-0.000 000 000 742 147 681 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 681(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
-0.000 000 000 742 147 681(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:

|-0.000 000 000 742 147 681| = 0.000 000 000 742 147 681


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: 0.000 000 000 742 147 681.

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) 0.000 000 000 742 147 681 × 2 = 0 + 0.000 000 001 484 295 362;
  • 2) 0.000 000 001 484 295 362 × 2 = 0 + 0.000 000 002 968 590 724;
  • 3) 0.000 000 002 968 590 724 × 2 = 0 + 0.000 000 005 937 181 448;
  • 4) 0.000 000 005 937 181 448 × 2 = 0 + 0.000 000 011 874 362 896;
  • 5) 0.000 000 011 874 362 896 × 2 = 0 + 0.000 000 023 748 725 792;
  • 6) 0.000 000 023 748 725 792 × 2 = 0 + 0.000 000 047 497 451 584;
  • 7) 0.000 000 047 497 451 584 × 2 = 0 + 0.000 000 094 994 903 168;
  • 8) 0.000 000 094 994 903 168 × 2 = 0 + 0.000 000 189 989 806 336;
  • 9) 0.000 000 189 989 806 336 × 2 = 0 + 0.000 000 379 979 612 672;
  • 10) 0.000 000 379 979 612 672 × 2 = 0 + 0.000 000 759 959 225 344;
  • 11) 0.000 000 759 959 225 344 × 2 = 0 + 0.000 001 519 918 450 688;
  • 12) 0.000 001 519 918 450 688 × 2 = 0 + 0.000 003 039 836 901 376;
  • 13) 0.000 003 039 836 901 376 × 2 = 0 + 0.000 006 079 673 802 752;
  • 14) 0.000 006 079 673 802 752 × 2 = 0 + 0.000 012 159 347 605 504;
  • 15) 0.000 012 159 347 605 504 × 2 = 0 + 0.000 024 318 695 211 008;
  • 16) 0.000 024 318 695 211 008 × 2 = 0 + 0.000 048 637 390 422 016;
  • 17) 0.000 048 637 390 422 016 × 2 = 0 + 0.000 097 274 780 844 032;
  • 18) 0.000 097 274 780 844 032 × 2 = 0 + 0.000 194 549 561 688 064;
  • 19) 0.000 194 549 561 688 064 × 2 = 0 + 0.000 389 099 123 376 128;
  • 20) 0.000 389 099 123 376 128 × 2 = 0 + 0.000 778 198 246 752 256;
  • 21) 0.000 778 198 246 752 256 × 2 = 0 + 0.001 556 396 493 504 512;
  • 22) 0.001 556 396 493 504 512 × 2 = 0 + 0.003 112 792 987 009 024;
  • 23) 0.003 112 792 987 009 024 × 2 = 0 + 0.006 225 585 974 018 048;
  • 24) 0.006 225 585 974 018 048 × 2 = 0 + 0.012 451 171 948 036 096;
  • 25) 0.012 451 171 948 036 096 × 2 = 0 + 0.024 902 343 896 072 192;
  • 26) 0.024 902 343 896 072 192 × 2 = 0 + 0.049 804 687 792 144 384;
  • 27) 0.049 804 687 792 144 384 × 2 = 0 + 0.099 609 375 584 288 768;
  • 28) 0.099 609 375 584 288 768 × 2 = 0 + 0.199 218 751 168 577 536;
  • 29) 0.199 218 751 168 577 536 × 2 = 0 + 0.398 437 502 337 155 072;
  • 30) 0.398 437 502 337 155 072 × 2 = 0 + 0.796 875 004 674 310 144;
  • 31) 0.796 875 004 674 310 144 × 2 = 1 + 0.593 750 009 348 620 288;
  • 32) 0.593 750 009 348 620 288 × 2 = 1 + 0.187 500 018 697 240 576;
  • 33) 0.187 500 018 697 240 576 × 2 = 0 + 0.375 000 037 394 481 152;
  • 34) 0.375 000 037 394 481 152 × 2 = 0 + 0.750 000 074 788 962 304;
  • 35) 0.750 000 074 788 962 304 × 2 = 1 + 0.500 000 149 577 924 608;
  • 36) 0.500 000 149 577 924 608 × 2 = 1 + 0.000 000 299 155 849 216;
  • 37) 0.000 000 299 155 849 216 × 2 = 0 + 0.000 000 598 311 698 432;
  • 38) 0.000 000 598 311 698 432 × 2 = 0 + 0.000 001 196 623 396 864;
  • 39) 0.000 001 196 623 396 864 × 2 = 0 + 0.000 002 393 246 793 728;
  • 40) 0.000 002 393 246 793 728 × 2 = 0 + 0.000 004 786 493 587 456;
  • 41) 0.000 004 786 493 587 456 × 2 = 0 + 0.000 009 572 987 174 912;
  • 42) 0.000 009 572 987 174 912 × 2 = 0 + 0.000 019 145 974 349 824;
  • 43) 0.000 019 145 974 349 824 × 2 = 0 + 0.000 038 291 948 699 648;
  • 44) 0.000 038 291 948 699 648 × 2 = 0 + 0.000 076 583 897 399 296;
  • 45) 0.000 076 583 897 399 296 × 2 = 0 + 0.000 153 167 794 798 592;
  • 46) 0.000 153 167 794 798 592 × 2 = 0 + 0.000 306 335 589 597 184;
  • 47) 0.000 306 335 589 597 184 × 2 = 0 + 0.000 612 671 179 194 368;
  • 48) 0.000 612 671 179 194 368 × 2 = 0 + 0.001 225 342 358 388 736;
  • 49) 0.001 225 342 358 388 736 × 2 = 0 + 0.002 450 684 716 777 472;
  • 50) 0.002 450 684 716 777 472 × 2 = 0 + 0.004 901 369 433 554 944;
  • 51) 0.004 901 369 433 554 944 × 2 = 0 + 0.009 802 738 867 109 888;
  • 52) 0.009 802 738 867 109 888 × 2 = 0 + 0.019 605 477 734 219 776;
  • 53) 0.019 605 477 734 219 776 × 2 = 0 + 0.039 210 955 468 439 552;
  • 54) 0.039 210 955 468 439 552 × 2 = 0 + 0.078 421 910 936 879 104;

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:


0.000 000 000 742 147 681(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 681(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:


0.000 000 000 742 147 681(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 -0.000 000 000 742 147 681 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