32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -0.000 000 000 000 151 582 441 260 46 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -0.000 000 000 000 151 582 441 260 46(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:

|-0.000 000 000 000 151 582 441 260 46| = 0.000 000 000 000 151 582 441 260 46

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: 0.000 000 000 000 151 582 441 260 46.

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) 0.000 000 000 000 151 582 441 260 46 × 2 = 0 + 0.000 000 000 000 303 164 882 520 92;
  • 2) 0.000 000 000 000 303 164 882 520 92 × 2 = 0 + 0.000 000 000 000 606 329 765 041 84;
  • 3) 0.000 000 000 000 606 329 765 041 84 × 2 = 0 + 0.000 000 000 001 212 659 530 083 68;
  • 4) 0.000 000 000 001 212 659 530 083 68 × 2 = 0 + 0.000 000 000 002 425 319 060 167 36;
  • 5) 0.000 000 000 002 425 319 060 167 36 × 2 = 0 + 0.000 000 000 004 850 638 120 334 72;
  • 6) 0.000 000 000 004 850 638 120 334 72 × 2 = 0 + 0.000 000 000 009 701 276 240 669 44;
  • 7) 0.000 000 000 009 701 276 240 669 44 × 2 = 0 + 0.000 000 000 019 402 552 481 338 88;
  • 8) 0.000 000 000 019 402 552 481 338 88 × 2 = 0 + 0.000 000 000 038 805 104 962 677 76;
  • 9) 0.000 000 000 038 805 104 962 677 76 × 2 = 0 + 0.000 000 000 077 610 209 925 355 52;
  • 10) 0.000 000 000 077 610 209 925 355 52 × 2 = 0 + 0.000 000 000 155 220 419 850 711 04;
  • 11) 0.000 000 000 155 220 419 850 711 04 × 2 = 0 + 0.000 000 000 310 440 839 701 422 08;
  • 12) 0.000 000 000 310 440 839 701 422 08 × 2 = 0 + 0.000 000 000 620 881 679 402 844 16;
  • 13) 0.000 000 000 620 881 679 402 844 16 × 2 = 0 + 0.000 000 001 241 763 358 805 688 32;
  • 14) 0.000 000 001 241 763 358 805 688 32 × 2 = 0 + 0.000 000 002 483 526 717 611 376 64;
  • 15) 0.000 000 002 483 526 717 611 376 64 × 2 = 0 + 0.000 000 004 967 053 435 222 753 28;
  • 16) 0.000 000 004 967 053 435 222 753 28 × 2 = 0 + 0.000 000 009 934 106 870 445 506 56;
  • 17) 0.000 000 009 934 106 870 445 506 56 × 2 = 0 + 0.000 000 019 868 213 740 891 013 12;
  • 18) 0.000 000 019 868 213 740 891 013 12 × 2 = 0 + 0.000 000 039 736 427 481 782 026 24;
  • 19) 0.000 000 039 736 427 481 782 026 24 × 2 = 0 + 0.000 000 079 472 854 963 564 052 48;
  • 20) 0.000 000 079 472 854 963 564 052 48 × 2 = 0 + 0.000 000 158 945 709 927 128 104 96;
  • 21) 0.000 000 158 945 709 927 128 104 96 × 2 = 0 + 0.000 000 317 891 419 854 256 209 92;
  • 22) 0.000 000 317 891 419 854 256 209 92 × 2 = 0 + 0.000 000 635 782 839 708 512 419 84;
  • 23) 0.000 000 635 782 839 708 512 419 84 × 2 = 0 + 0.000 001 271 565 679 417 024 839 68;
  • 24) 0.000 001 271 565 679 417 024 839 68 × 2 = 0 + 0.000 002 543 131 358 834 049 679 36;
  • 25) 0.000 002 543 131 358 834 049 679 36 × 2 = 0 + 0.000 005 086 262 717 668 099 358 72;
  • 26) 0.000 005 086 262 717 668 099 358 72 × 2 = 0 + 0.000 010 172 525 435 336 198 717 44;
  • 27) 0.000 010 172 525 435 336 198 717 44 × 2 = 0 + 0.000 020 345 050 870 672 397 434 88;
  • 28) 0.000 020 345 050 870 672 397 434 88 × 2 = 0 + 0.000 040 690 101 741 344 794 869 76;
  • 29) 0.000 040 690 101 741 344 794 869 76 × 2 = 0 + 0.000 081 380 203 482 689 589 739 52;
  • 30) 0.000 081 380 203 482 689 589 739 52 × 2 = 0 + 0.000 162 760 406 965 379 179 479 04;
  • 31) 0.000 162 760 406 965 379 179 479 04 × 2 = 0 + 0.000 325 520 813 930 758 358 958 08;
  • 32) 0.000 325 520 813 930 758 358 958 08 × 2 = 0 + 0.000 651 041 627 861 516 717 916 16;
  • 33) 0.000 651 041 627 861 516 717 916 16 × 2 = 0 + 0.001 302 083 255 723 033 435 832 32;
  • 34) 0.001 302 083 255 723 033 435 832 32 × 2 = 0 + 0.002 604 166 511 446 066 871 664 64;
  • 35) 0.002 604 166 511 446 066 871 664 64 × 2 = 0 + 0.005 208 333 022 892 133 743 329 28;
  • 36) 0.005 208 333 022 892 133 743 329 28 × 2 = 0 + 0.010 416 666 045 784 267 486 658 56;
  • 37) 0.010 416 666 045 784 267 486 658 56 × 2 = 0 + 0.020 833 332 091 568 534 973 317 12;
  • 38) 0.020 833 332 091 568 534 973 317 12 × 2 = 0 + 0.041 666 664 183 137 069 946 634 24;
  • 39) 0.041 666 664 183 137 069 946 634 24 × 2 = 0 + 0.083 333 328 366 274 139 893 268 48;
  • 40) 0.083 333 328 366 274 139 893 268 48 × 2 = 0 + 0.166 666 656 732 548 279 786 536 96;
  • 41) 0.166 666 656 732 548 279 786 536 96 × 2 = 0 + 0.333 333 313 465 096 559 573 073 92;
  • 42) 0.333 333 313 465 096 559 573 073 92 × 2 = 0 + 0.666 666 626 930 193 119 146 147 84;
  • 43) 0.666 666 626 930 193 119 146 147 84 × 2 = 1 + 0.333 333 253 860 386 238 292 295 68;
  • 44) 0.333 333 253 860 386 238 292 295 68 × 2 = 0 + 0.666 666 507 720 772 476 584 591 36;
  • 45) 0.666 666 507 720 772 476 584 591 36 × 2 = 1 + 0.333 333 015 441 544 953 169 182 72;
  • 46) 0.333 333 015 441 544 953 169 182 72 × 2 = 0 + 0.666 666 030 883 089 906 338 365 44;
  • 47) 0.666 666 030 883 089 906 338 365 44 × 2 = 1 + 0.333 332 061 766 179 812 676 730 88;
  • 48) 0.333 332 061 766 179 812 676 730 88 × 2 = 0 + 0.666 664 123 532 359 625 353 461 76;
  • 49) 0.666 664 123 532 359 625 353 461 76 × 2 = 1 + 0.333 328 247 064 719 250 706 923 52;
  • 50) 0.333 328 247 064 719 250 706 923 52 × 2 = 0 + 0.666 656 494 129 438 501 413 847 04;
  • 51) 0.666 656 494 129 438 501 413 847 04 × 2 = 1 + 0.333 312 988 258 877 002 827 694 08;
  • 52) 0.333 312 988 258 877 002 827 694 08 × 2 = 0 + 0.666 625 976 517 754 005 655 388 16;
  • 53) 0.666 625 976 517 754 005 655 388 16 × 2 = 1 + 0.333 251 953 035 508 011 310 776 32;
  • 54) 0.333 251 953 035 508 011 310 776 32 × 2 = 0 + 0.666 503 906 071 016 022 621 552 64;
  • 55) 0.666 503 906 071 016 022 621 552 64 × 2 = 1 + 0.333 007 812 142 032 045 243 105 28;
  • 56) 0.333 007 812 142 032 045 243 105 28 × 2 = 0 + 0.666 015 624 284 064 090 486 210 56;
  • 57) 0.666 015 624 284 064 090 486 210 56 × 2 = 1 + 0.332 031 248 568 128 180 972 421 12;
  • 58) 0.332 031 248 568 128 180 972 421 12 × 2 = 0 + 0.664 062 497 136 256 361 944 842 24;
  • 59) 0.664 062 497 136 256 361 944 842 24 × 2 = 1 + 0.328 124 994 272 512 723 889 684 48;
  • 60) 0.328 124 994 272 512 723 889 684 48 × 2 = 0 + 0.656 249 988 545 025 447 779 368 96;
  • 61) 0.656 249 988 545 025 447 779 368 96 × 2 = 1 + 0.312 499 977 090 050 895 558 737 92;
  • 62) 0.312 499 977 090 050 895 558 737 92 × 2 = 0 + 0.624 999 954 180 101 791 117 475 84;
  • 63) 0.624 999 954 180 101 791 117 475 84 × 2 = 1 + 0.249 999 908 360 203 582 234 951 68;
  • 64) 0.249 999 908 360 203 582 234 951 68 × 2 = 0 + 0.499 999 816 720 407 164 469 903 36;
  • 65) 0.499 999 816 720 407 164 469 903 36 × 2 = 0 + 0.999 999 633 440 814 328 939 806 72;
  • 66) 0.999 999 633 440 814 328 939 806 72 × 2 = 1 + 0.999 999 266 881 628 657 879 613 44;

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:


0.000 000 000 000 151 582 441 260 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 1010 1010 01(2)


6. Positive number before normalization:

0.000 000 000 000 151 582 441 260 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 1010 1010 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 151 582 441 260 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 1010 1010 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 1010 1010 01(2) × 20 =

1.0101 0101 0101 0101 0101 001(2) × 2-43


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): -43

Mantissa (not normalized):
1.0101 0101 0101 0101 0101 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(8-1) - 1 =

(-43 + 127)(10) =

84(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;
  • 84 ÷ 2 = 42 + 0;
  • 42 ÷ 2 = 21 + 0;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 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) =

84(10) =

0101 0100(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. 010 1010 1010 1010 1010 1001 =

010 1010 1010 1010 1010 1001


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) =
0101 0100

Mantissa (23 bits) =
010 1010 1010 1010 1010 1001


The base ten decimal number -0.000 000 000 000 151 582 441 260 46 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 0101 0100 - 010 1010 1010 1010 1010 1001

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

» Number -0.000 000 000 000 151 582 441 260 47 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

» Number -0.000 000 000 000 151 582 441 260 45 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

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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