-0.000 000 000 742 147 676 888 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 888(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
-0.000 000 000 742 147 676 888(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:

|-0.000 000 000 742 147 676 888| = 0.000 000 000 742 147 676 888


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: 0.000 000 000 742 147 676 888.

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) 0.000 000 000 742 147 676 888 × 2 = 0 + 0.000 000 001 484 295 353 776;
  • 2) 0.000 000 001 484 295 353 776 × 2 = 0 + 0.000 000 002 968 590 707 552;
  • 3) 0.000 000 002 968 590 707 552 × 2 = 0 + 0.000 000 005 937 181 415 104;
  • 4) 0.000 000 005 937 181 415 104 × 2 = 0 + 0.000 000 011 874 362 830 208;
  • 5) 0.000 000 011 874 362 830 208 × 2 = 0 + 0.000 000 023 748 725 660 416;
  • 6) 0.000 000 023 748 725 660 416 × 2 = 0 + 0.000 000 047 497 451 320 832;
  • 7) 0.000 000 047 497 451 320 832 × 2 = 0 + 0.000 000 094 994 902 641 664;
  • 8) 0.000 000 094 994 902 641 664 × 2 = 0 + 0.000 000 189 989 805 283 328;
  • 9) 0.000 000 189 989 805 283 328 × 2 = 0 + 0.000 000 379 979 610 566 656;
  • 10) 0.000 000 379 979 610 566 656 × 2 = 0 + 0.000 000 759 959 221 133 312;
  • 11) 0.000 000 759 959 221 133 312 × 2 = 0 + 0.000 001 519 918 442 266 624;
  • 12) 0.000 001 519 918 442 266 624 × 2 = 0 + 0.000 003 039 836 884 533 248;
  • 13) 0.000 003 039 836 884 533 248 × 2 = 0 + 0.000 006 079 673 769 066 496;
  • 14) 0.000 006 079 673 769 066 496 × 2 = 0 + 0.000 012 159 347 538 132 992;
  • 15) 0.000 012 159 347 538 132 992 × 2 = 0 + 0.000 024 318 695 076 265 984;
  • 16) 0.000 024 318 695 076 265 984 × 2 = 0 + 0.000 048 637 390 152 531 968;
  • 17) 0.000 048 637 390 152 531 968 × 2 = 0 + 0.000 097 274 780 305 063 936;
  • 18) 0.000 097 274 780 305 063 936 × 2 = 0 + 0.000 194 549 560 610 127 872;
  • 19) 0.000 194 549 560 610 127 872 × 2 = 0 + 0.000 389 099 121 220 255 744;
  • 20) 0.000 389 099 121 220 255 744 × 2 = 0 + 0.000 778 198 242 440 511 488;
  • 21) 0.000 778 198 242 440 511 488 × 2 = 0 + 0.001 556 396 484 881 022 976;
  • 22) 0.001 556 396 484 881 022 976 × 2 = 0 + 0.003 112 792 969 762 045 952;
  • 23) 0.003 112 792 969 762 045 952 × 2 = 0 + 0.006 225 585 939 524 091 904;
  • 24) 0.006 225 585 939 524 091 904 × 2 = 0 + 0.012 451 171 879 048 183 808;
  • 25) 0.012 451 171 879 048 183 808 × 2 = 0 + 0.024 902 343 758 096 367 616;
  • 26) 0.024 902 343 758 096 367 616 × 2 = 0 + 0.049 804 687 516 192 735 232;
  • 27) 0.049 804 687 516 192 735 232 × 2 = 0 + 0.099 609 375 032 385 470 464;
  • 28) 0.099 609 375 032 385 470 464 × 2 = 0 + 0.199 218 750 064 770 940 928;
  • 29) 0.199 218 750 064 770 940 928 × 2 = 0 + 0.398 437 500 129 541 881 856;
  • 30) 0.398 437 500 129 541 881 856 × 2 = 0 + 0.796 875 000 259 083 763 712;
  • 31) 0.796 875 000 259 083 763 712 × 2 = 1 + 0.593 750 000 518 167 527 424;
  • 32) 0.593 750 000 518 167 527 424 × 2 = 1 + 0.187 500 001 036 335 054 848;
  • 33) 0.187 500 001 036 335 054 848 × 2 = 0 + 0.375 000 002 072 670 109 696;
  • 34) 0.375 000 002 072 670 109 696 × 2 = 0 + 0.750 000 004 145 340 219 392;
  • 35) 0.750 000 004 145 340 219 392 × 2 = 1 + 0.500 000 008 290 680 438 784;
  • 36) 0.500 000 008 290 680 438 784 × 2 = 1 + 0.000 000 016 581 360 877 568;
  • 37) 0.000 000 016 581 360 877 568 × 2 = 0 + 0.000 000 033 162 721 755 136;
  • 38) 0.000 000 033 162 721 755 136 × 2 = 0 + 0.000 000 066 325 443 510 272;
  • 39) 0.000 000 066 325 443 510 272 × 2 = 0 + 0.000 000 132 650 887 020 544;
  • 40) 0.000 000 132 650 887 020 544 × 2 = 0 + 0.000 000 265 301 774 041 088;
  • 41) 0.000 000 265 301 774 041 088 × 2 = 0 + 0.000 000 530 603 548 082 176;
  • 42) 0.000 000 530 603 548 082 176 × 2 = 0 + 0.000 001 061 207 096 164 352;
  • 43) 0.000 001 061 207 096 164 352 × 2 = 0 + 0.000 002 122 414 192 328 704;
  • 44) 0.000 002 122 414 192 328 704 × 2 = 0 + 0.000 004 244 828 384 657 408;
  • 45) 0.000 004 244 828 384 657 408 × 2 = 0 + 0.000 008 489 656 769 314 816;
  • 46) 0.000 008 489 656 769 314 816 × 2 = 0 + 0.000 016 979 313 538 629 632;
  • 47) 0.000 016 979 313 538 629 632 × 2 = 0 + 0.000 033 958 627 077 259 264;
  • 48) 0.000 033 958 627 077 259 264 × 2 = 0 + 0.000 067 917 254 154 518 528;
  • 49) 0.000 067 917 254 154 518 528 × 2 = 0 + 0.000 135 834 508 309 037 056;
  • 50) 0.000 135 834 508 309 037 056 × 2 = 0 + 0.000 271 669 016 618 074 112;
  • 51) 0.000 271 669 016 618 074 112 × 2 = 0 + 0.000 543 338 033 236 148 224;
  • 52) 0.000 543 338 033 236 148 224 × 2 = 0 + 0.001 086 676 066 472 296 448;
  • 53) 0.001 086 676 066 472 296 448 × 2 = 0 + 0.002 173 352 132 944 592 896;
  • 54) 0.002 173 352 132 944 592 896 × 2 = 0 + 0.004 346 704 265 889 185 792;

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:


0.000 000 000 742 147 676 888(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 888(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:


0.000 000 000 742 147 676 888(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 -0.000 000 000 742 147 676 888 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