867 141 12;
• 24) 0.316 661 142 069 628 409 867 141 12 × 2 = 0 + 0.633 322 284 139 256 819 734 282 24;
• 25) 0.633 322 284 139 256 819 734 282 24 × 2 = 1 + 0.266 644 568 278 513 639 468 564 48;
• 26) 0.266 644 568 278 513 639 468 564 48 × 2 = 0 + 0.533 289 136 557 027 278 937 128 96;
• 27) 0.533 289 136 557 027 278 937 128 96 × 2 = 1 + 0.066 578 273 114 054 557 874 257 92;
• 28) 0.066 578 273 114 054 557 874 257 92 × 2 = 0 + 0.133 156 546 228 109 115 748 515 84;
• 29) 0.133 156 546 228 109 115 748 515 84 × 2 = 0 + 0.266 313 092 456 218 231 497 031 68;
• 30) 0.266 313 092 456 218 231 497 031 68 × 2 = 0 + 0.532 626 184 912 436 462 994 063 36;
• 31) 0.532 626 184 912 436 462 994 063 36 × 2 = 1 + 0.065 252 369 824 872 925 988 126 72;
• 32) 0.065 252 369 824 872 925 988 126 72 × 2 = 0 + 0.130 504 739 649 745 851 976 253 44;
• 33) 0.130 504 739 649 745 851 976 253 44 × 2 = 0 + 0.261 009 479 299 491 703 952 506 88;
• 34) 0.261 009 479 299 491 703 952 506 88 × 2 = 0 + 0.522 018 958 598 983 407 905 013 76;
• 35) 0.522 018 958 598 983 407 905 013 76 × 2 = 1 + 0.044 037 917 197 966 815 810 027 52;
• 36) 0.044 037 917 197 966 815 810 027 52 × 2 = 0 + 0.088 075 834 395 933 631 620 055 04;
• 37) 0.088 075 834 395 933 631 620 055 04 × 2 = 0 + 0.176 151 668 791 867 263 240 110 08;
• 38) 0.176 151 668 791 867 263 240 110 08 × 2 = 0 + 0.352 303 337 583 734 526 480 220 16;
• 39) 0.352 303 337 583 734 526 480 220 16 × 2 = 0 + 0.704 606 675 167 469 052 960 440 32;
• 40) 0.704 606 675 167 469 052 960 440 32 × 2 = 1 + 0.409 213 350 334 938 105 920 880 64;
• 41) 0.409 213 350 334 938 105 920 880 64 × 2 = 0 + 0.818 426 700 669 876 211 841 761 28;
• 42) 0.818 426 700 669 876 211 841 761 28 × 2 = 1 + 0.636 853 401 339 752 423 683 522 56;
• 43) 0.636 853 401 339 752 423 683 522 56 × 2 = 1 + 0.273 706 802 679 504 847 367 045 12;
• 44) 0.273 706 802 679 504 847 367 045 12 × 2 = 0 + 0.547 413 605 359 009 694 734 090 24;
• 45) 0.547 413 605 359 009 694 734 090 24 × 2 = 1 + 0.094 827 210 718 019 389 468 180 48;
• 46) 0.094 827 210 718 019 389 468 180 48 × 2 = 0 + 0.189 654 421 436 038 778 936 360 96;
• 47) 0.189 654 421 436 038 778 936 360 96 × 2 = 0 + 0.379 308 842 872 077 557 872 721 92;
• 48) 0.379 308 842 872 077 557 872 721 92 × 2 = 0 + 0.758 617 685 744 155 115 745 443 84;
• 49) 0.758 617 685 744 155 115 745 443 84 × 2 = 1 + 0.517 235 371 488 310 231 490 887 68;
• 50) 0.517 235 371 488 310 231 490 887 68 × 2 = 1 + 0.034 470 742 976 620 462 981 775 36;
• 51) 0.034 470 742 976 620 462 981 775 36 × 2 = 0 + 0.068 941 485 953 240 925 963 550 72;
• 52) 0.068 941 485 953 240 925 963 550 72 × 2 = 0 + 0.137 882 971 906 481 851 927 101 44;
• 53) 0.137 882 971 906 481 851 927 101 44 × 2 = 0 + 0.275 765 943 812 963 703 854 202 88;

### 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;
• 1 022 ÷ 2 = 511 + 0;
• 511 ÷ 2 = 255 + 1;
• 255 ÷ 2 = 127 + 1;
• 127 ÷ 2 = 63 + 1;
• 63 ÷ 2 = 31 + 1;
• 31 ÷ 2 = 15 + 1;
• 15 ÷ 2 = 7 + 1;
• 7 ÷ 2 = 3 + 1;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;

## Number 0.785 398 163 397 448 309 615 660 89 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point: 0 - 011 1111 1110 - 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000

(64 bits IEEE 754)

• 0

63

• 0

62
• 1

61
• 1

60
• 1

59
• 1

58
• 1

57
• 1

56
• 1

55
• 1

54
• 1

53
• 0

52

• 1

51
• 0

50
• 0

49
• 1

48
• 0

47
• 0

46
• 1

45
• 0

44
• 0

43
• 0

42
• 0

41
• 1

40
• 1

39
• 1

38
• 1

37
• 1

36
• 1

35
• 0

34
• 1

33
• 1

32
• 0

31
• 1

30
• 0

29
• 1

28
• 0

27
• 1

26
• 0

25
• 0

24
• 0

23
• 1

22
• 0

21
• 0

20
• 0

19
• 1

18
• 0

17
• 0

16
• 0

15
• 0

14
• 1

13
• 0

12
• 1

11
• 1

10
• 0

9
• 1

8
• 0

7
• 0

6
• 0

5
• 1

4
• 1

3
• 0

2
• 0

1
• 0

0

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

 0.785 398 163 397 448 309 615 660 89 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:08 UTC (GMT) 55 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:08 UTC (GMT) -267 276 187 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:07 UTC (GMT) 92.299 999 999 999 997 157 829 056 959 599 256 515 502 929 66 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:07 UTC (GMT) 6 864 421 736 393 290 841 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:07 UTC (GMT) 2.381 2 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:07 UTC (GMT) -0.001 010 1 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:07 UTC (GMT) 0.766 03 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:07 UTC (GMT) -120.208 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:07 UTC (GMT) -5.125 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:07 UTC (GMT) 2 019.092 4 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:07 UTC (GMT) -4.876 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:07 UTC (GMT) -0.016 738 891 601 562 527 755 575 615 4 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 11:06 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