-0.000 000 000 742 147 466 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 466(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
-0.000 000 000 742 147 466(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:

|-0.000 000 000 742 147 466| = 0.000 000 000 742 147 466


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: 0.000 000 000 742 147 466.

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) 0.000 000 000 742 147 466 × 2 = 0 + 0.000 000 001 484 294 932;
  • 2) 0.000 000 001 484 294 932 × 2 = 0 + 0.000 000 002 968 589 864;
  • 3) 0.000 000 002 968 589 864 × 2 = 0 + 0.000 000 005 937 179 728;
  • 4) 0.000 000 005 937 179 728 × 2 = 0 + 0.000 000 011 874 359 456;
  • 5) 0.000 000 011 874 359 456 × 2 = 0 + 0.000 000 023 748 718 912;
  • 6) 0.000 000 023 748 718 912 × 2 = 0 + 0.000 000 047 497 437 824;
  • 7) 0.000 000 047 497 437 824 × 2 = 0 + 0.000 000 094 994 875 648;
  • 8) 0.000 000 094 994 875 648 × 2 = 0 + 0.000 000 189 989 751 296;
  • 9) 0.000 000 189 989 751 296 × 2 = 0 + 0.000 000 379 979 502 592;
  • 10) 0.000 000 379 979 502 592 × 2 = 0 + 0.000 000 759 959 005 184;
  • 11) 0.000 000 759 959 005 184 × 2 = 0 + 0.000 001 519 918 010 368;
  • 12) 0.000 001 519 918 010 368 × 2 = 0 + 0.000 003 039 836 020 736;
  • 13) 0.000 003 039 836 020 736 × 2 = 0 + 0.000 006 079 672 041 472;
  • 14) 0.000 006 079 672 041 472 × 2 = 0 + 0.000 012 159 344 082 944;
  • 15) 0.000 012 159 344 082 944 × 2 = 0 + 0.000 024 318 688 165 888;
  • 16) 0.000 024 318 688 165 888 × 2 = 0 + 0.000 048 637 376 331 776;
  • 17) 0.000 048 637 376 331 776 × 2 = 0 + 0.000 097 274 752 663 552;
  • 18) 0.000 097 274 752 663 552 × 2 = 0 + 0.000 194 549 505 327 104;
  • 19) 0.000 194 549 505 327 104 × 2 = 0 + 0.000 389 099 010 654 208;
  • 20) 0.000 389 099 010 654 208 × 2 = 0 + 0.000 778 198 021 308 416;
  • 21) 0.000 778 198 021 308 416 × 2 = 0 + 0.001 556 396 042 616 832;
  • 22) 0.001 556 396 042 616 832 × 2 = 0 + 0.003 112 792 085 233 664;
  • 23) 0.003 112 792 085 233 664 × 2 = 0 + 0.006 225 584 170 467 328;
  • 24) 0.006 225 584 170 467 328 × 2 = 0 + 0.012 451 168 340 934 656;
  • 25) 0.012 451 168 340 934 656 × 2 = 0 + 0.024 902 336 681 869 312;
  • 26) 0.024 902 336 681 869 312 × 2 = 0 + 0.049 804 673 363 738 624;
  • 27) 0.049 804 673 363 738 624 × 2 = 0 + 0.099 609 346 727 477 248;
  • 28) 0.099 609 346 727 477 248 × 2 = 0 + 0.199 218 693 454 954 496;
  • 29) 0.199 218 693 454 954 496 × 2 = 0 + 0.398 437 386 909 908 992;
  • 30) 0.398 437 386 909 908 992 × 2 = 0 + 0.796 874 773 819 817 984;
  • 31) 0.796 874 773 819 817 984 × 2 = 1 + 0.593 749 547 639 635 968;
  • 32) 0.593 749 547 639 635 968 × 2 = 1 + 0.187 499 095 279 271 936;
  • 33) 0.187 499 095 279 271 936 × 2 = 0 + 0.374 998 190 558 543 872;
  • 34) 0.374 998 190 558 543 872 × 2 = 0 + 0.749 996 381 117 087 744;
  • 35) 0.749 996 381 117 087 744 × 2 = 1 + 0.499 992 762 234 175 488;
  • 36) 0.499 992 762 234 175 488 × 2 = 0 + 0.999 985 524 468 350 976;
  • 37) 0.999 985 524 468 350 976 × 2 = 1 + 0.999 971 048 936 701 952;
  • 38) 0.999 971 048 936 701 952 × 2 = 1 + 0.999 942 097 873 403 904;
  • 39) 0.999 942 097 873 403 904 × 2 = 1 + 0.999 884 195 746 807 808;
  • 40) 0.999 884 195 746 807 808 × 2 = 1 + 0.999 768 391 493 615 616;
  • 41) 0.999 768 391 493 615 616 × 2 = 1 + 0.999 536 782 987 231 232;
  • 42) 0.999 536 782 987 231 232 × 2 = 1 + 0.999 073 565 974 462 464;
  • 43) 0.999 073 565 974 462 464 × 2 = 1 + 0.998 147 131 948 924 928;
  • 44) 0.998 147 131 948 924 928 × 2 = 1 + 0.996 294 263 897 849 856;
  • 45) 0.996 294 263 897 849 856 × 2 = 1 + 0.992 588 527 795 699 712;
  • 46) 0.992 588 527 795 699 712 × 2 = 1 + 0.985 177 055 591 399 424;
  • 47) 0.985 177 055 591 399 424 × 2 = 1 + 0.970 354 111 182 798 848;
  • 48) 0.970 354 111 182 798 848 × 2 = 1 + 0.940 708 222 365 597 696;
  • 49) 0.940 708 222 365 597 696 × 2 = 1 + 0.881 416 444 731 195 392;
  • 50) 0.881 416 444 731 195 392 × 2 = 1 + 0.762 832 889 462 390 784;
  • 51) 0.762 832 889 462 390 784 × 2 = 1 + 0.525 665 778 924 781 568;
  • 52) 0.525 665 778 924 781 568 × 2 = 1 + 0.051 331 557 849 563 136;
  • 53) 0.051 331 557 849 563 136 × 2 = 0 + 0.102 663 115 699 126 272;
  • 54) 0.102 663 115 699 126 272 × 2 = 0 + 0.205 326 231 398 252 544;

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:


0.000 000 000 742 147 466(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 466(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 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:


0.000 000 000 742 147 466(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) × 20 =

1.1001 0111 1111 1111 1111 100(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 0111 1111 1111 1111 100


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 1011 1111 1111 1111 1100 =

100 1011 1111 1111 1111 1100


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 1011 1111 1111 1111 1100


Decimal number -0.000 000 000 742 147 466 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1100

Share

» Convert decimal -0.000 000 000 742 147 478 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 Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 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