-0.000 000 000 742 147 676 646 721 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 721 4(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
-0.000 000 000 742 147 676 646 721 4(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:

|-0.000 000 000 742 147 676 646 721 4| = 0.000 000 000 742 147 676 646 721 4


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: 0.000 000 000 742 147 676 646 721 4.

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) 0.000 000 000 742 147 676 646 721 4 × 2 = 0 + 0.000 000 001 484 295 353 293 442 8;
  • 2) 0.000 000 001 484 295 353 293 442 8 × 2 = 0 + 0.000 000 002 968 590 706 586 885 6;
  • 3) 0.000 000 002 968 590 706 586 885 6 × 2 = 0 + 0.000 000 005 937 181 413 173 771 2;
  • 4) 0.000 000 005 937 181 413 173 771 2 × 2 = 0 + 0.000 000 011 874 362 826 347 542 4;
  • 5) 0.000 000 011 874 362 826 347 542 4 × 2 = 0 + 0.000 000 023 748 725 652 695 084 8;
  • 6) 0.000 000 023 748 725 652 695 084 8 × 2 = 0 + 0.000 000 047 497 451 305 390 169 6;
  • 7) 0.000 000 047 497 451 305 390 169 6 × 2 = 0 + 0.000 000 094 994 902 610 780 339 2;
  • 8) 0.000 000 094 994 902 610 780 339 2 × 2 = 0 + 0.000 000 189 989 805 221 560 678 4;
  • 9) 0.000 000 189 989 805 221 560 678 4 × 2 = 0 + 0.000 000 379 979 610 443 121 356 8;
  • 10) 0.000 000 379 979 610 443 121 356 8 × 2 = 0 + 0.000 000 759 959 220 886 242 713 6;
  • 11) 0.000 000 759 959 220 886 242 713 6 × 2 = 0 + 0.000 001 519 918 441 772 485 427 2;
  • 12) 0.000 001 519 918 441 772 485 427 2 × 2 = 0 + 0.000 003 039 836 883 544 970 854 4;
  • 13) 0.000 003 039 836 883 544 970 854 4 × 2 = 0 + 0.000 006 079 673 767 089 941 708 8;
  • 14) 0.000 006 079 673 767 089 941 708 8 × 2 = 0 + 0.000 012 159 347 534 179 883 417 6;
  • 15) 0.000 012 159 347 534 179 883 417 6 × 2 = 0 + 0.000 024 318 695 068 359 766 835 2;
  • 16) 0.000 024 318 695 068 359 766 835 2 × 2 = 0 + 0.000 048 637 390 136 719 533 670 4;
  • 17) 0.000 048 637 390 136 719 533 670 4 × 2 = 0 + 0.000 097 274 780 273 439 067 340 8;
  • 18) 0.000 097 274 780 273 439 067 340 8 × 2 = 0 + 0.000 194 549 560 546 878 134 681 6;
  • 19) 0.000 194 549 560 546 878 134 681 6 × 2 = 0 + 0.000 389 099 121 093 756 269 363 2;
  • 20) 0.000 389 099 121 093 756 269 363 2 × 2 = 0 + 0.000 778 198 242 187 512 538 726 4;
  • 21) 0.000 778 198 242 187 512 538 726 4 × 2 = 0 + 0.001 556 396 484 375 025 077 452 8;
  • 22) 0.001 556 396 484 375 025 077 452 8 × 2 = 0 + 0.003 112 792 968 750 050 154 905 6;
  • 23) 0.003 112 792 968 750 050 154 905 6 × 2 = 0 + 0.006 225 585 937 500 100 309 811 2;
  • 24) 0.006 225 585 937 500 100 309 811 2 × 2 = 0 + 0.012 451 171 875 000 200 619 622 4;
  • 25) 0.012 451 171 875 000 200 619 622 4 × 2 = 0 + 0.024 902 343 750 000 401 239 244 8;
  • 26) 0.024 902 343 750 000 401 239 244 8 × 2 = 0 + 0.049 804 687 500 000 802 478 489 6;
  • 27) 0.049 804 687 500 000 802 478 489 6 × 2 = 0 + 0.099 609 375 000 001 604 956 979 2;
  • 28) 0.099 609 375 000 001 604 956 979 2 × 2 = 0 + 0.199 218 750 000 003 209 913 958 4;
  • 29) 0.199 218 750 000 003 209 913 958 4 × 2 = 0 + 0.398 437 500 000 006 419 827 916 8;
  • 30) 0.398 437 500 000 006 419 827 916 8 × 2 = 0 + 0.796 875 000 000 012 839 655 833 6;
  • 31) 0.796 875 000 000 012 839 655 833 6 × 2 = 1 + 0.593 750 000 000 025 679 311 667 2;
  • 32) 0.593 750 000 000 025 679 311 667 2 × 2 = 1 + 0.187 500 000 000 051 358 623 334 4;
  • 33) 0.187 500 000 000 051 358 623 334 4 × 2 = 0 + 0.375 000 000 000 102 717 246 668 8;
  • 34) 0.375 000 000 000 102 717 246 668 8 × 2 = 0 + 0.750 000 000 000 205 434 493 337 6;
  • 35) 0.750 000 000 000 205 434 493 337 6 × 2 = 1 + 0.500 000 000 000 410 868 986 675 2;
  • 36) 0.500 000 000 000 410 868 986 675 2 × 2 = 1 + 0.000 000 000 000 821 737 973 350 4;
  • 37) 0.000 000 000 000 821 737 973 350 4 × 2 = 0 + 0.000 000 000 001 643 475 946 700 8;
  • 38) 0.000 000 000 001 643 475 946 700 8 × 2 = 0 + 0.000 000 000 003 286 951 893 401 6;
  • 39) 0.000 000 000 003 286 951 893 401 6 × 2 = 0 + 0.000 000 000 006 573 903 786 803 2;
  • 40) 0.000 000 000 006 573 903 786 803 2 × 2 = 0 + 0.000 000 000 013 147 807 573 606 4;
  • 41) 0.000 000 000 013 147 807 573 606 4 × 2 = 0 + 0.000 000 000 026 295 615 147 212 8;
  • 42) 0.000 000 000 026 295 615 147 212 8 × 2 = 0 + 0.000 000 000 052 591 230 294 425 6;
  • 43) 0.000 000 000 052 591 230 294 425 6 × 2 = 0 + 0.000 000 000 105 182 460 588 851 2;
  • 44) 0.000 000 000 105 182 460 588 851 2 × 2 = 0 + 0.000 000 000 210 364 921 177 702 4;
  • 45) 0.000 000 000 210 364 921 177 702 4 × 2 = 0 + 0.000 000 000 420 729 842 355 404 8;
  • 46) 0.000 000 000 420 729 842 355 404 8 × 2 = 0 + 0.000 000 000 841 459 684 710 809 6;
  • 47) 0.000 000 000 841 459 684 710 809 6 × 2 = 0 + 0.000 000 001 682 919 369 421 619 2;
  • 48) 0.000 000 001 682 919 369 421 619 2 × 2 = 0 + 0.000 000 003 365 838 738 843 238 4;
  • 49) 0.000 000 003 365 838 738 843 238 4 × 2 = 0 + 0.000 000 006 731 677 477 686 476 8;
  • 50) 0.000 000 006 731 677 477 686 476 8 × 2 = 0 + 0.000 000 013 463 354 955 372 953 6;
  • 51) 0.000 000 013 463 354 955 372 953 6 × 2 = 0 + 0.000 000 026 926 709 910 745 907 2;
  • 52) 0.000 000 026 926 709 910 745 907 2 × 2 = 0 + 0.000 000 053 853 419 821 491 814 4;
  • 53) 0.000 000 053 853 419 821 491 814 4 × 2 = 0 + 0.000 000 107 706 839 642 983 628 8;
  • 54) 0.000 000 107 706 839 642 983 628 8 × 2 = 0 + 0.000 000 215 413 679 285 967 257 6;

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:


0.000 000 000 742 147 676 646 721 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 721 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 646 721 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 1000 0000 0000 0000 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

96(10) =

0110 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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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) =
0110 0000

Mantissa (23 bits) =
100 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 721 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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