-0.000 000 000 742 147 692 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 692(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
-0.000 000 000 742 147 692(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:

|-0.000 000 000 742 147 692| = 0.000 000 000 742 147 692


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: 0.000 000 000 742 147 692.

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) 0.000 000 000 742 147 692 × 2 = 0 + 0.000 000 001 484 295 384;
  • 2) 0.000 000 001 484 295 384 × 2 = 0 + 0.000 000 002 968 590 768;
  • 3) 0.000 000 002 968 590 768 × 2 = 0 + 0.000 000 005 937 181 536;
  • 4) 0.000 000 005 937 181 536 × 2 = 0 + 0.000 000 011 874 363 072;
  • 5) 0.000 000 011 874 363 072 × 2 = 0 + 0.000 000 023 748 726 144;
  • 6) 0.000 000 023 748 726 144 × 2 = 0 + 0.000 000 047 497 452 288;
  • 7) 0.000 000 047 497 452 288 × 2 = 0 + 0.000 000 094 994 904 576;
  • 8) 0.000 000 094 994 904 576 × 2 = 0 + 0.000 000 189 989 809 152;
  • 9) 0.000 000 189 989 809 152 × 2 = 0 + 0.000 000 379 979 618 304;
  • 10) 0.000 000 379 979 618 304 × 2 = 0 + 0.000 000 759 959 236 608;
  • 11) 0.000 000 759 959 236 608 × 2 = 0 + 0.000 001 519 918 473 216;
  • 12) 0.000 001 519 918 473 216 × 2 = 0 + 0.000 003 039 836 946 432;
  • 13) 0.000 003 039 836 946 432 × 2 = 0 + 0.000 006 079 673 892 864;
  • 14) 0.000 006 079 673 892 864 × 2 = 0 + 0.000 012 159 347 785 728;
  • 15) 0.000 012 159 347 785 728 × 2 = 0 + 0.000 024 318 695 571 456;
  • 16) 0.000 024 318 695 571 456 × 2 = 0 + 0.000 048 637 391 142 912;
  • 17) 0.000 048 637 391 142 912 × 2 = 0 + 0.000 097 274 782 285 824;
  • 18) 0.000 097 274 782 285 824 × 2 = 0 + 0.000 194 549 564 571 648;
  • 19) 0.000 194 549 564 571 648 × 2 = 0 + 0.000 389 099 129 143 296;
  • 20) 0.000 389 099 129 143 296 × 2 = 0 + 0.000 778 198 258 286 592;
  • 21) 0.000 778 198 258 286 592 × 2 = 0 + 0.001 556 396 516 573 184;
  • 22) 0.001 556 396 516 573 184 × 2 = 0 + 0.003 112 793 033 146 368;
  • 23) 0.003 112 793 033 146 368 × 2 = 0 + 0.006 225 586 066 292 736;
  • 24) 0.006 225 586 066 292 736 × 2 = 0 + 0.012 451 172 132 585 472;
  • 25) 0.012 451 172 132 585 472 × 2 = 0 + 0.024 902 344 265 170 944;
  • 26) 0.024 902 344 265 170 944 × 2 = 0 + 0.049 804 688 530 341 888;
  • 27) 0.049 804 688 530 341 888 × 2 = 0 + 0.099 609 377 060 683 776;
  • 28) 0.099 609 377 060 683 776 × 2 = 0 + 0.199 218 754 121 367 552;
  • 29) 0.199 218 754 121 367 552 × 2 = 0 + 0.398 437 508 242 735 104;
  • 30) 0.398 437 508 242 735 104 × 2 = 0 + 0.796 875 016 485 470 208;
  • 31) 0.796 875 016 485 470 208 × 2 = 1 + 0.593 750 032 970 940 416;
  • 32) 0.593 750 032 970 940 416 × 2 = 1 + 0.187 500 065 941 880 832;
  • 33) 0.187 500 065 941 880 832 × 2 = 0 + 0.375 000 131 883 761 664;
  • 34) 0.375 000 131 883 761 664 × 2 = 0 + 0.750 000 263 767 523 328;
  • 35) 0.750 000 263 767 523 328 × 2 = 1 + 0.500 000 527 535 046 656;
  • 36) 0.500 000 527 535 046 656 × 2 = 1 + 0.000 001 055 070 093 312;
  • 37) 0.000 001 055 070 093 312 × 2 = 0 + 0.000 002 110 140 186 624;
  • 38) 0.000 002 110 140 186 624 × 2 = 0 + 0.000 004 220 280 373 248;
  • 39) 0.000 004 220 280 373 248 × 2 = 0 + 0.000 008 440 560 746 496;
  • 40) 0.000 008 440 560 746 496 × 2 = 0 + 0.000 016 881 121 492 992;
  • 41) 0.000 016 881 121 492 992 × 2 = 0 + 0.000 033 762 242 985 984;
  • 42) 0.000 033 762 242 985 984 × 2 = 0 + 0.000 067 524 485 971 968;
  • 43) 0.000 067 524 485 971 968 × 2 = 0 + 0.000 135 048 971 943 936;
  • 44) 0.000 135 048 971 943 936 × 2 = 0 + 0.000 270 097 943 887 872;
  • 45) 0.000 270 097 943 887 872 × 2 = 0 + 0.000 540 195 887 775 744;
  • 46) 0.000 540 195 887 775 744 × 2 = 0 + 0.001 080 391 775 551 488;
  • 47) 0.001 080 391 775 551 488 × 2 = 0 + 0.002 160 783 551 102 976;
  • 48) 0.002 160 783 551 102 976 × 2 = 0 + 0.004 321 567 102 205 952;
  • 49) 0.004 321 567 102 205 952 × 2 = 0 + 0.008 643 134 204 411 904;
  • 50) 0.008 643 134 204 411 904 × 2 = 0 + 0.017 286 268 408 823 808;
  • 51) 0.017 286 268 408 823 808 × 2 = 0 + 0.034 572 536 817 647 616;
  • 52) 0.034 572 536 817 647 616 × 2 = 0 + 0.069 145 073 635 295 232;
  • 53) 0.069 145 073 635 295 232 × 2 = 0 + 0.138 290 147 270 590 464;
  • 54) 0.138 290 147 270 590 464 × 2 = 0 + 0.276 580 294 541 180 928;

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:


0.000 000 000 742 147 692(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 692(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:


0.000 000 000 742 147 692(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 -0.000 000 000 742 147 692 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