-0.000 000 000 000 000 000 176 177 62 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 000 176 177 62(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 000 000 000 176 177 62(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 000 000 000 176 177 62| = 0.000 000 000 000 000 000 176 177 62


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

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


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

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 000 176 177 62.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

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


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 000 000 000 176 177 62 × 2 = 0 + 0.000 000 000 000 000 000 352 355 24;
  • 2) 0.000 000 000 000 000 000 352 355 24 × 2 = 0 + 0.000 000 000 000 000 000 704 710 48;
  • 3) 0.000 000 000 000 000 000 704 710 48 × 2 = 0 + 0.000 000 000 000 000 001 409 420 96;
  • 4) 0.000 000 000 000 000 001 409 420 96 × 2 = 0 + 0.000 000 000 000 000 002 818 841 92;
  • 5) 0.000 000 000 000 000 002 818 841 92 × 2 = 0 + 0.000 000 000 000 000 005 637 683 84;
  • 6) 0.000 000 000 000 000 005 637 683 84 × 2 = 0 + 0.000 000 000 000 000 011 275 367 68;
  • 7) 0.000 000 000 000 000 011 275 367 68 × 2 = 0 + 0.000 000 000 000 000 022 550 735 36;
  • 8) 0.000 000 000 000 000 022 550 735 36 × 2 = 0 + 0.000 000 000 000 000 045 101 470 72;
  • 9) 0.000 000 000 000 000 045 101 470 72 × 2 = 0 + 0.000 000 000 000 000 090 202 941 44;
  • 10) 0.000 000 000 000 000 090 202 941 44 × 2 = 0 + 0.000 000 000 000 000 180 405 882 88;
  • 11) 0.000 000 000 000 000 180 405 882 88 × 2 = 0 + 0.000 000 000 000 000 360 811 765 76;
  • 12) 0.000 000 000 000 000 360 811 765 76 × 2 = 0 + 0.000 000 000 000 000 721 623 531 52;
  • 13) 0.000 000 000 000 000 721 623 531 52 × 2 = 0 + 0.000 000 000 000 001 443 247 063 04;
  • 14) 0.000 000 000 000 001 443 247 063 04 × 2 = 0 + 0.000 000 000 000 002 886 494 126 08;
  • 15) 0.000 000 000 000 002 886 494 126 08 × 2 = 0 + 0.000 000 000 000 005 772 988 252 16;
  • 16) 0.000 000 000 000 005 772 988 252 16 × 2 = 0 + 0.000 000 000 000 011 545 976 504 32;
  • 17) 0.000 000 000 000 011 545 976 504 32 × 2 = 0 + 0.000 000 000 000 023 091 953 008 64;
  • 18) 0.000 000 000 000 023 091 953 008 64 × 2 = 0 + 0.000 000 000 000 046 183 906 017 28;
  • 19) 0.000 000 000 000 046 183 906 017 28 × 2 = 0 + 0.000 000 000 000 092 367 812 034 56;
  • 20) 0.000 000 000 000 092 367 812 034 56 × 2 = 0 + 0.000 000 000 000 184 735 624 069 12;
  • 21) 0.000 000 000 000 184 735 624 069 12 × 2 = 0 + 0.000 000 000 000 369 471 248 138 24;
  • 22) 0.000 000 000 000 369 471 248 138 24 × 2 = 0 + 0.000 000 000 000 738 942 496 276 48;
  • 23) 0.000 000 000 000 738 942 496 276 48 × 2 = 0 + 0.000 000 000 001 477 884 992 552 96;
  • 24) 0.000 000 000 001 477 884 992 552 96 × 2 = 0 + 0.000 000 000 002 955 769 985 105 92;
  • 25) 0.000 000 000 002 955 769 985 105 92 × 2 = 0 + 0.000 000 000 005 911 539 970 211 84;
  • 26) 0.000 000 000 005 911 539 970 211 84 × 2 = 0 + 0.000 000 000 011 823 079 940 423 68;
  • 27) 0.000 000 000 011 823 079 940 423 68 × 2 = 0 + 0.000 000 000 023 646 159 880 847 36;
  • 28) 0.000 000 000 023 646 159 880 847 36 × 2 = 0 + 0.000 000 000 047 292 319 761 694 72;
  • 29) 0.000 000 000 047 292 319 761 694 72 × 2 = 0 + 0.000 000 000 094 584 639 523 389 44;
  • 30) 0.000 000 000 094 584 639 523 389 44 × 2 = 0 + 0.000 000 000 189 169 279 046 778 88;
  • 31) 0.000 000 000 189 169 279 046 778 88 × 2 = 0 + 0.000 000 000 378 338 558 093 557 76;
  • 32) 0.000 000 000 378 338 558 093 557 76 × 2 = 0 + 0.000 000 000 756 677 116 187 115 52;
  • 33) 0.000 000 000 756 677 116 187 115 52 × 2 = 0 + 0.000 000 001 513 354 232 374 231 04;
  • 34) 0.000 000 001 513 354 232 374 231 04 × 2 = 0 + 0.000 000 003 026 708 464 748 462 08;
  • 35) 0.000 000 003 026 708 464 748 462 08 × 2 = 0 + 0.000 000 006 053 416 929 496 924 16;
  • 36) 0.000 000 006 053 416 929 496 924 16 × 2 = 0 + 0.000 000 012 106 833 858 993 848 32;
  • 37) 0.000 000 012 106 833 858 993 848 32 × 2 = 0 + 0.000 000 024 213 667 717 987 696 64;
  • 38) 0.000 000 024 213 667 717 987 696 64 × 2 = 0 + 0.000 000 048 427 335 435 975 393 28;
  • 39) 0.000 000 048 427 335 435 975 393 28 × 2 = 0 + 0.000 000 096 854 670 871 950 786 56;
  • 40) 0.000 000 096 854 670 871 950 786 56 × 2 = 0 + 0.000 000 193 709 341 743 901 573 12;
  • 41) 0.000 000 193 709 341 743 901 573 12 × 2 = 0 + 0.000 000 387 418 683 487 803 146 24;
  • 42) 0.000 000 387 418 683 487 803 146 24 × 2 = 0 + 0.000 000 774 837 366 975 606 292 48;
  • 43) 0.000 000 774 837 366 975 606 292 48 × 2 = 0 + 0.000 001 549 674 733 951 212 584 96;
  • 44) 0.000 001 549 674 733 951 212 584 96 × 2 = 0 + 0.000 003 099 349 467 902 425 169 92;
  • 45) 0.000 003 099 349 467 902 425 169 92 × 2 = 0 + 0.000 006 198 698 935 804 850 339 84;
  • 46) 0.000 006 198 698 935 804 850 339 84 × 2 = 0 + 0.000 012 397 397 871 609 700 679 68;
  • 47) 0.000 012 397 397 871 609 700 679 68 × 2 = 0 + 0.000 024 794 795 743 219 401 359 36;
  • 48) 0.000 024 794 795 743 219 401 359 36 × 2 = 0 + 0.000 049 589 591 486 438 802 718 72;
  • 49) 0.000 049 589 591 486 438 802 718 72 × 2 = 0 + 0.000 099 179 182 972 877 605 437 44;
  • 50) 0.000 099 179 182 972 877 605 437 44 × 2 = 0 + 0.000 198 358 365 945 755 210 874 88;
  • 51) 0.000 198 358 365 945 755 210 874 88 × 2 = 0 + 0.000 396 716 731 891 510 421 749 76;
  • 52) 0.000 396 716 731 891 510 421 749 76 × 2 = 0 + 0.000 793 433 463 783 020 843 499 52;
  • 53) 0.000 793 433 463 783 020 843 499 52 × 2 = 0 + 0.001 586 866 927 566 041 686 999 04;
  • 54) 0.001 586 866 927 566 041 686 999 04 × 2 = 0 + 0.003 173 733 855 132 083 373 998 08;
  • 55) 0.003 173 733 855 132 083 373 998 08 × 2 = 0 + 0.006 347 467 710 264 166 747 996 16;
  • 56) 0.006 347 467 710 264 166 747 996 16 × 2 = 0 + 0.012 694 935 420 528 333 495 992 32;
  • 57) 0.012 694 935 420 528 333 495 992 32 × 2 = 0 + 0.025 389 870 841 056 666 991 984 64;
  • 58) 0.025 389 870 841 056 666 991 984 64 × 2 = 0 + 0.050 779 741 682 113 333 983 969 28;
  • 59) 0.050 779 741 682 113 333 983 969 28 × 2 = 0 + 0.101 559 483 364 226 667 967 938 56;
  • 60) 0.101 559 483 364 226 667 967 938 56 × 2 = 0 + 0.203 118 966 728 453 335 935 877 12;
  • 61) 0.203 118 966 728 453 335 935 877 12 × 2 = 0 + 0.406 237 933 456 906 671 871 754 24;
  • 62) 0.406 237 933 456 906 671 871 754 24 × 2 = 0 + 0.812 475 866 913 813 343 743 508 48;
  • 63) 0.812 475 866 913 813 343 743 508 48 × 2 = 1 + 0.624 951 733 827 626 687 487 016 96;
  • 64) 0.624 951 733 827 626 687 487 016 96 × 2 = 1 + 0.249 903 467 655 253 374 974 033 92;
  • 65) 0.249 903 467 655 253 374 974 033 92 × 2 = 0 + 0.499 806 935 310 506 749 948 067 84;
  • 66) 0.499 806 935 310 506 749 948 067 84 × 2 = 0 + 0.999 613 870 621 013 499 896 135 68;
  • 67) 0.999 613 870 621 013 499 896 135 68 × 2 = 1 + 0.999 227 741 242 026 999 792 271 36;
  • 68) 0.999 227 741 242 026 999 792 271 36 × 2 = 1 + 0.998 455 482 484 053 999 584 542 72;
  • 69) 0.998 455 482 484 053 999 584 542 72 × 2 = 1 + 0.996 910 964 968 107 999 169 085 44;
  • 70) 0.996 910 964 968 107 999 169 085 44 × 2 = 1 + 0.993 821 929 936 215 998 338 170 88;
  • 71) 0.993 821 929 936 215 998 338 170 88 × 2 = 1 + 0.987 643 859 872 431 996 676 341 76;
  • 72) 0.987 643 859 872 431 996 676 341 76 × 2 = 1 + 0.975 287 719 744 863 993 352 683 52;
  • 73) 0.975 287 719 744 863 993 352 683 52 × 2 = 1 + 0.950 575 439 489 727 986 705 367 04;
  • 74) 0.950 575 439 489 727 986 705 367 04 × 2 = 1 + 0.901 150 878 979 455 973 410 734 08;
  • 75) 0.901 150 878 979 455 973 410 734 08 × 2 = 1 + 0.802 301 757 958 911 946 821 468 16;
  • 76) 0.802 301 757 958 911 946 821 468 16 × 2 = 1 + 0.604 603 515 917 823 893 642 936 32;
  • 77) 0.604 603 515 917 823 893 642 936 32 × 2 = 1 + 0.209 207 031 835 647 787 285 872 64;
  • 78) 0.209 207 031 835 647 787 285 872 64 × 2 = 0 + 0.418 414 063 671 295 574 571 745 28;
  • 79) 0.418 414 063 671 295 574 571 745 28 × 2 = 0 + 0.836 828 127 342 591 149 143 490 56;
  • 80) 0.836 828 127 342 591 149 143 490 56 × 2 = 1 + 0.673 656 254 685 182 298 286 981 12;
  • 81) 0.673 656 254 685 182 298 286 981 12 × 2 = 1 + 0.347 312 509 370 364 596 573 962 24;
  • 82) 0.347 312 509 370 364 596 573 962 24 × 2 = 0 + 0.694 625 018 740 729 193 147 924 48;
  • 83) 0.694 625 018 740 729 193 147 924 48 × 2 = 1 + 0.389 250 037 481 458 386 295 848 96;
  • 84) 0.389 250 037 481 458 386 295 848 96 × 2 = 0 + 0.778 500 074 962 916 772 591 697 92;
  • 85) 0.778 500 074 962 916 772 591 697 92 × 2 = 1 + 0.557 000 149 925 833 545 183 395 84;
  • 86) 0.557 000 149 925 833 545 183 395 84 × 2 = 1 + 0.114 000 299 851 667 090 366 791 68;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 000 000 000 176 177 62(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1001 1010 11(2)

6. Positive number before normalization:

0.000 000 000 000 000 000 176 177 62(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1001 1010 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 63 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 176 177 62(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1001 1010 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1001 1010 11(2) × 20 =

1.1001 1111 1111 1100 1101 011(2) × 2-63


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -63

Mantissa (not normalized):
1.1001 1111 1111 1100 1101 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-63 + 2(8-1) - 1 =

(-63 + 127)(10) =

64(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

64(10) =

0100 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1111 1111 1110 0110 1011 =

100 1111 1111 1110 0110 1011


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

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

Exponent (8 bits) =
0100 0000

Mantissa (23 bits) =
100 1111 1111 1110 0110 1011


Decimal number -0.000 000 000 000 000 000 176 177 62 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0100 0000 - 100 1111 1111 1110 0110 1011

» Convert decimal -0.000 000 000 000 000 000 176 177 96 to 32 bit single precision IEEE 754 binary floating point representation standard

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

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

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

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

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111