-0.000 000 000 742 147 676 646 710 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 710 1(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
-0.000 000 000 742 147 676 646 710 1(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:

|-0.000 000 000 742 147 676 646 710 1| = 0.000 000 000 742 147 676 646 710 1


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: 0.000 000 000 742 147 676 646 710 1.

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) 0.000 000 000 742 147 676 646 710 1 × 2 = 0 + 0.000 000 001 484 295 353 293 420 2;
  • 2) 0.000 000 001 484 295 353 293 420 2 × 2 = 0 + 0.000 000 002 968 590 706 586 840 4;
  • 3) 0.000 000 002 968 590 706 586 840 4 × 2 = 0 + 0.000 000 005 937 181 413 173 680 8;
  • 4) 0.000 000 005 937 181 413 173 680 8 × 2 = 0 + 0.000 000 011 874 362 826 347 361 6;
  • 5) 0.000 000 011 874 362 826 347 361 6 × 2 = 0 + 0.000 000 023 748 725 652 694 723 2;
  • 6) 0.000 000 023 748 725 652 694 723 2 × 2 = 0 + 0.000 000 047 497 451 305 389 446 4;
  • 7) 0.000 000 047 497 451 305 389 446 4 × 2 = 0 + 0.000 000 094 994 902 610 778 892 8;
  • 8) 0.000 000 094 994 902 610 778 892 8 × 2 = 0 + 0.000 000 189 989 805 221 557 785 6;
  • 9) 0.000 000 189 989 805 221 557 785 6 × 2 = 0 + 0.000 000 379 979 610 443 115 571 2;
  • 10) 0.000 000 379 979 610 443 115 571 2 × 2 = 0 + 0.000 000 759 959 220 886 231 142 4;
  • 11) 0.000 000 759 959 220 886 231 142 4 × 2 = 0 + 0.000 001 519 918 441 772 462 284 8;
  • 12) 0.000 001 519 918 441 772 462 284 8 × 2 = 0 + 0.000 003 039 836 883 544 924 569 6;
  • 13) 0.000 003 039 836 883 544 924 569 6 × 2 = 0 + 0.000 006 079 673 767 089 849 139 2;
  • 14) 0.000 006 079 673 767 089 849 139 2 × 2 = 0 + 0.000 012 159 347 534 179 698 278 4;
  • 15) 0.000 012 159 347 534 179 698 278 4 × 2 = 0 + 0.000 024 318 695 068 359 396 556 8;
  • 16) 0.000 024 318 695 068 359 396 556 8 × 2 = 0 + 0.000 048 637 390 136 718 793 113 6;
  • 17) 0.000 048 637 390 136 718 793 113 6 × 2 = 0 + 0.000 097 274 780 273 437 586 227 2;
  • 18) 0.000 097 274 780 273 437 586 227 2 × 2 = 0 + 0.000 194 549 560 546 875 172 454 4;
  • 19) 0.000 194 549 560 546 875 172 454 4 × 2 = 0 + 0.000 389 099 121 093 750 344 908 8;
  • 20) 0.000 389 099 121 093 750 344 908 8 × 2 = 0 + 0.000 778 198 242 187 500 689 817 6;
  • 21) 0.000 778 198 242 187 500 689 817 6 × 2 = 0 + 0.001 556 396 484 375 001 379 635 2;
  • 22) 0.001 556 396 484 375 001 379 635 2 × 2 = 0 + 0.003 112 792 968 750 002 759 270 4;
  • 23) 0.003 112 792 968 750 002 759 270 4 × 2 = 0 + 0.006 225 585 937 500 005 518 540 8;
  • 24) 0.006 225 585 937 500 005 518 540 8 × 2 = 0 + 0.012 451 171 875 000 011 037 081 6;
  • 25) 0.012 451 171 875 000 011 037 081 6 × 2 = 0 + 0.024 902 343 750 000 022 074 163 2;
  • 26) 0.024 902 343 750 000 022 074 163 2 × 2 = 0 + 0.049 804 687 500 000 044 148 326 4;
  • 27) 0.049 804 687 500 000 044 148 326 4 × 2 = 0 + 0.099 609 375 000 000 088 296 652 8;
  • 28) 0.099 609 375 000 000 088 296 652 8 × 2 = 0 + 0.199 218 750 000 000 176 593 305 6;
  • 29) 0.199 218 750 000 000 176 593 305 6 × 2 = 0 + 0.398 437 500 000 000 353 186 611 2;
  • 30) 0.398 437 500 000 000 353 186 611 2 × 2 = 0 + 0.796 875 000 000 000 706 373 222 4;
  • 31) 0.796 875 000 000 000 706 373 222 4 × 2 = 1 + 0.593 750 000 000 001 412 746 444 8;
  • 32) 0.593 750 000 000 001 412 746 444 8 × 2 = 1 + 0.187 500 000 000 002 825 492 889 6;
  • 33) 0.187 500 000 000 002 825 492 889 6 × 2 = 0 + 0.375 000 000 000 005 650 985 779 2;
  • 34) 0.375 000 000 000 005 650 985 779 2 × 2 = 0 + 0.750 000 000 000 011 301 971 558 4;
  • 35) 0.750 000 000 000 011 301 971 558 4 × 2 = 1 + 0.500 000 000 000 022 603 943 116 8;
  • 36) 0.500 000 000 000 022 603 943 116 8 × 2 = 1 + 0.000 000 000 000 045 207 886 233 6;
  • 37) 0.000 000 000 000 045 207 886 233 6 × 2 = 0 + 0.000 000 000 000 090 415 772 467 2;
  • 38) 0.000 000 000 000 090 415 772 467 2 × 2 = 0 + 0.000 000 000 000 180 831 544 934 4;
  • 39) 0.000 000 000 000 180 831 544 934 4 × 2 = 0 + 0.000 000 000 000 361 663 089 868 8;
  • 40) 0.000 000 000 000 361 663 089 868 8 × 2 = 0 + 0.000 000 000 000 723 326 179 737 6;
  • 41) 0.000 000 000 000 723 326 179 737 6 × 2 = 0 + 0.000 000 000 001 446 652 359 475 2;
  • 42) 0.000 000 000 001 446 652 359 475 2 × 2 = 0 + 0.000 000 000 002 893 304 718 950 4;
  • 43) 0.000 000 000 002 893 304 718 950 4 × 2 = 0 + 0.000 000 000 005 786 609 437 900 8;
  • 44) 0.000 000 000 005 786 609 437 900 8 × 2 = 0 + 0.000 000 000 011 573 218 875 801 6;
  • 45) 0.000 000 000 011 573 218 875 801 6 × 2 = 0 + 0.000 000 000 023 146 437 751 603 2;
  • 46) 0.000 000 000 023 146 437 751 603 2 × 2 = 0 + 0.000 000 000 046 292 875 503 206 4;
  • 47) 0.000 000 000 046 292 875 503 206 4 × 2 = 0 + 0.000 000 000 092 585 751 006 412 8;
  • 48) 0.000 000 000 092 585 751 006 412 8 × 2 = 0 + 0.000 000 000 185 171 502 012 825 6;
  • 49) 0.000 000 000 185 171 502 012 825 6 × 2 = 0 + 0.000 000 000 370 343 004 025 651 2;
  • 50) 0.000 000 000 370 343 004 025 651 2 × 2 = 0 + 0.000 000 000 740 686 008 051 302 4;
  • 51) 0.000 000 000 740 686 008 051 302 4 × 2 = 0 + 0.000 000 001 481 372 016 102 604 8;
  • 52) 0.000 000 001 481 372 016 102 604 8 × 2 = 0 + 0.000 000 002 962 744 032 205 209 6;
  • 53) 0.000 000 002 962 744 032 205 209 6 × 2 = 0 + 0.000 000 005 925 488 064 410 419 2;
  • 54) 0.000 000 005 925 488 064 410 419 2 × 2 = 0 + 0.000 000 011 850 976 128 820 838 4;

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:


0.000 000 000 742 147 676 646 710 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 710 1(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:


0.000 000 000 742 147 676 646 710 1(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 -0.000 000 000 742 147 676 646 710 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 646 711 3 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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