-0.000 000 000 742 147 787 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 787(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
-0.000 000 000 742 147 787(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:

|-0.000 000 000 742 147 787| = 0.000 000 000 742 147 787


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: 0.000 000 000 742 147 787.

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) 0.000 000 000 742 147 787 × 2 = 0 + 0.000 000 001 484 295 574;
  • 2) 0.000 000 001 484 295 574 × 2 = 0 + 0.000 000 002 968 591 148;
  • 3) 0.000 000 002 968 591 148 × 2 = 0 + 0.000 000 005 937 182 296;
  • 4) 0.000 000 005 937 182 296 × 2 = 0 + 0.000 000 011 874 364 592;
  • 5) 0.000 000 011 874 364 592 × 2 = 0 + 0.000 000 023 748 729 184;
  • 6) 0.000 000 023 748 729 184 × 2 = 0 + 0.000 000 047 497 458 368;
  • 7) 0.000 000 047 497 458 368 × 2 = 0 + 0.000 000 094 994 916 736;
  • 8) 0.000 000 094 994 916 736 × 2 = 0 + 0.000 000 189 989 833 472;
  • 9) 0.000 000 189 989 833 472 × 2 = 0 + 0.000 000 379 979 666 944;
  • 10) 0.000 000 379 979 666 944 × 2 = 0 + 0.000 000 759 959 333 888;
  • 11) 0.000 000 759 959 333 888 × 2 = 0 + 0.000 001 519 918 667 776;
  • 12) 0.000 001 519 918 667 776 × 2 = 0 + 0.000 003 039 837 335 552;
  • 13) 0.000 003 039 837 335 552 × 2 = 0 + 0.000 006 079 674 671 104;
  • 14) 0.000 006 079 674 671 104 × 2 = 0 + 0.000 012 159 349 342 208;
  • 15) 0.000 012 159 349 342 208 × 2 = 0 + 0.000 024 318 698 684 416;
  • 16) 0.000 024 318 698 684 416 × 2 = 0 + 0.000 048 637 397 368 832;
  • 17) 0.000 048 637 397 368 832 × 2 = 0 + 0.000 097 274 794 737 664;
  • 18) 0.000 097 274 794 737 664 × 2 = 0 + 0.000 194 549 589 475 328;
  • 19) 0.000 194 549 589 475 328 × 2 = 0 + 0.000 389 099 178 950 656;
  • 20) 0.000 389 099 178 950 656 × 2 = 0 + 0.000 778 198 357 901 312;
  • 21) 0.000 778 198 357 901 312 × 2 = 0 + 0.001 556 396 715 802 624;
  • 22) 0.001 556 396 715 802 624 × 2 = 0 + 0.003 112 793 431 605 248;
  • 23) 0.003 112 793 431 605 248 × 2 = 0 + 0.006 225 586 863 210 496;
  • 24) 0.006 225 586 863 210 496 × 2 = 0 + 0.012 451 173 726 420 992;
  • 25) 0.012 451 173 726 420 992 × 2 = 0 + 0.024 902 347 452 841 984;
  • 26) 0.024 902 347 452 841 984 × 2 = 0 + 0.049 804 694 905 683 968;
  • 27) 0.049 804 694 905 683 968 × 2 = 0 + 0.099 609 389 811 367 936;
  • 28) 0.099 609 389 811 367 936 × 2 = 0 + 0.199 218 779 622 735 872;
  • 29) 0.199 218 779 622 735 872 × 2 = 0 + 0.398 437 559 245 471 744;
  • 30) 0.398 437 559 245 471 744 × 2 = 0 + 0.796 875 118 490 943 488;
  • 31) 0.796 875 118 490 943 488 × 2 = 1 + 0.593 750 236 981 886 976;
  • 32) 0.593 750 236 981 886 976 × 2 = 1 + 0.187 500 473 963 773 952;
  • 33) 0.187 500 473 963 773 952 × 2 = 0 + 0.375 000 947 927 547 904;
  • 34) 0.375 000 947 927 547 904 × 2 = 0 + 0.750 001 895 855 095 808;
  • 35) 0.750 001 895 855 095 808 × 2 = 1 + 0.500 003 791 710 191 616;
  • 36) 0.500 003 791 710 191 616 × 2 = 1 + 0.000 007 583 420 383 232;
  • 37) 0.000 007 583 420 383 232 × 2 = 0 + 0.000 015 166 840 766 464;
  • 38) 0.000 015 166 840 766 464 × 2 = 0 + 0.000 030 333 681 532 928;
  • 39) 0.000 030 333 681 532 928 × 2 = 0 + 0.000 060 667 363 065 856;
  • 40) 0.000 060 667 363 065 856 × 2 = 0 + 0.000 121 334 726 131 712;
  • 41) 0.000 121 334 726 131 712 × 2 = 0 + 0.000 242 669 452 263 424;
  • 42) 0.000 242 669 452 263 424 × 2 = 0 + 0.000 485 338 904 526 848;
  • 43) 0.000 485 338 904 526 848 × 2 = 0 + 0.000 970 677 809 053 696;
  • 44) 0.000 970 677 809 053 696 × 2 = 0 + 0.001 941 355 618 107 392;
  • 45) 0.001 941 355 618 107 392 × 2 = 0 + 0.003 882 711 236 214 784;
  • 46) 0.003 882 711 236 214 784 × 2 = 0 + 0.007 765 422 472 429 568;
  • 47) 0.007 765 422 472 429 568 × 2 = 0 + 0.015 530 844 944 859 136;
  • 48) 0.015 530 844 944 859 136 × 2 = 0 + 0.031 061 689 889 718 272;
  • 49) 0.031 061 689 889 718 272 × 2 = 0 + 0.062 123 379 779 436 544;
  • 50) 0.062 123 379 779 436 544 × 2 = 0 + 0.124 246 759 558 873 088;
  • 51) 0.124 246 759 558 873 088 × 2 = 0 + 0.248 493 519 117 746 176;
  • 52) 0.248 493 519 117 746 176 × 2 = 0 + 0.496 987 038 235 492 352;
  • 53) 0.496 987 038 235 492 352 × 2 = 0 + 0.993 974 076 470 984 704;
  • 54) 0.993 974 076 470 984 704 × 2 = 1 + 0.987 948 152 941 969 408;

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:


0.000 000 000 742 147 787(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 787(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 01(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:


0.000 000 000 742 147 787(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 01(2) × 20 =

1.1001 1000 0000 0000 0000 001(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 001


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 0001 =

100 1100 0000 0000 0000 0001


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 0001


Decimal number -0.000 000 000 742 147 787 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0001

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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