-0.000 000 000 742 147 676 728 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 728(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
-0.000 000 000 742 147 676 728(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:

|-0.000 000 000 742 147 676 728| = 0.000 000 000 742 147 676 728


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: 0.000 000 000 742 147 676 728.

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) 0.000 000 000 742 147 676 728 × 2 = 0 + 0.000 000 001 484 295 353 456;
  • 2) 0.000 000 001 484 295 353 456 × 2 = 0 + 0.000 000 002 968 590 706 912;
  • 3) 0.000 000 002 968 590 706 912 × 2 = 0 + 0.000 000 005 937 181 413 824;
  • 4) 0.000 000 005 937 181 413 824 × 2 = 0 + 0.000 000 011 874 362 827 648;
  • 5) 0.000 000 011 874 362 827 648 × 2 = 0 + 0.000 000 023 748 725 655 296;
  • 6) 0.000 000 023 748 725 655 296 × 2 = 0 + 0.000 000 047 497 451 310 592;
  • 7) 0.000 000 047 497 451 310 592 × 2 = 0 + 0.000 000 094 994 902 621 184;
  • 8) 0.000 000 094 994 902 621 184 × 2 = 0 + 0.000 000 189 989 805 242 368;
  • 9) 0.000 000 189 989 805 242 368 × 2 = 0 + 0.000 000 379 979 610 484 736;
  • 10) 0.000 000 379 979 610 484 736 × 2 = 0 + 0.000 000 759 959 220 969 472;
  • 11) 0.000 000 759 959 220 969 472 × 2 = 0 + 0.000 001 519 918 441 938 944;
  • 12) 0.000 001 519 918 441 938 944 × 2 = 0 + 0.000 003 039 836 883 877 888;
  • 13) 0.000 003 039 836 883 877 888 × 2 = 0 + 0.000 006 079 673 767 755 776;
  • 14) 0.000 006 079 673 767 755 776 × 2 = 0 + 0.000 012 159 347 535 511 552;
  • 15) 0.000 012 159 347 535 511 552 × 2 = 0 + 0.000 024 318 695 071 023 104;
  • 16) 0.000 024 318 695 071 023 104 × 2 = 0 + 0.000 048 637 390 142 046 208;
  • 17) 0.000 048 637 390 142 046 208 × 2 = 0 + 0.000 097 274 780 284 092 416;
  • 18) 0.000 097 274 780 284 092 416 × 2 = 0 + 0.000 194 549 560 568 184 832;
  • 19) 0.000 194 549 560 568 184 832 × 2 = 0 + 0.000 389 099 121 136 369 664;
  • 20) 0.000 389 099 121 136 369 664 × 2 = 0 + 0.000 778 198 242 272 739 328;
  • 21) 0.000 778 198 242 272 739 328 × 2 = 0 + 0.001 556 396 484 545 478 656;
  • 22) 0.001 556 396 484 545 478 656 × 2 = 0 + 0.003 112 792 969 090 957 312;
  • 23) 0.003 112 792 969 090 957 312 × 2 = 0 + 0.006 225 585 938 181 914 624;
  • 24) 0.006 225 585 938 181 914 624 × 2 = 0 + 0.012 451 171 876 363 829 248;
  • 25) 0.012 451 171 876 363 829 248 × 2 = 0 + 0.024 902 343 752 727 658 496;
  • 26) 0.024 902 343 752 727 658 496 × 2 = 0 + 0.049 804 687 505 455 316 992;
  • 27) 0.049 804 687 505 455 316 992 × 2 = 0 + 0.099 609 375 010 910 633 984;
  • 28) 0.099 609 375 010 910 633 984 × 2 = 0 + 0.199 218 750 021 821 267 968;
  • 29) 0.199 218 750 021 821 267 968 × 2 = 0 + 0.398 437 500 043 642 535 936;
  • 30) 0.398 437 500 043 642 535 936 × 2 = 0 + 0.796 875 000 087 285 071 872;
  • 31) 0.796 875 000 087 285 071 872 × 2 = 1 + 0.593 750 000 174 570 143 744;
  • 32) 0.593 750 000 174 570 143 744 × 2 = 1 + 0.187 500 000 349 140 287 488;
  • 33) 0.187 500 000 349 140 287 488 × 2 = 0 + 0.375 000 000 698 280 574 976;
  • 34) 0.375 000 000 698 280 574 976 × 2 = 0 + 0.750 000 001 396 561 149 952;
  • 35) 0.750 000 001 396 561 149 952 × 2 = 1 + 0.500 000 002 793 122 299 904;
  • 36) 0.500 000 002 793 122 299 904 × 2 = 1 + 0.000 000 005 586 244 599 808;
  • 37) 0.000 000 005 586 244 599 808 × 2 = 0 + 0.000 000 011 172 489 199 616;
  • 38) 0.000 000 011 172 489 199 616 × 2 = 0 + 0.000 000 022 344 978 399 232;
  • 39) 0.000 000 022 344 978 399 232 × 2 = 0 + 0.000 000 044 689 956 798 464;
  • 40) 0.000 000 044 689 956 798 464 × 2 = 0 + 0.000 000 089 379 913 596 928;
  • 41) 0.000 000 089 379 913 596 928 × 2 = 0 + 0.000 000 178 759 827 193 856;
  • 42) 0.000 000 178 759 827 193 856 × 2 = 0 + 0.000 000 357 519 654 387 712;
  • 43) 0.000 000 357 519 654 387 712 × 2 = 0 + 0.000 000 715 039 308 775 424;
  • 44) 0.000 000 715 039 308 775 424 × 2 = 0 + 0.000 001 430 078 617 550 848;
  • 45) 0.000 001 430 078 617 550 848 × 2 = 0 + 0.000 002 860 157 235 101 696;
  • 46) 0.000 002 860 157 235 101 696 × 2 = 0 + 0.000 005 720 314 470 203 392;
  • 47) 0.000 005 720 314 470 203 392 × 2 = 0 + 0.000 011 440 628 940 406 784;
  • 48) 0.000 011 440 628 940 406 784 × 2 = 0 + 0.000 022 881 257 880 813 568;
  • 49) 0.000 022 881 257 880 813 568 × 2 = 0 + 0.000 045 762 515 761 627 136;
  • 50) 0.000 045 762 515 761 627 136 × 2 = 0 + 0.000 091 525 031 523 254 272;
  • 51) 0.000 091 525 031 523 254 272 × 2 = 0 + 0.000 183 050 063 046 508 544;
  • 52) 0.000 183 050 063 046 508 544 × 2 = 0 + 0.000 366 100 126 093 017 088;
  • 53) 0.000 366 100 126 093 017 088 × 2 = 0 + 0.000 732 200 252 186 034 176;
  • 54) 0.000 732 200 252 186 034 176 × 2 = 0 + 0.001 464 400 504 372 068 352;

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:


0.000 000 000 742 147 676 728(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 728(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:


0.000 000 000 742 147 676 728(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 -0.000 000 000 742 147 676 728 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