-0.000 000 000 742 147 834 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 834(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
-0.000 000 000 742 147 834(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:

|-0.000 000 000 742 147 834| = 0.000 000 000 742 147 834


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: 0.000 000 000 742 147 834.

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) 0.000 000 000 742 147 834 × 2 = 0 + 0.000 000 001 484 295 668;
  • 2) 0.000 000 001 484 295 668 × 2 = 0 + 0.000 000 002 968 591 336;
  • 3) 0.000 000 002 968 591 336 × 2 = 0 + 0.000 000 005 937 182 672;
  • 4) 0.000 000 005 937 182 672 × 2 = 0 + 0.000 000 011 874 365 344;
  • 5) 0.000 000 011 874 365 344 × 2 = 0 + 0.000 000 023 748 730 688;
  • 6) 0.000 000 023 748 730 688 × 2 = 0 + 0.000 000 047 497 461 376;
  • 7) 0.000 000 047 497 461 376 × 2 = 0 + 0.000 000 094 994 922 752;
  • 8) 0.000 000 094 994 922 752 × 2 = 0 + 0.000 000 189 989 845 504;
  • 9) 0.000 000 189 989 845 504 × 2 = 0 + 0.000 000 379 979 691 008;
  • 10) 0.000 000 379 979 691 008 × 2 = 0 + 0.000 000 759 959 382 016;
  • 11) 0.000 000 759 959 382 016 × 2 = 0 + 0.000 001 519 918 764 032;
  • 12) 0.000 001 519 918 764 032 × 2 = 0 + 0.000 003 039 837 528 064;
  • 13) 0.000 003 039 837 528 064 × 2 = 0 + 0.000 006 079 675 056 128;
  • 14) 0.000 006 079 675 056 128 × 2 = 0 + 0.000 012 159 350 112 256;
  • 15) 0.000 012 159 350 112 256 × 2 = 0 + 0.000 024 318 700 224 512;
  • 16) 0.000 024 318 700 224 512 × 2 = 0 + 0.000 048 637 400 449 024;
  • 17) 0.000 048 637 400 449 024 × 2 = 0 + 0.000 097 274 800 898 048;
  • 18) 0.000 097 274 800 898 048 × 2 = 0 + 0.000 194 549 601 796 096;
  • 19) 0.000 194 549 601 796 096 × 2 = 0 + 0.000 389 099 203 592 192;
  • 20) 0.000 389 099 203 592 192 × 2 = 0 + 0.000 778 198 407 184 384;
  • 21) 0.000 778 198 407 184 384 × 2 = 0 + 0.001 556 396 814 368 768;
  • 22) 0.001 556 396 814 368 768 × 2 = 0 + 0.003 112 793 628 737 536;
  • 23) 0.003 112 793 628 737 536 × 2 = 0 + 0.006 225 587 257 475 072;
  • 24) 0.006 225 587 257 475 072 × 2 = 0 + 0.012 451 174 514 950 144;
  • 25) 0.012 451 174 514 950 144 × 2 = 0 + 0.024 902 349 029 900 288;
  • 26) 0.024 902 349 029 900 288 × 2 = 0 + 0.049 804 698 059 800 576;
  • 27) 0.049 804 698 059 800 576 × 2 = 0 + 0.099 609 396 119 601 152;
  • 28) 0.099 609 396 119 601 152 × 2 = 0 + 0.199 218 792 239 202 304;
  • 29) 0.199 218 792 239 202 304 × 2 = 0 + 0.398 437 584 478 404 608;
  • 30) 0.398 437 584 478 404 608 × 2 = 0 + 0.796 875 168 956 809 216;
  • 31) 0.796 875 168 956 809 216 × 2 = 1 + 0.593 750 337 913 618 432;
  • 32) 0.593 750 337 913 618 432 × 2 = 1 + 0.187 500 675 827 236 864;
  • 33) 0.187 500 675 827 236 864 × 2 = 0 + 0.375 001 351 654 473 728;
  • 34) 0.375 001 351 654 473 728 × 2 = 0 + 0.750 002 703 308 947 456;
  • 35) 0.750 002 703 308 947 456 × 2 = 1 + 0.500 005 406 617 894 912;
  • 36) 0.500 005 406 617 894 912 × 2 = 1 + 0.000 010 813 235 789 824;
  • 37) 0.000 010 813 235 789 824 × 2 = 0 + 0.000 021 626 471 579 648;
  • 38) 0.000 021 626 471 579 648 × 2 = 0 + 0.000 043 252 943 159 296;
  • 39) 0.000 043 252 943 159 296 × 2 = 0 + 0.000 086 505 886 318 592;
  • 40) 0.000 086 505 886 318 592 × 2 = 0 + 0.000 173 011 772 637 184;
  • 41) 0.000 173 011 772 637 184 × 2 = 0 + 0.000 346 023 545 274 368;
  • 42) 0.000 346 023 545 274 368 × 2 = 0 + 0.000 692 047 090 548 736;
  • 43) 0.000 692 047 090 548 736 × 2 = 0 + 0.001 384 094 181 097 472;
  • 44) 0.001 384 094 181 097 472 × 2 = 0 + 0.002 768 188 362 194 944;
  • 45) 0.002 768 188 362 194 944 × 2 = 0 + 0.005 536 376 724 389 888;
  • 46) 0.005 536 376 724 389 888 × 2 = 0 + 0.011 072 753 448 779 776;
  • 47) 0.011 072 753 448 779 776 × 2 = 0 + 0.022 145 506 897 559 552;
  • 48) 0.022 145 506 897 559 552 × 2 = 0 + 0.044 291 013 795 119 104;
  • 49) 0.044 291 013 795 119 104 × 2 = 0 + 0.088 582 027 590 238 208;
  • 50) 0.088 582 027 590 238 208 × 2 = 0 + 0.177 164 055 180 476 416;
  • 51) 0.177 164 055 180 476 416 × 2 = 0 + 0.354 328 110 360 952 832;
  • 52) 0.354 328 110 360 952 832 × 2 = 0 + 0.708 656 220 721 905 664;
  • 53) 0.708 656 220 721 905 664 × 2 = 1 + 0.417 312 441 443 811 328;
  • 54) 0.417 312 441 443 811 328 × 2 = 0 + 0.834 624 882 887 622 656;

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:


0.000 000 000 742 147 834(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 834(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(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:


0.000 000 000 742 147 834(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) × 20 =

1.1001 1000 0000 0000 0000 010(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 010


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 0010 =

100 1100 0000 0000 0000 0010


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 0010


Decimal number -0.000 000 000 742 147 834 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0010

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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