-0.000 000 000 742 147 676 646 718 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 718 6(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
-0.000 000 000 742 147 676 646 718 6(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:

|-0.000 000 000 742 147 676 646 718 6| = 0.000 000 000 742 147 676 646 718 6


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: 0.000 000 000 742 147 676 646 718 6.

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) 0.000 000 000 742 147 676 646 718 6 × 2 = 0 + 0.000 000 001 484 295 353 293 437 2;
  • 2) 0.000 000 001 484 295 353 293 437 2 × 2 = 0 + 0.000 000 002 968 590 706 586 874 4;
  • 3) 0.000 000 002 968 590 706 586 874 4 × 2 = 0 + 0.000 000 005 937 181 413 173 748 8;
  • 4) 0.000 000 005 937 181 413 173 748 8 × 2 = 0 + 0.000 000 011 874 362 826 347 497 6;
  • 5) 0.000 000 011 874 362 826 347 497 6 × 2 = 0 + 0.000 000 023 748 725 652 694 995 2;
  • 6) 0.000 000 023 748 725 652 694 995 2 × 2 = 0 + 0.000 000 047 497 451 305 389 990 4;
  • 7) 0.000 000 047 497 451 305 389 990 4 × 2 = 0 + 0.000 000 094 994 902 610 779 980 8;
  • 8) 0.000 000 094 994 902 610 779 980 8 × 2 = 0 + 0.000 000 189 989 805 221 559 961 6;
  • 9) 0.000 000 189 989 805 221 559 961 6 × 2 = 0 + 0.000 000 379 979 610 443 119 923 2;
  • 10) 0.000 000 379 979 610 443 119 923 2 × 2 = 0 + 0.000 000 759 959 220 886 239 846 4;
  • 11) 0.000 000 759 959 220 886 239 846 4 × 2 = 0 + 0.000 001 519 918 441 772 479 692 8;
  • 12) 0.000 001 519 918 441 772 479 692 8 × 2 = 0 + 0.000 003 039 836 883 544 959 385 6;
  • 13) 0.000 003 039 836 883 544 959 385 6 × 2 = 0 + 0.000 006 079 673 767 089 918 771 2;
  • 14) 0.000 006 079 673 767 089 918 771 2 × 2 = 0 + 0.000 012 159 347 534 179 837 542 4;
  • 15) 0.000 012 159 347 534 179 837 542 4 × 2 = 0 + 0.000 024 318 695 068 359 675 084 8;
  • 16) 0.000 024 318 695 068 359 675 084 8 × 2 = 0 + 0.000 048 637 390 136 719 350 169 6;
  • 17) 0.000 048 637 390 136 719 350 169 6 × 2 = 0 + 0.000 097 274 780 273 438 700 339 2;
  • 18) 0.000 097 274 780 273 438 700 339 2 × 2 = 0 + 0.000 194 549 560 546 877 400 678 4;
  • 19) 0.000 194 549 560 546 877 400 678 4 × 2 = 0 + 0.000 389 099 121 093 754 801 356 8;
  • 20) 0.000 389 099 121 093 754 801 356 8 × 2 = 0 + 0.000 778 198 242 187 509 602 713 6;
  • 21) 0.000 778 198 242 187 509 602 713 6 × 2 = 0 + 0.001 556 396 484 375 019 205 427 2;
  • 22) 0.001 556 396 484 375 019 205 427 2 × 2 = 0 + 0.003 112 792 968 750 038 410 854 4;
  • 23) 0.003 112 792 968 750 038 410 854 4 × 2 = 0 + 0.006 225 585 937 500 076 821 708 8;
  • 24) 0.006 225 585 937 500 076 821 708 8 × 2 = 0 + 0.012 451 171 875 000 153 643 417 6;
  • 25) 0.012 451 171 875 000 153 643 417 6 × 2 = 0 + 0.024 902 343 750 000 307 286 835 2;
  • 26) 0.024 902 343 750 000 307 286 835 2 × 2 = 0 + 0.049 804 687 500 000 614 573 670 4;
  • 27) 0.049 804 687 500 000 614 573 670 4 × 2 = 0 + 0.099 609 375 000 001 229 147 340 8;
  • 28) 0.099 609 375 000 001 229 147 340 8 × 2 = 0 + 0.199 218 750 000 002 458 294 681 6;
  • 29) 0.199 218 750 000 002 458 294 681 6 × 2 = 0 + 0.398 437 500 000 004 916 589 363 2;
  • 30) 0.398 437 500 000 004 916 589 363 2 × 2 = 0 + 0.796 875 000 000 009 833 178 726 4;
  • 31) 0.796 875 000 000 009 833 178 726 4 × 2 = 1 + 0.593 750 000 000 019 666 357 452 8;
  • 32) 0.593 750 000 000 019 666 357 452 8 × 2 = 1 + 0.187 500 000 000 039 332 714 905 6;
  • 33) 0.187 500 000 000 039 332 714 905 6 × 2 = 0 + 0.375 000 000 000 078 665 429 811 2;
  • 34) 0.375 000 000 000 078 665 429 811 2 × 2 = 0 + 0.750 000 000 000 157 330 859 622 4;
  • 35) 0.750 000 000 000 157 330 859 622 4 × 2 = 1 + 0.500 000 000 000 314 661 719 244 8;
  • 36) 0.500 000 000 000 314 661 719 244 8 × 2 = 1 + 0.000 000 000 000 629 323 438 489 6;
  • 37) 0.000 000 000 000 629 323 438 489 6 × 2 = 0 + 0.000 000 000 001 258 646 876 979 2;
  • 38) 0.000 000 000 001 258 646 876 979 2 × 2 = 0 + 0.000 000 000 002 517 293 753 958 4;
  • 39) 0.000 000 000 002 517 293 753 958 4 × 2 = 0 + 0.000 000 000 005 034 587 507 916 8;
  • 40) 0.000 000 000 005 034 587 507 916 8 × 2 = 0 + 0.000 000 000 010 069 175 015 833 6;
  • 41) 0.000 000 000 010 069 175 015 833 6 × 2 = 0 + 0.000 000 000 020 138 350 031 667 2;
  • 42) 0.000 000 000 020 138 350 031 667 2 × 2 = 0 + 0.000 000 000 040 276 700 063 334 4;
  • 43) 0.000 000 000 040 276 700 063 334 4 × 2 = 0 + 0.000 000 000 080 553 400 126 668 8;
  • 44) 0.000 000 000 080 553 400 126 668 8 × 2 = 0 + 0.000 000 000 161 106 800 253 337 6;
  • 45) 0.000 000 000 161 106 800 253 337 6 × 2 = 0 + 0.000 000 000 322 213 600 506 675 2;
  • 46) 0.000 000 000 322 213 600 506 675 2 × 2 = 0 + 0.000 000 000 644 427 201 013 350 4;
  • 47) 0.000 000 000 644 427 201 013 350 4 × 2 = 0 + 0.000 000 001 288 854 402 026 700 8;
  • 48) 0.000 000 001 288 854 402 026 700 8 × 2 = 0 + 0.000 000 002 577 708 804 053 401 6;
  • 49) 0.000 000 002 577 708 804 053 401 6 × 2 = 0 + 0.000 000 005 155 417 608 106 803 2;
  • 50) 0.000 000 005 155 417 608 106 803 2 × 2 = 0 + 0.000 000 010 310 835 216 213 606 4;
  • 51) 0.000 000 010 310 835 216 213 606 4 × 2 = 0 + 0.000 000 020 621 670 432 427 212 8;
  • 52) 0.000 000 020 621 670 432 427 212 8 × 2 = 0 + 0.000 000 041 243 340 864 854 425 6;
  • 53) 0.000 000 041 243 340 864 854 425 6 × 2 = 0 + 0.000 000 082 486 681 729 708 851 2;
  • 54) 0.000 000 082 486 681 729 708 851 2 × 2 = 0 + 0.000 000 164 973 363 459 417 702 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:


0.000 000 000 742 147 676 646 718 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 718 6(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:


0.000 000 000 742 147 676 646 718 6(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 -0.000 000 000 742 147 676 646 718 6 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