-0.000 000 000 742 147 676 699 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 699(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
-0.000 000 000 742 147 676 699(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:

|-0.000 000 000 742 147 676 699| = 0.000 000 000 742 147 676 699


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: 0.000 000 000 742 147 676 699.

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) 0.000 000 000 742 147 676 699 × 2 = 0 + 0.000 000 001 484 295 353 398;
  • 2) 0.000 000 001 484 295 353 398 × 2 = 0 + 0.000 000 002 968 590 706 796;
  • 3) 0.000 000 002 968 590 706 796 × 2 = 0 + 0.000 000 005 937 181 413 592;
  • 4) 0.000 000 005 937 181 413 592 × 2 = 0 + 0.000 000 011 874 362 827 184;
  • 5) 0.000 000 011 874 362 827 184 × 2 = 0 + 0.000 000 023 748 725 654 368;
  • 6) 0.000 000 023 748 725 654 368 × 2 = 0 + 0.000 000 047 497 451 308 736;
  • 7) 0.000 000 047 497 451 308 736 × 2 = 0 + 0.000 000 094 994 902 617 472;
  • 8) 0.000 000 094 994 902 617 472 × 2 = 0 + 0.000 000 189 989 805 234 944;
  • 9) 0.000 000 189 989 805 234 944 × 2 = 0 + 0.000 000 379 979 610 469 888;
  • 10) 0.000 000 379 979 610 469 888 × 2 = 0 + 0.000 000 759 959 220 939 776;
  • 11) 0.000 000 759 959 220 939 776 × 2 = 0 + 0.000 001 519 918 441 879 552;
  • 12) 0.000 001 519 918 441 879 552 × 2 = 0 + 0.000 003 039 836 883 759 104;
  • 13) 0.000 003 039 836 883 759 104 × 2 = 0 + 0.000 006 079 673 767 518 208;
  • 14) 0.000 006 079 673 767 518 208 × 2 = 0 + 0.000 012 159 347 535 036 416;
  • 15) 0.000 012 159 347 535 036 416 × 2 = 0 + 0.000 024 318 695 070 072 832;
  • 16) 0.000 024 318 695 070 072 832 × 2 = 0 + 0.000 048 637 390 140 145 664;
  • 17) 0.000 048 637 390 140 145 664 × 2 = 0 + 0.000 097 274 780 280 291 328;
  • 18) 0.000 097 274 780 280 291 328 × 2 = 0 + 0.000 194 549 560 560 582 656;
  • 19) 0.000 194 549 560 560 582 656 × 2 = 0 + 0.000 389 099 121 121 165 312;
  • 20) 0.000 389 099 121 121 165 312 × 2 = 0 + 0.000 778 198 242 242 330 624;
  • 21) 0.000 778 198 242 242 330 624 × 2 = 0 + 0.001 556 396 484 484 661 248;
  • 22) 0.001 556 396 484 484 661 248 × 2 = 0 + 0.003 112 792 968 969 322 496;
  • 23) 0.003 112 792 968 969 322 496 × 2 = 0 + 0.006 225 585 937 938 644 992;
  • 24) 0.006 225 585 937 938 644 992 × 2 = 0 + 0.012 451 171 875 877 289 984;
  • 25) 0.012 451 171 875 877 289 984 × 2 = 0 + 0.024 902 343 751 754 579 968;
  • 26) 0.024 902 343 751 754 579 968 × 2 = 0 + 0.049 804 687 503 509 159 936;
  • 27) 0.049 804 687 503 509 159 936 × 2 = 0 + 0.099 609 375 007 018 319 872;
  • 28) 0.099 609 375 007 018 319 872 × 2 = 0 + 0.199 218 750 014 036 639 744;
  • 29) 0.199 218 750 014 036 639 744 × 2 = 0 + 0.398 437 500 028 073 279 488;
  • 30) 0.398 437 500 028 073 279 488 × 2 = 0 + 0.796 875 000 056 146 558 976;
  • 31) 0.796 875 000 056 146 558 976 × 2 = 1 + 0.593 750 000 112 293 117 952;
  • 32) 0.593 750 000 112 293 117 952 × 2 = 1 + 0.187 500 000 224 586 235 904;
  • 33) 0.187 500 000 224 586 235 904 × 2 = 0 + 0.375 000 000 449 172 471 808;
  • 34) 0.375 000 000 449 172 471 808 × 2 = 0 + 0.750 000 000 898 344 943 616;
  • 35) 0.750 000 000 898 344 943 616 × 2 = 1 + 0.500 000 001 796 689 887 232;
  • 36) 0.500 000 001 796 689 887 232 × 2 = 1 + 0.000 000 003 593 379 774 464;
  • 37) 0.000 000 003 593 379 774 464 × 2 = 0 + 0.000 000 007 186 759 548 928;
  • 38) 0.000 000 007 186 759 548 928 × 2 = 0 + 0.000 000 014 373 519 097 856;
  • 39) 0.000 000 014 373 519 097 856 × 2 = 0 + 0.000 000 028 747 038 195 712;
  • 40) 0.000 000 028 747 038 195 712 × 2 = 0 + 0.000 000 057 494 076 391 424;
  • 41) 0.000 000 057 494 076 391 424 × 2 = 0 + 0.000 000 114 988 152 782 848;
  • 42) 0.000 000 114 988 152 782 848 × 2 = 0 + 0.000 000 229 976 305 565 696;
  • 43) 0.000 000 229 976 305 565 696 × 2 = 0 + 0.000 000 459 952 611 131 392;
  • 44) 0.000 000 459 952 611 131 392 × 2 = 0 + 0.000 000 919 905 222 262 784;
  • 45) 0.000 000 919 905 222 262 784 × 2 = 0 + 0.000 001 839 810 444 525 568;
  • 46) 0.000 001 839 810 444 525 568 × 2 = 0 + 0.000 003 679 620 889 051 136;
  • 47) 0.000 003 679 620 889 051 136 × 2 = 0 + 0.000 007 359 241 778 102 272;
  • 48) 0.000 007 359 241 778 102 272 × 2 = 0 + 0.000 014 718 483 556 204 544;
  • 49) 0.000 014 718 483 556 204 544 × 2 = 0 + 0.000 029 436 967 112 409 088;
  • 50) 0.000 029 436 967 112 409 088 × 2 = 0 + 0.000 058 873 934 224 818 176;
  • 51) 0.000 058 873 934 224 818 176 × 2 = 0 + 0.000 117 747 868 449 636 352;
  • 52) 0.000 117 747 868 449 636 352 × 2 = 0 + 0.000 235 495 736 899 272 704;
  • 53) 0.000 235 495 736 899 272 704 × 2 = 0 + 0.000 470 991 473 798 545 408;
  • 54) 0.000 470 991 473 798 545 408 × 2 = 0 + 0.000 941 982 947 597 090 816;

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:


0.000 000 000 742 147 676 699(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 699(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:


0.000 000 000 742 147 676 699(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 -0.000 000 000 742 147 676 699 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 744 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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