32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -0.000 000 000 000 000 000 176 182 8 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -0.000 000 000 000 000 000 176 182 8(10) converted and written in 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:

|-0.000 000 000 000 000 000 176 182 8| = 0.000 000 000 000 000 000 176 182 8

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: 0.000 000 000 000 000 000 176 182 8.

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) 0.000 000 000 000 000 000 176 182 8 × 2 = 0 + 0.000 000 000 000 000 000 352 365 6;
  • 2) 0.000 000 000 000 000 000 352 365 6 × 2 = 0 + 0.000 000 000 000 000 000 704 731 2;
  • 3) 0.000 000 000 000 000 000 704 731 2 × 2 = 0 + 0.000 000 000 000 000 001 409 462 4;
  • 4) 0.000 000 000 000 000 001 409 462 4 × 2 = 0 + 0.000 000 000 000 000 002 818 924 8;
  • 5) 0.000 000 000 000 000 002 818 924 8 × 2 = 0 + 0.000 000 000 000 000 005 637 849 6;
  • 6) 0.000 000 000 000 000 005 637 849 6 × 2 = 0 + 0.000 000 000 000 000 011 275 699 2;
  • 7) 0.000 000 000 000 000 011 275 699 2 × 2 = 0 + 0.000 000 000 000 000 022 551 398 4;
  • 8) 0.000 000 000 000 000 022 551 398 4 × 2 = 0 + 0.000 000 000 000 000 045 102 796 8;
  • 9) 0.000 000 000 000 000 045 102 796 8 × 2 = 0 + 0.000 000 000 000 000 090 205 593 6;
  • 10) 0.000 000 000 000 000 090 205 593 6 × 2 = 0 + 0.000 000 000 000 000 180 411 187 2;
  • 11) 0.000 000 000 000 000 180 411 187 2 × 2 = 0 + 0.000 000 000 000 000 360 822 374 4;
  • 12) 0.000 000 000 000 000 360 822 374 4 × 2 = 0 + 0.000 000 000 000 000 721 644 748 8;
  • 13) 0.000 000 000 000 000 721 644 748 8 × 2 = 0 + 0.000 000 000 000 001 443 289 497 6;
  • 14) 0.000 000 000 000 001 443 289 497 6 × 2 = 0 + 0.000 000 000 000 002 886 578 995 2;
  • 15) 0.000 000 000 000 002 886 578 995 2 × 2 = 0 + 0.000 000 000 000 005 773 157 990 4;
  • 16) 0.000 000 000 000 005 773 157 990 4 × 2 = 0 + 0.000 000 000 000 011 546 315 980 8;
  • 17) 0.000 000 000 000 011 546 315 980 8 × 2 = 0 + 0.000 000 000 000 023 092 631 961 6;
  • 18) 0.000 000 000 000 023 092 631 961 6 × 2 = 0 + 0.000 000 000 000 046 185 263 923 2;
  • 19) 0.000 000 000 000 046 185 263 923 2 × 2 = 0 + 0.000 000 000 000 092 370 527 846 4;
  • 20) 0.000 000 000 000 092 370 527 846 4 × 2 = 0 + 0.000 000 000 000 184 741 055 692 8;
  • 21) 0.000 000 000 000 184 741 055 692 8 × 2 = 0 + 0.000 000 000 000 369 482 111 385 6;
  • 22) 0.000 000 000 000 369 482 111 385 6 × 2 = 0 + 0.000 000 000 000 738 964 222 771 2;
  • 23) 0.000 000 000 000 738 964 222 771 2 × 2 = 0 + 0.000 000 000 001 477 928 445 542 4;
  • 24) 0.000 000 000 001 477 928 445 542 4 × 2 = 0 + 0.000 000 000 002 955 856 891 084 8;
  • 25) 0.000 000 000 002 955 856 891 084 8 × 2 = 0 + 0.000 000 000 005 911 713 782 169 6;
  • 26) 0.000 000 000 005 911 713 782 169 6 × 2 = 0 + 0.000 000 000 011 823 427 564 339 2;
  • 27) 0.000 000 000 011 823 427 564 339 2 × 2 = 0 + 0.000 000 000 023 646 855 128 678 4;
  • 28) 0.000 000 000 023 646 855 128 678 4 × 2 = 0 + 0.000 000 000 047 293 710 257 356 8;
  • 29) 0.000 000 000 047 293 710 257 356 8 × 2 = 0 + 0.000 000 000 094 587 420 514 713 6;
  • 30) 0.000 000 000 094 587 420 514 713 6 × 2 = 0 + 0.000 000 000 189 174 841 029 427 2;
  • 31) 0.000 000 000 189 174 841 029 427 2 × 2 = 0 + 0.000 000 000 378 349 682 058 854 4;
  • 32) 0.000 000 000 378 349 682 058 854 4 × 2 = 0 + 0.000 000 000 756 699 364 117 708 8;
  • 33) 0.000 000 000 756 699 364 117 708 8 × 2 = 0 + 0.000 000 001 513 398 728 235 417 6;
  • 34) 0.000 000 001 513 398 728 235 417 6 × 2 = 0 + 0.000 000 003 026 797 456 470 835 2;
  • 35) 0.000 000 003 026 797 456 470 835 2 × 2 = 0 + 0.000 000 006 053 594 912 941 670 4;
  • 36) 0.000 000 006 053 594 912 941 670 4 × 2 = 0 + 0.000 000 012 107 189 825 883 340 8;
  • 37) 0.000 000 012 107 189 825 883 340 8 × 2 = 0 + 0.000 000 024 214 379 651 766 681 6;
  • 38) 0.000 000 024 214 379 651 766 681 6 × 2 = 0 + 0.000 000 048 428 759 303 533 363 2;
  • 39) 0.000 000 048 428 759 303 533 363 2 × 2 = 0 + 0.000 000 096 857 518 607 066 726 4;
  • 40) 0.000 000 096 857 518 607 066 726 4 × 2 = 0 + 0.000 000 193 715 037 214 133 452 8;
  • 41) 0.000 000 193 715 037 214 133 452 8 × 2 = 0 + 0.000 000 387 430 074 428 266 905 6;
  • 42) 0.000 000 387 430 074 428 266 905 6 × 2 = 0 + 0.000 000 774 860 148 856 533 811 2;
  • 43) 0.000 000 774 860 148 856 533 811 2 × 2 = 0 + 0.000 001 549 720 297 713 067 622 4;
  • 44) 0.000 001 549 720 297 713 067 622 4 × 2 = 0 + 0.000 003 099 440 595 426 135 244 8;
  • 45) 0.000 003 099 440 595 426 135 244 8 × 2 = 0 + 0.000 006 198 881 190 852 270 489 6;
  • 46) 0.000 006 198 881 190 852 270 489 6 × 2 = 0 + 0.000 012 397 762 381 704 540 979 2;
  • 47) 0.000 012 397 762 381 704 540 979 2 × 2 = 0 + 0.000 024 795 524 763 409 081 958 4;
  • 48) 0.000 024 795 524 763 409 081 958 4 × 2 = 0 + 0.000 049 591 049 526 818 163 916 8;
  • 49) 0.000 049 591 049 526 818 163 916 8 × 2 = 0 + 0.000 099 182 099 053 636 327 833 6;
  • 50) 0.000 099 182 099 053 636 327 833 6 × 2 = 0 + 0.000 198 364 198 107 272 655 667 2;
  • 51) 0.000 198 364 198 107 272 655 667 2 × 2 = 0 + 0.000 396 728 396 214 545 311 334 4;
  • 52) 0.000 396 728 396 214 545 311 334 4 × 2 = 0 + 0.000 793 456 792 429 090 622 668 8;
  • 53) 0.000 793 456 792 429 090 622 668 8 × 2 = 0 + 0.001 586 913 584 858 181 245 337 6;
  • 54) 0.001 586 913 584 858 181 245 337 6 × 2 = 0 + 0.003 173 827 169 716 362 490 675 2;
  • 55) 0.003 173 827 169 716 362 490 675 2 × 2 = 0 + 0.006 347 654 339 432 724 981 350 4;
  • 56) 0.006 347 654 339 432 724 981 350 4 × 2 = 0 + 0.012 695 308 678 865 449 962 700 8;
  • 57) 0.012 695 308 678 865 449 962 700 8 × 2 = 0 + 0.025 390 617 357 730 899 925 401 6;
  • 58) 0.025 390 617 357 730 899 925 401 6 × 2 = 0 + 0.050 781 234 715 461 799 850 803 2;
  • 59) 0.050 781 234 715 461 799 850 803 2 × 2 = 0 + 0.101 562 469 430 923 599 701 606 4;
  • 60) 0.101 562 469 430 923 599 701 606 4 × 2 = 0 + 0.203 124 938 861 847 199 403 212 8;
  • 61) 0.203 124 938 861 847 199 403 212 8 × 2 = 0 + 0.406 249 877 723 694 398 806 425 6;
  • 62) 0.406 249 877 723 694 398 806 425 6 × 2 = 0 + 0.812 499 755 447 388 797 612 851 2;
  • 63) 0.812 499 755 447 388 797 612 851 2 × 2 = 1 + 0.624 999 510 894 777 595 225 702 4;
  • 64) 0.624 999 510 894 777 595 225 702 4 × 2 = 1 + 0.249 999 021 789 555 190 451 404 8;
  • 65) 0.249 999 021 789 555 190 451 404 8 × 2 = 0 + 0.499 998 043 579 110 380 902 809 6;
  • 66) 0.499 998 043 579 110 380 902 809 6 × 2 = 0 + 0.999 996 087 158 220 761 805 619 2;
  • 67) 0.999 996 087 158 220 761 805 619 2 × 2 = 1 + 0.999 992 174 316 441 523 611 238 4;
  • 68) 0.999 992 174 316 441 523 611 238 4 × 2 = 1 + 0.999 984 348 632 883 047 222 476 8;
  • 69) 0.999 984 348 632 883 047 222 476 8 × 2 = 1 + 0.999 968 697 265 766 094 444 953 6;
  • 70) 0.999 968 697 265 766 094 444 953 6 × 2 = 1 + 0.999 937 394 531 532 188 889 907 2;
  • 71) 0.999 937 394 531 532 188 889 907 2 × 2 = 1 + 0.999 874 789 063 064 377 779 814 4;
  • 72) 0.999 874 789 063 064 377 779 814 4 × 2 = 1 + 0.999 749 578 126 128 755 559 628 8;
  • 73) 0.999 749 578 126 128 755 559 628 8 × 2 = 1 + 0.999 499 156 252 257 511 119 257 6;
  • 74) 0.999 499 156 252 257 511 119 257 6 × 2 = 1 + 0.998 998 312 504 515 022 238 515 2;
  • 75) 0.998 998 312 504 515 022 238 515 2 × 2 = 1 + 0.997 996 625 009 030 044 477 030 4;
  • 76) 0.997 996 625 009 030 044 477 030 4 × 2 = 1 + 0.995 993 250 018 060 088 954 060 8;
  • 77) 0.995 993 250 018 060 088 954 060 8 × 2 = 1 + 0.991 986 500 036 120 177 908 121 6;
  • 78) 0.991 986 500 036 120 177 908 121 6 × 2 = 1 + 0.983 973 000 072 240 355 816 243 2;
  • 79) 0.983 973 000 072 240 355 816 243 2 × 2 = 1 + 0.967 946 000 144 480 711 632 486 4;
  • 80) 0.967 946 000 144 480 711 632 486 4 × 2 = 1 + 0.935 892 000 288 961 423 264 972 8;
  • 81) 0.935 892 000 288 961 423 264 972 8 × 2 = 1 + 0.871 784 000 577 922 846 529 945 6;
  • 82) 0.871 784 000 577 922 846 529 945 6 × 2 = 1 + 0.743 568 001 155 845 693 059 891 2;
  • 83) 0.743 568 001 155 845 693 059 891 2 × 2 = 1 + 0.487 136 002 311 691 386 119 782 4;
  • 84) 0.487 136 002 311 691 386 119 782 4 × 2 = 0 + 0.974 272 004 623 382 772 239 564 8;
  • 85) 0.974 272 004 623 382 772 239 564 8 × 2 = 1 + 0.948 544 009 246 765 544 479 129 6;
  • 86) 0.948 544 009 246 765 544 479 129 6 × 2 = 1 + 0.897 088 018 493 531 088 958 259 2;

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...)


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:


0.000 000 000 000 000 000 176 182 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1111 1110 11(2)


6. Positive number before normalization:

0.000 000 000 000 000 000 176 182 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1111 1110 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:


0.000 000 000 000 000 000 176 182 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1111 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 1111 1111 1111 1110 11(2) × 20 =

1.1001 1111 1111 1111 1111 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 1111 1111 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 1111 1111 1011 =

100 1111 1111 1111 1111 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 1111 1111 1011


The base ten decimal number -0.000 000 000 000 000 000 176 182 8 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 0100 0000 - 100 1111 1111 1111 1111 1011

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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