15.065 970 697 666 566 408 388 561 202 549 041 756 564 890 81 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 15.065 970 697 666 566 408 388 561 202 549 041 756 564 890 81(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
15.065 970 697 666 566 408 388 561 202 549 041 756 564 890 81(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. First, convert to binary (in base 2) the integer part: 15.
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;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. 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.

15(10) =

1111(2)


3. Convert to binary (base 2) the fractional part: 0.065 970 697 666 566 408 388 561 202 549 041 756 564 890 81.

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) 0.065 970 697 666 566 408 388 561 202 549 041 756 564 890 81 × 2 = 0 + 0.131 941 395 333 132 816 777 122 405 098 083 513 129 781 62;
  • 2) 0.131 941 395 333 132 816 777 122 405 098 083 513 129 781 62 × 2 = 0 + 0.263 882 790 666 265 633 554 244 810 196 167 026 259 563 24;
  • 3) 0.263 882 790 666 265 633 554 244 810 196 167 026 259 563 24 × 2 = 0 + 0.527 765 581 332 531 267 108 489 620 392 334 052 519 126 48;
  • 4) 0.527 765 581 332 531 267 108 489 620 392 334 052 519 126 48 × 2 = 1 + 0.055 531 162 665 062 534 216 979 240 784 668 105 038 252 96;
  • 5) 0.055 531 162 665 062 534 216 979 240 784 668 105 038 252 96 × 2 = 0 + 0.111 062 325 330 125 068 433 958 481 569 336 210 076 505 92;
  • 6) 0.111 062 325 330 125 068 433 958 481 569 336 210 076 505 92 × 2 = 0 + 0.222 124 650 660 250 136 867 916 963 138 672 420 153 011 84;
  • 7) 0.222 124 650 660 250 136 867 916 963 138 672 420 153 011 84 × 2 = 0 + 0.444 249 301 320 500 273 735 833 926 277 344 840 306 023 68;
  • 8) 0.444 249 301 320 500 273 735 833 926 277 344 840 306 023 68 × 2 = 0 + 0.888 498 602 641 000 547 471 667 852 554 689 680 612 047 36;
  • 9) 0.888 498 602 641 000 547 471 667 852 554 689 680 612 047 36 × 2 = 1 + 0.776 997 205 282 001 094 943 335 705 109 379 361 224 094 72;
  • 10) 0.776 997 205 282 001 094 943 335 705 109 379 361 224 094 72 × 2 = 1 + 0.553 994 410 564 002 189 886 671 410 218 758 722 448 189 44;
  • 11) 0.553 994 410 564 002 189 886 671 410 218 758 722 448 189 44 × 2 = 1 + 0.107 988 821 128 004 379 773 342 820 437 517 444 896 378 88;
  • 12) 0.107 988 821 128 004 379 773 342 820 437 517 444 896 378 88 × 2 = 0 + 0.215 977 642 256 008 759 546 685 640 875 034 889 792 757 76;
  • 13) 0.215 977 642 256 008 759 546 685 640 875 034 889 792 757 76 × 2 = 0 + 0.431 955 284 512 017 519 093 371 281 750 069 779 585 515 52;
  • 14) 0.431 955 284 512 017 519 093 371 281 750 069 779 585 515 52 × 2 = 0 + 0.863 910 569 024 035 038 186 742 563 500 139 559 171 031 04;
  • 15) 0.863 910 569 024 035 038 186 742 563 500 139 559 171 031 04 × 2 = 1 + 0.727 821 138 048 070 076 373 485 127 000 279 118 342 062 08;
  • 16) 0.727 821 138 048 070 076 373 485 127 000 279 118 342 062 08 × 2 = 1 + 0.455 642 276 096 140 152 746 970 254 000 558 236 684 124 16;
  • 17) 0.455 642 276 096 140 152 746 970 254 000 558 236 684 124 16 × 2 = 0 + 0.911 284 552 192 280 305 493 940 508 001 116 473 368 248 32;
  • 18) 0.911 284 552 192 280 305 493 940 508 001 116 473 368 248 32 × 2 = 1 + 0.822 569 104 384 560 610 987 881 016 002 232 946 736 496 64;
  • 19) 0.822 569 104 384 560 610 987 881 016 002 232 946 736 496 64 × 2 = 1 + 0.645 138 208 769 121 221 975 762 032 004 465 893 472 993 28;
  • 20) 0.645 138 208 769 121 221 975 762 032 004 465 893 472 993 28 × 2 = 1 + 0.290 276 417 538 242 443 951 524 064 008 931 786 945 986 56;
  • 21) 0.290 276 417 538 242 443 951 524 064 008 931 786 945 986 56 × 2 = 0 + 0.580 552 835 076 484 887 903 048 128 017 863 573 891 973 12;
  • 22) 0.580 552 835 076 484 887 903 048 128 017 863 573 891 973 12 × 2 = 1 + 0.161 105 670 152 969 775 806 096 256 035 727 147 783 946 24;
  • 23) 0.161 105 670 152 969 775 806 096 256 035 727 147 783 946 24 × 2 = 0 + 0.322 211 340 305 939 551 612 192 512 071 454 295 567 892 48;
  • 24) 0.322 211 340 305 939 551 612 192 512 071 454 295 567 892 48 × 2 = 0 + 0.644 422 680 611 879 103 224 385 024 142 908 591 135 784 96;

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).


4. 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:


0.065 970 697 666 566 408 388 561 202 549 041 756 564 890 81(10) =

0.0001 0000 1110 0011 0111 0100(2)

5. Positive number before normalization:

15.065 970 697 666 566 408 388 561 202 549 041 756 564 890 81(10) =

1111.0001 0000 1110 0011 0111 0100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


15.065 970 697 666 566 408 388 561 202 549 041 756 564 890 81(10) =

1111.0001 0000 1110 0011 0111 0100(2) =

1111.0001 0000 1110 0011 0111 0100(2) × 20 =

1.1110 0010 0001 1100 0110 1110 100(2) × 23


7. 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 0 (a positive number)

Exponent (unadjusted): 3

Mantissa (not normalized):
1.1110 0010 0001 1100 0110 1110 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

3 + 2(8-1) - 1 =

(3 + 127)(10) =

130(10)


9. 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;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

130(10) =

1000 0010(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 111 0001 0000 1110 0011 0111 0100 =

111 0001 0000 1110 0011 0111


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
1000 0010

Mantissa (23 bits) =
111 0001 0000 1110 0011 0111


Decimal number 15.065 970 697 666 566 408 388 561 202 549 041 756 564 890 81 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0010 - 111 0001 0000 1110 0011 0111

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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