-0.000 000 000 742 147 769 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 769(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
-0.000 000 000 742 147 769(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:

|-0.000 000 000 742 147 769| = 0.000 000 000 742 147 769


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: 0.000 000 000 742 147 769.

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) 0.000 000 000 742 147 769 × 2 = 0 + 0.000 000 001 484 295 538;
  • 2) 0.000 000 001 484 295 538 × 2 = 0 + 0.000 000 002 968 591 076;
  • 3) 0.000 000 002 968 591 076 × 2 = 0 + 0.000 000 005 937 182 152;
  • 4) 0.000 000 005 937 182 152 × 2 = 0 + 0.000 000 011 874 364 304;
  • 5) 0.000 000 011 874 364 304 × 2 = 0 + 0.000 000 023 748 728 608;
  • 6) 0.000 000 023 748 728 608 × 2 = 0 + 0.000 000 047 497 457 216;
  • 7) 0.000 000 047 497 457 216 × 2 = 0 + 0.000 000 094 994 914 432;
  • 8) 0.000 000 094 994 914 432 × 2 = 0 + 0.000 000 189 989 828 864;
  • 9) 0.000 000 189 989 828 864 × 2 = 0 + 0.000 000 379 979 657 728;
  • 10) 0.000 000 379 979 657 728 × 2 = 0 + 0.000 000 759 959 315 456;
  • 11) 0.000 000 759 959 315 456 × 2 = 0 + 0.000 001 519 918 630 912;
  • 12) 0.000 001 519 918 630 912 × 2 = 0 + 0.000 003 039 837 261 824;
  • 13) 0.000 003 039 837 261 824 × 2 = 0 + 0.000 006 079 674 523 648;
  • 14) 0.000 006 079 674 523 648 × 2 = 0 + 0.000 012 159 349 047 296;
  • 15) 0.000 012 159 349 047 296 × 2 = 0 + 0.000 024 318 698 094 592;
  • 16) 0.000 024 318 698 094 592 × 2 = 0 + 0.000 048 637 396 189 184;
  • 17) 0.000 048 637 396 189 184 × 2 = 0 + 0.000 097 274 792 378 368;
  • 18) 0.000 097 274 792 378 368 × 2 = 0 + 0.000 194 549 584 756 736;
  • 19) 0.000 194 549 584 756 736 × 2 = 0 + 0.000 389 099 169 513 472;
  • 20) 0.000 389 099 169 513 472 × 2 = 0 + 0.000 778 198 339 026 944;
  • 21) 0.000 778 198 339 026 944 × 2 = 0 + 0.001 556 396 678 053 888;
  • 22) 0.001 556 396 678 053 888 × 2 = 0 + 0.003 112 793 356 107 776;
  • 23) 0.003 112 793 356 107 776 × 2 = 0 + 0.006 225 586 712 215 552;
  • 24) 0.006 225 586 712 215 552 × 2 = 0 + 0.012 451 173 424 431 104;
  • 25) 0.012 451 173 424 431 104 × 2 = 0 + 0.024 902 346 848 862 208;
  • 26) 0.024 902 346 848 862 208 × 2 = 0 + 0.049 804 693 697 724 416;
  • 27) 0.049 804 693 697 724 416 × 2 = 0 + 0.099 609 387 395 448 832;
  • 28) 0.099 609 387 395 448 832 × 2 = 0 + 0.199 218 774 790 897 664;
  • 29) 0.199 218 774 790 897 664 × 2 = 0 + 0.398 437 549 581 795 328;
  • 30) 0.398 437 549 581 795 328 × 2 = 0 + 0.796 875 099 163 590 656;
  • 31) 0.796 875 099 163 590 656 × 2 = 1 + 0.593 750 198 327 181 312;
  • 32) 0.593 750 198 327 181 312 × 2 = 1 + 0.187 500 396 654 362 624;
  • 33) 0.187 500 396 654 362 624 × 2 = 0 + 0.375 000 793 308 725 248;
  • 34) 0.375 000 793 308 725 248 × 2 = 0 + 0.750 001 586 617 450 496;
  • 35) 0.750 001 586 617 450 496 × 2 = 1 + 0.500 003 173 234 900 992;
  • 36) 0.500 003 173 234 900 992 × 2 = 1 + 0.000 006 346 469 801 984;
  • 37) 0.000 006 346 469 801 984 × 2 = 0 + 0.000 012 692 939 603 968;
  • 38) 0.000 012 692 939 603 968 × 2 = 0 + 0.000 025 385 879 207 936;
  • 39) 0.000 025 385 879 207 936 × 2 = 0 + 0.000 050 771 758 415 872;
  • 40) 0.000 050 771 758 415 872 × 2 = 0 + 0.000 101 543 516 831 744;
  • 41) 0.000 101 543 516 831 744 × 2 = 0 + 0.000 203 087 033 663 488;
  • 42) 0.000 203 087 033 663 488 × 2 = 0 + 0.000 406 174 067 326 976;
  • 43) 0.000 406 174 067 326 976 × 2 = 0 + 0.000 812 348 134 653 952;
  • 44) 0.000 812 348 134 653 952 × 2 = 0 + 0.001 624 696 269 307 904;
  • 45) 0.001 624 696 269 307 904 × 2 = 0 + 0.003 249 392 538 615 808;
  • 46) 0.003 249 392 538 615 808 × 2 = 0 + 0.006 498 785 077 231 616;
  • 47) 0.006 498 785 077 231 616 × 2 = 0 + 0.012 997 570 154 463 232;
  • 48) 0.012 997 570 154 463 232 × 2 = 0 + 0.025 995 140 308 926 464;
  • 49) 0.025 995 140 308 926 464 × 2 = 0 + 0.051 990 280 617 852 928;
  • 50) 0.051 990 280 617 852 928 × 2 = 0 + 0.103 980 561 235 705 856;
  • 51) 0.103 980 561 235 705 856 × 2 = 0 + 0.207 961 122 471 411 712;
  • 52) 0.207 961 122 471 411 712 × 2 = 0 + 0.415 922 244 942 823 424;
  • 53) 0.415 922 244 942 823 424 × 2 = 0 + 0.831 844 489 885 646 848;
  • 54) 0.831 844 489 885 646 848 × 2 = 1 + 0.663 688 979 771 293 696;

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:


0.000 000 000 742 147 769(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 769(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:


0.000 000 000 742 147 769(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 -0.000 000 000 742 147 769 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