-0.000 000 000 742 147 764 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 764(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
-0.000 000 000 742 147 764(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:

|-0.000 000 000 742 147 764| = 0.000 000 000 742 147 764


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: 0.000 000 000 742 147 764.

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) 0.000 000 000 742 147 764 × 2 = 0 + 0.000 000 001 484 295 528;
  • 2) 0.000 000 001 484 295 528 × 2 = 0 + 0.000 000 002 968 591 056;
  • 3) 0.000 000 002 968 591 056 × 2 = 0 + 0.000 000 005 937 182 112;
  • 4) 0.000 000 005 937 182 112 × 2 = 0 + 0.000 000 011 874 364 224;
  • 5) 0.000 000 011 874 364 224 × 2 = 0 + 0.000 000 023 748 728 448;
  • 6) 0.000 000 023 748 728 448 × 2 = 0 + 0.000 000 047 497 456 896;
  • 7) 0.000 000 047 497 456 896 × 2 = 0 + 0.000 000 094 994 913 792;
  • 8) 0.000 000 094 994 913 792 × 2 = 0 + 0.000 000 189 989 827 584;
  • 9) 0.000 000 189 989 827 584 × 2 = 0 + 0.000 000 379 979 655 168;
  • 10) 0.000 000 379 979 655 168 × 2 = 0 + 0.000 000 759 959 310 336;
  • 11) 0.000 000 759 959 310 336 × 2 = 0 + 0.000 001 519 918 620 672;
  • 12) 0.000 001 519 918 620 672 × 2 = 0 + 0.000 003 039 837 241 344;
  • 13) 0.000 003 039 837 241 344 × 2 = 0 + 0.000 006 079 674 482 688;
  • 14) 0.000 006 079 674 482 688 × 2 = 0 + 0.000 012 159 348 965 376;
  • 15) 0.000 012 159 348 965 376 × 2 = 0 + 0.000 024 318 697 930 752;
  • 16) 0.000 024 318 697 930 752 × 2 = 0 + 0.000 048 637 395 861 504;
  • 17) 0.000 048 637 395 861 504 × 2 = 0 + 0.000 097 274 791 723 008;
  • 18) 0.000 097 274 791 723 008 × 2 = 0 + 0.000 194 549 583 446 016;
  • 19) 0.000 194 549 583 446 016 × 2 = 0 + 0.000 389 099 166 892 032;
  • 20) 0.000 389 099 166 892 032 × 2 = 0 + 0.000 778 198 333 784 064;
  • 21) 0.000 778 198 333 784 064 × 2 = 0 + 0.001 556 396 667 568 128;
  • 22) 0.001 556 396 667 568 128 × 2 = 0 + 0.003 112 793 335 136 256;
  • 23) 0.003 112 793 335 136 256 × 2 = 0 + 0.006 225 586 670 272 512;
  • 24) 0.006 225 586 670 272 512 × 2 = 0 + 0.012 451 173 340 545 024;
  • 25) 0.012 451 173 340 545 024 × 2 = 0 + 0.024 902 346 681 090 048;
  • 26) 0.024 902 346 681 090 048 × 2 = 0 + 0.049 804 693 362 180 096;
  • 27) 0.049 804 693 362 180 096 × 2 = 0 + 0.099 609 386 724 360 192;
  • 28) 0.099 609 386 724 360 192 × 2 = 0 + 0.199 218 773 448 720 384;
  • 29) 0.199 218 773 448 720 384 × 2 = 0 + 0.398 437 546 897 440 768;
  • 30) 0.398 437 546 897 440 768 × 2 = 0 + 0.796 875 093 794 881 536;
  • 31) 0.796 875 093 794 881 536 × 2 = 1 + 0.593 750 187 589 763 072;
  • 32) 0.593 750 187 589 763 072 × 2 = 1 + 0.187 500 375 179 526 144;
  • 33) 0.187 500 375 179 526 144 × 2 = 0 + 0.375 000 750 359 052 288;
  • 34) 0.375 000 750 359 052 288 × 2 = 0 + 0.750 001 500 718 104 576;
  • 35) 0.750 001 500 718 104 576 × 2 = 1 + 0.500 003 001 436 209 152;
  • 36) 0.500 003 001 436 209 152 × 2 = 1 + 0.000 006 002 872 418 304;
  • 37) 0.000 006 002 872 418 304 × 2 = 0 + 0.000 012 005 744 836 608;
  • 38) 0.000 012 005 744 836 608 × 2 = 0 + 0.000 024 011 489 673 216;
  • 39) 0.000 024 011 489 673 216 × 2 = 0 + 0.000 048 022 979 346 432;
  • 40) 0.000 048 022 979 346 432 × 2 = 0 + 0.000 096 045 958 692 864;
  • 41) 0.000 096 045 958 692 864 × 2 = 0 + 0.000 192 091 917 385 728;
  • 42) 0.000 192 091 917 385 728 × 2 = 0 + 0.000 384 183 834 771 456;
  • 43) 0.000 384 183 834 771 456 × 2 = 0 + 0.000 768 367 669 542 912;
  • 44) 0.000 768 367 669 542 912 × 2 = 0 + 0.001 536 735 339 085 824;
  • 45) 0.001 536 735 339 085 824 × 2 = 0 + 0.003 073 470 678 171 648;
  • 46) 0.003 073 470 678 171 648 × 2 = 0 + 0.006 146 941 356 343 296;
  • 47) 0.006 146 941 356 343 296 × 2 = 0 + 0.012 293 882 712 686 592;
  • 48) 0.012 293 882 712 686 592 × 2 = 0 + 0.024 587 765 425 373 184;
  • 49) 0.024 587 765 425 373 184 × 2 = 0 + 0.049 175 530 850 746 368;
  • 50) 0.049 175 530 850 746 368 × 2 = 0 + 0.098 351 061 701 492 736;
  • 51) 0.098 351 061 701 492 736 × 2 = 0 + 0.196 702 123 402 985 472;
  • 52) 0.196 702 123 402 985 472 × 2 = 0 + 0.393 404 246 805 970 944;
  • 53) 0.393 404 246 805 970 944 × 2 = 0 + 0.786 808 493 611 941 888;
  • 54) 0.786 808 493 611 941 888 × 2 = 1 + 0.573 616 987 223 883 776;

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:


0.000 000 000 742 147 764(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 764(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:


0.000 000 000 742 147 764(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 -0.000 000 000 742 147 764 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